RS Aggarwal 2017 Solutions for Class 9 Math Chapter 2 Polynomials are provided here with simple step-by-step explanations. These solutions for Polynomials are extremely popular among class 9 students for Math Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2017 Book of class 9 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2017 Solutions. All RS Aggarwal 2017 Solutions for class 9 Math are prepared by experts and are 100% accurate.

Page No 51:

Question 1:

Which of the following expression are polynomials?
(i) x5 − 2x3 + x + 7
(ii) y3-3y
(iii) t2-25t+2
(iv) 5z-6
(v) x-1x
(vi) x108 − 1
(vii) x3-27
(viii) 12x2-2x+2
(ix) x-2+2x-1+3
(x) 1
(xi) -35
(xii) 2y23-8
In case of a polynomial, wrtie its degree.

Answer:

(i) x5-2x3+x+7 is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 5, so it is a polynomial of degree 5.

(ii) y3-3y is an expression having only non-negative integral powers of y. So, it is a polynomial. Also, the highest power of y is 3, so it is a polynomial of degree 3.

(iii) t2-25t+2 is an expression having only non-negative integral powers of t. So, it is a polynomial. Also, the highest power of t is 2, so it is a polynomial of degree 2.

(iv) 5z-6 is an expression in which 5z has a rational power of z. So, it is not a polynomial.

(v) The given expression may be written as x-x-1. It contains a term having a negative integral power of x. So, it is not a polynomial.

(vi) x108-1 is an expression having a non-negative integral power of x. So, it is a polynomial. Also, the power of x is 108, so it is a polynomial of degree 108.

(vii) The given expression may be written as x13-27. It contains a term having a rational power of x. So, it is not a polynomial.

(viii) 12x2-2 x+2 is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.

(ix) x-2+2x-1+3 is an expression having negative integral powers of x. So, it is not a polynomial.

(x) Clearly, 1 is a constant polynomial of degree 0.

(xi) Clearly, -35 is a constant polynomial of degree 0.

(xii) 23 y2-8 is an expression having only a non-negative integral power of x. So, it is a polynomial. Also, the highest power of y is 2, so it is a polynomial of degree 2.

Page No 51:

Question 2:

Write the degree of each of the following polynomials:
(i) 2x-5
(ii) 3 − x + x2− 6x3
(iii) 9
(iv) 8x4 − 36x + 5x7
(v) x9x5  + 3x10 + 8
(vi) 2 − 3x2

Answer:

(i) The highest power of x is 1. So, the degree of the given polynomial is 1.

(ii) The highest power of x is 3. So, the degree of the given polynomial is 3.

(iii) Clearly, 9 is a constant polynomial of degree 0.

(iv) The highest power of x is 7. So, the degree of the given polynomial is 7.

(v) The highest power of x is 10. So, the degree of the given polynomial is 10.

(vi) The highest power of x is 2. So, the degree of the given polynomial is 2.

Page No 51:

Question 3:

Write:
(i) Coefficient of x3 in 2x + x2 − 5x3 + x4
(ii) Coefficient of x in 3-22x+4x2.
(iii) Coefficient of x2 in π3x2+7x-3.
(iv) Coefficient of x2 in 3x − 5

Answer:

(i) The coefficient of x3 in the given polynomial is -5.

(ii) The coefficient of x in the given polynomial is -22.

(iii) The coefficient of x2 in the given polynomial is π3.

(iv) The coefficient of x2 in the given polynomial is 0.

Page No 51:

Question 4:

(i) Give an example of a binomial of degree 27.
(ii) Give an example of a monomial of degree 16.
(iii) Give an example of a trinomial of degree 3.

Answer:

(i) x27-5 is a binomial of degree 27.

(ii) x16 is a monomial of degree 16.

(iii) 4x3-2x+3 is a trinomial of degree 3.

Page No 51:

Question 5:

Classify the following as linear, quadratic and cubic polynomials:
(i) 2x2 + 4x
(ii) x x3
(iii) 2 − yy2
(iv) −7 + z
(v) 5t
(vi) p3

Answer:

(i) Clearly, 2x2+4x is a polynomial of degree 2. So, it is a quadratic polynomial.

(ii) Clearly, x-x3 is a polynomial of degree 3. So, it is a cubic polynomial.

(iii) Clearly, 2-y-y2 is a polynomial of degree 2. So, it is a quadratic polynomial.

(iv) Clearly, -7 + z is a polynomial of degree 1. So, it is a linear polynomial.

(v) Clearly, 5t is a polynomial of degree of 1. So, it is a linear polynomial.

(vi) Clearly, p3 is a polynomial of degree 3. So, it is a cubic polynomial.



Page No 52:

Question 1:

If p(x) = 5 − 4x + 2x2, find (i) p(0), (ii) p(3), (iii) p(−2)

Answer:

i px=5-4x+2x2p0=5-4×0+2×02
        =5-0+0=5

ii px=5-4x+2x2p3=5-4×3+2×32
        =5-12+18=11

iii px=5-4x+2x2p-2=5-4×-2+2×-22            
          =5+8+8=21

Page No 52:

Question 2:

If p(y) = 4 + 3yy2 + 5y3, find (i) p(0), (ii) p(2), (iii) p(−1).

Answer:

i py=4+3y-y2+5y3p0=4+3×0-02+5×03
        =4+0-0+0=4

ii py=4+3y-y2+5y3p2=4+3×2-22+5×23
        =4+6-4+40=46

iii py=4+3y-y2+5y3p-1=4+3×-1--12+5×-13
          =4-3-1-5=-5

Page No 52:

Question 3:

If f(t) = 4t2 − 3t + 6, find (i) f(0), (ii) f(4), (iii) f(−5).

Answer:

i ft=4t2-3t+6f0=4×02-3×0+6
       =0-0+6=6

ii ft=4t2-3t+6f4=4×42-3×4+6
        =64-12+6=58

iii ft=4t2-3t+6f-5=4×-52-3×-5+6
          =100+15+6=121

Page No 52:

Question 4:

Find the zero of the polynomial:
(i) p(x) = x − 5
(ii) q(x) = x + 4
(iii) p(t) = 2t − 3
(iv) f(x) = 3x + 1
(v) g(x) = 5 − 4x
(vi) h(x) = 6x − 1
(vii) p(x) = ax + b, a ≠ 0
(viii) q(x) = 4x
(ix) p(x) = ax, a ≠ 0

Answer:

i px=0x-5=0                x=5Hence, 5 is the zero of the polynomial px.ii qx=0x+4=0                x=-4Hence, -4 is the zero of the polynomial qx.iii pt=02t-3=0                 t=32Hence, 32 is the zero of the polynomial pt.

iv fx=03x+1=0                 x=-13Hence, -13 is the zero of the polynomial fx.v gx=05-4x=0                 x=54Hence, 54 is the zero of the polynomial gx.vi hx=06x-1=0                 x=16Hence, 16 is the zero of the polynomial hx.

vii px=0ax+b=0                   x=-baHence, -ba is the zero of the polynomial px.viii qx=04x=0                   x=0Hence, 0 is the zero of the polynomial qx.ix px=0ax=0                 x=0Hence, 0 is the zero of the polynomial px.



Page No 53:

Question 5:

Verify that:
(i) 4 is a zero of the polynomial p(x) = x − 4.
(ii) −3 is a zero of the polynomial p(x) = x − 3.
(iii) -12is a zero of the polynomial p(y) = 2y + 1.
(iv) 25is a zero of the polynomial p(x) = 2 − 5x.
(v) 1 and 2 are the zeros of the polynomial p(x) = (x − 1)(x − 2).
(vi) 0 and 3 are the zeros of the polynomial p(x) = x2 − 3x.
(vii) 2 and −3 are the zeros of the polynomial p(x) = x2 + x − 6.

Answer:

i px=x-4p4=4-4
         = 0
Hence, 4 is the zero of the given polynomial.

ii Disclaimer: In the question, instead of -3, 3 should be the zero of the given polynomial. px=x-3p3=3-3
        = 0
Hence, 3 is the zero of the given polynomial.


iii py=2y+1p-12=2×-12+1                =-1+1                =0

Hence, -12 is the zero of the given polynomial.

iv px=2-5xp25=2-5×25
          =2-2=0

Hence, 25 is the zero of the given polynomial.

v px=x-1x-2p1=1-1×1-2
        =0×-1=0
Also,
p2=2-12-2
     =-1×0=0

Hence, 1 and 2 are the zeroes of the given polynomial.

vi px=x2-3xp0=02-3×0

Also,
p3=32-3×3      =9-9      =0

Hence, 0 and 3 are the zeroes of the given polynomial.

vii px=x2+x-6p2=22+2-6
        =4-4=0
Also,
p-3=-32+-3-6        =9-9        =0

Hence, 2 and -3 are the zeroes of the given polynomial.



Page No 55:

Question 1:

Using remainder theorem, find the remainder when:
(x3 − 6x2 + 9x + 3) is divided by (x − 1)

Answer:

We know:
x-1=0x=1
By the remainder theorem, we know that when f(x) is divided by (x - 1), the remainder is f(1).
Thus, we have:
fx=x3-6x2+9x+3 f1=13-6×12+9×1+3          =1-6+9+3          =7
Hence, the required remainder is 7.

Page No 55:

Question 2:

Using remainder theorem, find the remainder when:
(2x3 − 5x2 + 9x − 8) is divided by (x − 3)

Answer:

We have:
x-3=0x=3
By the remainder theorem, we know that when f(x) is divided by (x - 3), the remainder is f(3).
Thus, we have:
fx=2x3-5x2+9x-8 f3=2×33-5×32+9×3-8          =54-45+27-8          =28
​Hence, the required remainder is 28.

Page No 55:

Question 3:

Using remainder theorem, find the remainder when:
(3x4 − 5x2 + 9x − 8) is divided by (x − 3)

Answer:

We have:
x-2=0x=2
By the remainder theorem, we know that when f(x) is divided by (x - 2), the remainder is f(2).
Thus, we have:
fx=3x4-6x2-8x+2 f2=3×24-6×22-8×2+2          =48-24-16+2          =10
​Hence, the required remainder is 10.

Page No 55:

Question 4:

Using remainder theorem, find the remainder when:
(x3 − 7x2 + 6x + 4) is divided by (x − 6)

Answer:

We have:
x-6=0x=6
By the remainder theorem, we know that when f(x) is divided by (x - 6), the remainder is f(6).
Thus, we have:
fx=x3-7x2+6x+4f6=63-7×62+6×6+4         =216-252+36+4         =4
​Hence, the required remainder is 4.

Page No 55:

Question 5:

Using remainder theorem, find the remainder when:
(x3 − 6x2 + 13x + 60) is divided by (x + 2)

Answer:

We have:
x+2=0x=-2
By the remainder theorem, we know that when f(x) is divided by (x + 2), the remainder is f(-2).
Thus, we have:
 fx=x3-6x2+13x+60 f-2=-23-6×-22+13×-2+60             =-8-24-26+60             =2

Hence, the required remainder is 2.



Page No 56:

Question 6:

Using remainder theorem, find the remainder when:
(2x4 + 6x3 + 2x2 + x − 8) is divided by (x + 3)

Answer:

We have:
x+3=0x=-3
By the remainder theorem, we know that when f(x) is divided by (x + 3), the remainder is f(-3).
Thus, we have:
fx=2x4+6x3+2x2+x-8 f-3=2×-34+6×-33+2×-32+-3-8             =162-162+18-11             =7
​Hence, the required remainder is 7.

Page No 56:

Question 7:

Using remainder theorem, find the remainder when:
(4x3 − 12x2 + 11x − 5) is divided by (2x − 1)

Answer:

We have:
2x-1=0x=12
By the remainder theorem, we know that when f(x) is divided by (2x - 1), the remainder is f(12).
Thus, we have:
fx=4x3-12x2+11x-5 f12= 4×123-12×122+11×12-5            = 12-3+112-5            = -2

Hence, the required remainder is -2.

Page No 56:

Question 8:

Using remainder theorem, find the remainder when:
(81x4 + 54x3 − 9x2 − 3x + 2) is divided by (3x + 2)

Answer:

We have:
3x+2=0x=-23
By the remainder theorem, we know that when f(x) is divided by (3x + 2), the remainder is f(-23).
Thus, we have:
fx=81x4+54x3-9x2-3x+2f-23=81×-234+54×-233-9×-232-3×-23+2              =16-16-4+2+2              =0
​Hence, the required remainder is 0.

Page No 56:

Question 9:

Using remainder theorem, find the remainder when:
x3ax2 + 2xa is divided by (xa)

Answer:

We have:
x-a=0x=a
By the remainder theorem, we know that when f(x) is divided by (x - a), the remainder is f(a).
Thus, we have:
    fx=x3-ax2+2x-a fa=a3-a×a2+2×a-a          =a3-a3+2a-a          =a
​Hence, the required remainder is a.

Page No 56:

Question 10:

The polynomials (ax3 + 3x2 − 3) and (2x3 − 5x + a) when divided by (x − 4) leave the same remainder. Find the value of a.

Answer:

Let:
fx=ax3+3x2-3 and gx=2x3-5x+a
Now,
When fx is divided by x-4, the remainder is f4.When gx is divided by x-4, the remainder is g4.
We now have:
 f4=a×43+3×42-3      =64a+48-3      =64a+45
And,
 g4=2×43-5×4+a       =128-20+a       =108+a
Thus, we have:
64a+45=108 + a 64a - a =108 -45  63a=63a=1

Page No 56:

Question 11:

The polynomial f(x) = x4 − 2x3 + 3x2ax = b when divided by (x − 1) and (x + 1) leaves the remainder 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when f(x) is divided by (x − 2).

Answer:

 Let:fx=x4-2x3+3x2-ax+b

Now,When fx is divided by x-1, the remainder is f1.When fx is divided by x+1, the remainder is f-1.
Thus, we have:
f1=14-2×13+3×12-a×1+b      =1-2+3-a+b      =2-a+b
And,
f-1=-14-2×-13+3×-12-a×-1+b         =1+2+3+a+b         =6+a+b

Now,

2-a+b=5    ...16+a+b=19  ...2

Adding (1) and (2), we get:8+2b=24
2b=16b=8
By putting the value of b, we get the value of a, i.e., 5.
∴ a = 5 and b = 8
Now,
fx = x4-2x3 + 3x2 - 5x + 8
Also,
When fx is divided by x-2, the remainder is f2.
Thus, we have:
f2=24-2×23+3×22-5×2+8   a=5 and b=8      =16-16+12-10+8      =10



Page No 59:

Question 1:

Using factor theorem, show that:
(x − 2) is a factor of (x3 − 8)

Answer:

Let:
fx=x3-8
Now,
x-2=0x=2
By the factor theorem, (x3 - 8) is a factor of the given polynomial if f(2) = 0.
Thus, we have:
f2=23-8     =0
Hence, (x - 2) is a factor of the given polynomial.

Page No 59:

Question 2:

Using factor theorem, show that:
(x − 3) is a factor of (2x3 + 7x2 − 24x − 45)

Answer:

Let:
fx=2x3+7x2-24x-45
Now,
x-3=0x=3
By the factor theorem, (x - 3) is a factor of the given polynomial if f(3) = 0.
Thus, we have:
f3=2×33-7×32-24×3-45      =54+63-72-45      =0
Hence, (x - 3) is a factor of the given polynomial.

Page No 59:

Question 3:

Using factor theorem, show that:
(x − 1) is a factor of (2x4 + 9x3 + 6x2 − 11x − 6)

Answer:

Let:
fx=2x4+9x3+6x2-11x-6
Here, 
x-1=0x=1
By the factor theorem, (x - 1) is a factor of the given polynomial if f(1) = 0.
Thus, we have:
f1=2×14+9×13+6×12-11×1-6      =2+9+6-11-6      =0
Hence, (x - 1) is a factor of the given polynomial.

Page No 59:

Question 4:

Using factor theorem, show that:
(x + 2) is a factor of (x4x2 − 12)

Answer:

Let:
fx=x4-x2-12
Here, 
 x+2=0x=-2
By the factor theorem, (x + 2) is a factor of the given polynomial if f (-2) = 0.
Thus, we have:
f-2=-24--22-12        =16-4-12        =0
Hence, (x + 2) is a factor of the given polynomial.

Page No 59:

Question 5:

Using factor theorem, show that:
(x + 5) is a factor of (2x3 + 9x2 − 11x − 30)

Answer:

Let:
fx=2x3+9x2-11x-30
Here, 
x+5=0x=-5
By the factor theorem, (x + 5) is a factor of the given polynomial if f (-5) = 0.
Thus, we have:
f-5=2×-53+9×-52-11×-5-30        =-250+225+55-30        =0
Hence, (x + 5) is a factor of the given polynomial.

Page No 59:

Question 6:

Using factor theorem, show that:
(2x − 3) is a factor of (2x4 + x3 − 8x2x + 6)

Answer:

Let:
fx=2x4+x3-8x2-x+6
Here, 
2x-3=0x=32
By the factor theorem, (2x - 3) is a factor of the given polynomial if f32=0.
Thus, we have:

f32=2×324+323-8×322-32+6        =818+278-18-32+6        =0
Hence, (2x - 3) is a factor of the given polynomial.

Page No 59:

Question 7:

Using factor theorem, show that:
x-2is a factor of 7x2-42x-6

Answer:

Let:
fx=7x2-42x-6
Here, 
x-2=0x=2
By the factor theorem, x-2 is a factor of the given polynomial if f2=0
Thus, we have:
f2=7×22-42×2-6         =14-8-6         =0
Hence, x-2 is a factor of the given polynomial.

Page No 59:

Question 8:

Using factor theorem, show that:
x + 2 is a factor of 22x2+5x +2

Answer:

Let:
fx=22x2+5x+2
Here,
x+2=0x=-2
By the factor theorem, x+2 will be a factor of the given polynomial if f-2 = 0.
Thus, we have:
f-2=22×-22+5×-2+2            =42-52+2            =0
Hence, x+2 is a factor of the given polynomial.

Page No 59:

Question 9:

Find the value of k for which (x − 1) is a factor of (2x3 + 9x2 + x + k).

Answer:

 Let:
fx=2x3+9x2+x+k
x-1 is a factor of fx=2x3+9x2+x+k.f1=02×13+9×12+1+k=012+k=0k=-12
Hence, the required value of k is -12.

Page No 59:

Question 10:

Find the value of a for which (x − 4) is a factor of (2x3 − 3x2 − 18x + a).

Answer:

Let:
fx=2x3-3x2-18x+a
x-4 is a factor of fx=2x3-3x2-18x+a.f4=02×43-3×42-18×4+a = 08+a=0a=-8
Hence, the required value of a is -8.

Page No 59:

Question 11:

Find the value of a for which the polynomial (x4x3 − 11x2x + a) is divisible by (x + 3).

Answer:

Let:
fx=x4-x3-11x2-x+a
Now, 
x+3=0x=-3
By the factor theorem, fx is exactly divisible by x+3 if f-3=0.
Thus, we have:
f-3=-34--33-11×-32--3+a        =81+27-99+3+a        =12+a
Also,
   f-3=012+a=0a=-12
Hence, fx is exactly divisible by x+3 when a is -12.

Page No 59:

Question 12:

For what value of a is the polynomial (2x3 + ax2 + 11x + a + 3) exactly divisible by (2x − 1)?

Answer:

Let:
fx=2x3+ax2+11x+a+3
Now,
2x-1=0x=12
By the factor theorem, fx is exactly divisible by 2x-1 if f12=0.
Thus, we have:
f12=2×123+a×122+11×12+a+3        =14+a4+112+a+3        =354+5a4
Also,
f12=0354+5a4=05a4=- 354a=-7

Hence, fx is exactly divisible by 2x-1 when a is -7.

Page No 59:

Question 13:

Find the values of a and b so that the polynomial (x3 − 10x2 + ax + b) is exactly divisible by (x −1) as well as (x − 2).

Answer:

Let:
fx=x3-10x2+ax+b
Now,
x-1=0x=1
By the factor theorem, we can say:
fx will be exactly divisible by x-1 if f1=0.
Thus, we have:
f1=13-10×12+a×1+b      =1-10+a+b      =-9+a+b
∴ f1=0a+b=9            ...1
Also,
x-2=0x=2
By the factor theorem, we can say:
fx will be exactly divisible by x-2 if f2=0.
Thus, we have:
f2=23-10×22+a×2+b      =8-40+2a+b      =-32+2a+b
f2=02a+b=32       ...2

Subtracting (1) from (2), we get:a=23
Putting the value of a, we get the value of b, i.e., -14.
∴ a = 23 and b = -14

Page No 59:

Question 14:

Find the values of a and b so that the polynomial (x4 + ax3 − 7x2 − 8x + b) is exactly divisible by (x + 2) as well as (x + 3).

Answer:

Let:
fx=x4+ax3-7x2-8x+b
Now,
x+2=0x=-2
By the factor theorem, we can say:
fx will be exactly divisible by x+2 if f-2=0.
Thus, we have:
f-2=-24+a×-23-7×-22-8×-2+b        =16-8a-28+16+b        =4-8a+b
∴ f-2=08a-b=4      ...1
Also,
x+3=0x=-3
By the factor theorem, we can say:
fx will be exactly divisible by x+3 if f-3=0.
Thus, we have:
f-3=-34+a×-33-7×-32-8×-3+b        =81-27a-63+24+b        =42-27a+b
∴ f-3=027a-b=42   ...2
Subtracting 1 from 2, we get:19a=38a=2
Putting the value of a, we get the value of b, i.e., 12.
∴ a = 2 and b = 12

Page No 59:

Question 15:

Without actual division, show that (x3 − 3x2 − 13x + 15) is exactly divisible by (x2 + 2x − 3).

Answer:

Let:
fx=x3-3x2-13x+15
And,
gx=x2+2x-3
     =x2+x-3x-3=xx-1+3x-1=x-1x+3
Now, fx will be exactly divisible by gx if it is exactly divisible by x-1 as well as x+3.
For this, we must have:
f1=0 and f-3=0
Thus, we have:
f1=13-3×12-13×1+15      =1-3-13+15      =0
And,
f-3=-33-3×-32-13×-3+15         =-27-27+39+15         =0
fx is exactly divisible by x-1 as well as x+3. So, fx is exactly divisible by x-1x+3.
Hence, fx is exactly divisible by x2+2x-3.

Page No 59:

Question 16:

If (x3 + ax2 + bx + 6) has (x − 2) as a factor and leaves a remainder 3 when divided by (x − 3), find the values of a and b.

Answer:

Let:
fx=x3+ax2+bx+6
x-2 is a factor of fx=x3+ax2+bx+6.f2=023+a×22+b×2+6=014+4a+2b=04a+2b=-142a+b=-7         ...1
Now, 
x-3=0x=3
By the factor theorem, we can say:
When fx will be divided by x-3, 3 will be its remainder.f3=3

Now,
f3=33+a×32+b×3+6     =27+9a+3b+6     =33+9a+3b
Thus, we have:
    f3=333+9a+3b=39a+3b=-303a+b=-10      ...2
Subtracting (1) from (2), we get:
a = -3
By putting the value of a in (1), we get the value of b, i.e., -1.
 a = -3 and b = -1



Page No 62:

Question 1:

Factorize:
9x2 + 12xy

Answer:

We have:
9x2+12xy=3x3x+4y

Page No 62:

Question 2:

Factorize:
18x2y − 24xyz

Answer:

We have:
18x2y-24xyz=6xy3y-4z

Page No 62:

Question 3:

Factorize:
27a3b3 − 45a4b2

Answer:

We have:
27a3b3-45a4b2=9a3b23b-5a

Page No 62:

Question 4:

Factorize:
2a(x + y) − 3b(x + y)

Answer:

We have:
2ax+y-3bx+y=x+y2a-3b

Page No 62:

Question 5:

Factorize:
2x(p2 + q2) + 4y(p2 + q2)

Answer:

We have:
2xp2+q2+4yp2+q2=2xp2+q2+2yp2+q2=2p2+q2x+2y
                            

Page No 62:

Question 6:

Factorize:
x(a − 5) + y(5 − a)

Answer:

We have:
xa-5+y5-a=xa-5-ya-5
                    =a-5x-y

Page No 62:

Question 7:

Factorize:
4(a + b) − 6(a + b)2

Answer:

We have:
4a+b-6a+b2=2a+b2-3a+b
                     =2a+b2-3a-3b

Page No 62:

Question 8:

Factorize:
8(3a − 2b)2 − 10(3a − 2b)

Answer:

We have:
83a-2b2-103a-2b=23a-2b43a-2b-5
                           =23a-2b12a-8b-5

Page No 62:

Question 9:

Factorize:
x(x + y)3 − 3x2y(x + y)

Answer:

We have:
xx+y3-3x2yx+y=xx+yx+y2-3xy
                        =xx+yx2+y2+2xy-3xy=xx+yx2+y2-xy

Page No 62:

Question 10:

Factorize:
x3 + 2x2 + 5x + 10

Answer:

We have:
x3+2x2+5x+10=x3+2x2+5x+10
                   =x2x+2+5x+2=x+2x2+5

Page No 62:

Question 11:

Factorize:
x2 + xy − 2xz − 2yz

Answer:

We have:
x2+xy-2xz-2yz=x2+xy-2xz+2yz                           =xx+y-2zx+y                           =x+yx-2z

Page No 62:

Question 12:

Factorize:
a3ba2b + 5ab − 5b

Answer:

We have:
a3b-a2b+5ab-5b=ba3-a2+5a-5                              =ba3-a2+5a-5

                      =ba2a-1+5a-1=ba-1a2+5

Page No 62:

Question 13:

Factorize:
8 − 4a − 2a3 + a4

Answer:

We have:
8-4a-2a3+a4= 8-4a-2a3-a4                         = 42-a- a32-a                         = 2-a 4 - a3
                     

Page No 62:

Question 14:

Factorize:
x3 − 2x2y + 3xy2 − 6y3

Answer:

We have:
x3-2x2y+3xy2-6y3=x3-2x2y+3xy2-6y3
                       =x2x-2y+3y2x-2y=x-2yx2+3y2

Page No 62:

Question 15:

Factorize:
px − 5q + pq − 5x

Answer:

We have:
px-5q+pq-5x=px-5x+pq-5q
                   =xp-5+qp-5=p-5x+q

Page No 62:

Question 16:

Factorize:
x2 + yxyx

Answer:

We have:
x2+y-xy-x=x2-xy-x-y
               =xx-y-1x-y=x-yx-1

Page No 62:

Question 17:

Factorize:
(3a − 1)2 − 6a + 2

Answer:

We have:
3a-12-6a+2=3a-12-23a-1
                   =3a-13a-1-2=3a-13a-1-2=3a-13a-3=33a-1a-1

Page No 62:

Question 18:

Factorize:
(2x − 3)2 − 8x + 12

Answer:

We have:
2x-32-8x+12=2x-32-42x-3
                    =2x-32x-3-4=2x-32x-3-4=2x-32x-7

Page No 62:

Question 19:

Factorize:
a3 + a − 3a2 − 3

Answer:

We have:
a3+a-3a2-3=a3-3a2+a-3
                =a2a-3+1a-3=a-3a2+1

Page No 62:

Question 20:

Factorize:
3ax − 6ay − 8by + 4bx

Answer:

We have:
3ax-6ay-8by+4bx=3ax-6ay+4bx-8by
                       =3ax-2y+4bx-2y=x-2y3a+4b

Page No 62:

Question 21:

Factorize:
abx2 + a2x + b2x + ab

Answer:

We have:
abx2+a2x+b2x+ab=abx2+b2x+a2x+ab
                       =bxax+b+aax+b=ax+bbx+a

Page No 62:

Question 22:

Factorize:
x3x2 + ax + xa − 1

Answer:

We have:
x3-x2+ax+x-a-1=x3-x2+ax-a+x-1
                        =x2x-1+ax-1+1x-1=x-1x2+a+1

Page No 62:

Question 23:

Factorize:
2x + 4y − 8xy − 1

Answer:

We have:
2x+4y-8xy-1=2x-8xy-1-4y
                  =2x1-4y-11-4y=1-4y2x-1

Page No 62:

Question 24:

Factorize:
ab(x2 + y2) − xy(a2 + b2)

Answer:

We have:
abx2+y2-xya2+b2=abx2+aby2-a2xy-b2xy
                          =abx2-a2xy-b2xy-aby2=axbx-ay-bybx-ay=bx-ayax-by

Page No 62:

Question 25:

Factorize:
a2 + ab(b + 1) + b3

Answer:

We have:
a2+abb+1+b3=a2+ab2+ab+b3
                    =a2+ab2+ab+b3=aa+b2+ba+b2=a+b2a+b

Page No 62:

Question 26:

Factorize:
a3 + ab(1 − 2a) − 2b2

Answer:

We have:
a3+ab1-2a-2b2=a3+ab-2a2b-2b2
                      =a3-2a2b+ab-2b2=a2a-2b+ba-2b=a-2ba2+b

Page No 62:

Question 27:

Factorize:
2a2 + bc − 2abac2

Answer:

We have:
2a2+bc-2ab-ac=2a2-2ab-ac-bc
                    =2aa-b-ca-b=a-b2a-c

Page No 62:

Question 28:

Factorize:
(ax + by)2 + (bxay)2

Answer:

We have:
ax+by2+bx-ay2=ax2+2×ax×by+by2+bx2-2×bx×ay+ay2
                        =a2x2+2abxy+b2y2+b2x2-2abxy+a2y2=a2x2+b2y2+b2x2+a2y2=a2x2+b2x2+a2y2+b2y2=x2a2+b2+y2a2+b2=a2+b2x2+y2

Page No 62:

Question 29:

Factorize:
a(a + bc) − bc

Answer:

We have:
aa+b-c-bc=a2+ab-ac-bc
                  =a2-ac+ab-bc=aa-c+ba-c=a-ca+b

Page No 62:

Question 30:

Factorize:
a(a − 2bc) + 2bc

Answer:

We have:
aa-2b-c+2bc=a2-2ab-ac+2bc
                    =a2-2ab-ac-2bc=aa-2b-ca-2b=a-2ba-c

Page No 62:

Question 31:

Factorize:
a2x2 + (ax2 + 1)x + a

Answer:

We have:
a2x2+ax2+1x+a=ax2+1x+a2x2+a
                      =xax2+1+aax2+1=ax2+1x+a

Page No 62:

Question 32:

Factorize:
ab(x2 + 1) + x(a2 + b2)

Answer:

We have:
abx2+1+xa2+b2=abx2+ab+a2x+b2x
                        =abx2+a2x+b2x+ab=axbx+a+bbx+a=bx+aax+b

Page No 62:

Question 33:

Factorize:
x2 − (a + b)x + ab

Answer:

We have:
x2-a+bx+ab=x2-ax-bx+ab
                   =x2-ax-bx-ab=xx-a-bx-a=x-ax-b

Page No 62:

Question 34:

Factorize:
x2+1x2-2-3x+3x

Answer:

We have: x2+1x2-2-3x+3x= x2-2+1x2-3x+3x=x2-2×x×1x+1x2-3x-1x=x-1x2-3x-1x=x-1xx-1x-3



Page No 64:

Question 1:

Factorize:
25x2 − 64y2

Answer:

25x2-64y2=5x2-8y2                    =(5x+8y)(5x-8y)

Page No 64:

Question 2:

Factorize:
100 − 9x2

Answer:

100-9x2=102-3x2                =(10-3x)(10+3x)

Page No 64:

Question 3:

Factorize:
5x2 − 7y2

Answer:

5x2-7y2=5x2-7y2               =5x-7y5x+7y

Page No 64:

Question 4:

Factorize:
(3x + 5y2) − 4z2

Answer:

3x+5y2-4z2=3x +5y2-2z2                          =(3x+5y-2z)(3x+5y+2z)

Page No 64:

Question 5:

Factorize:
150 − 6x2

Answer:

150-6x2=625-x2                 =652-x2                 =65-x5+x

Page No 64:

Question 6:

Factorize:
20x2 − 45

Answer:

20x2-45=54x2-9                =52x2-32                =52x-32x+3

Page No 64:

Question 7:

Factorize:
3x3 − 48x

Answer:

3x3-48x=3xx2-16                =3xx2-42                =3xx-4 x+4

Page No 64:

Question 8:

Factorize:
2 − 50x2

Answer:

2-50x2=21-25x2              =212-5x2              =21-5x1+5x

Page No 64:

Question 9:

Factorize:
27a2 − 48b2

Answer:

27a2-48b2=39a2-16b2                    =33a2-4b2                    =33a-4b3a+4b

Page No 64:

Question 10:

Factorize:
x − 64x3

Answer:

x-64x3=x1-64x2              =x1-8x2              =x1-8x 1+8x

Page No 64:

Question 11:

Factorize:
8ab2 − 18a3

Answer:

8ab2-18a3=2a4b2-9a2                    =2a2b2-3a2                    =2a2b-3a2b+3a

Page No 64:

Question 12:

Factorize:
3a3b − 243ab3

Answer:

3a3b-243ab3=3aba2-81b2                         =3aba2-9b2                         =3aba-9ba+9b                         

Page No 64:

Question 13:

Factorize:
(a + b)3ab

Answer:

a+b3-a-b=a+b3-a+b                        =a+ba+b2-1                        =a+ba+b2-12                        =a+ba+b-1a+b+1

Page No 64:

Question 14:

Factorize:
108a2 − 3(bc)2

Answer:

108a2-3b-c2=336a2-b-c2                            =36a2-b-c2                            =36a-b+c6a+b-c

Page No 64:

Question 15:

Factorize:
x3 − 5x2x + 5

Answer:

x3-5x2-x+5=x2x-5-1x-5                         =x-5x2-1                         =x-5x2-12                         =x-5x-1x+1

Page No 64:

Question 16:

Factorize:
a2 + 2ab + b2 − 9c2

Answer:

a2+2ab+b2-9c2=a+b2-3c2                               =a+b-3ca+b+3c

Page No 64:

Question 17:

Factorize:
9 − a2 + 2abb2

Answer:

9-a2+2ab-b2=9-a2-2ab+b2                            =32-a-b2                            =3-a-b3+a-b                            =3-a+b3+a-b

Page No 64:

Question 18:

Factorize:
a2b2 − 4ac + 4c2

Answer:

a2-b2-4ac+4c2=a2-4ac+4c2-b2                              =a2-2×2a×c +2c2-b2                              =a-2c2-b2                              =a-2c+ba-2c-b

Page No 64:

Question 19:

Factorize:
9a2 + 3a − 8b − 64b2

Answer:

9a2+3a-8b-64b2=9a2-64b2+3a-8b                                   =3a2-8b2+3a-8b                                   =3a-8b3a+8b+13a-8b                                   =3a-8b3a+8b+1

Page No 64:

Question 20:

Factorize:
x2y2 + 6y − 9

Answer:

 x2-y2+6y-9=x2-y2-6y+9                          =x2-y2-2×y×3 +32                          =x2-y-32                                                    =x+y-3x-y-3                          =x+y-3x-y+3

Page No 64:

Question 21:

Factorize:
4x2 − 9y2 − 2x − 3y

Answer:

4x2-9y2-2x-3y=4x2-9y2-2x+3y                               =2x2-3y2-2x+3y                               =2x-3y2x+3y-12x+3y                               =2x+3y2x-3y-1                                

Page No 64:

Question 22:

Factorize:
x4 − 1

Answer:

x4-1=x22-1         =x2-1x2+1         =x2-12x2+1         =x-1x+1x2+1

Page No 64:

Question 23:

Factorize:
aba2 + b2

Answer:

a-b-a2+b2=a-b-a2-b2                       =a-b-a-ba+b                       =a-b1-a+b                       =a-b1-a-b

Page No 64:

Question 24:

Factorize:
x4 − 625

Answer:

x4-625=x22-252               = x2-25x2+25               =x2-52x2+25               =x-5x+5x2+25



Page No 69:

Question 1:

Factorize:
x2 + 11x + 30

Answer:

We have:
x2+11x+30
We have to split 11 into two numbers such that their sum of is 11 and their product is 30.
Clearly, 5+6=11 and 5×6=30.

 x2+11x+30 = x2+5x+6x+30                             = x(x+5)+6(x+5)                             =(x+5)(x+6)

Page No 69:

Question 2:

Factorize:
x2 + 18x + 32

Answer:

We have:
x2+18x+32
We have to split 18 into two numbers such that their sum is 18 and their product is 32.
Clearly, 16+2=18 and 16×2=32.

x2+18x+32=x2+16x+2x+32                           =x(x+16)+2(x+16)                           =(x+16)(x+2)

Page No 69:

Question 3:

Factorize:
x2 + 7x − 18

Answer:

We have:
x2+7x-18
We have to split 7 into two numbers such that their sum is 7 and their product is (-18).
Clearly, 9+-2=7 and 9×-2=-18.

x2+7x-18=x2+9x-2x-18                        =x(x+9)-2(x+9)                        =(x+9)(x-2)

Page No 69:

Question 4:

Factorize:
x2 + 5x − 6

Answer:

We have:
x2+5x-6
We have to split 5 into two numbers such that their sum is 5 and their product is (-6).
Clearly, 6+-1=5 and 6×-1=-6.

x2+5x-6=x2+6x-x-6                      =x(x+6)-1(x+6)                      =(x+6)(x-1)

Page No 69:

Question 5:

Factorize:
y2 − 4y + 3

Answer:

We have:
y2-4y+3
We have to split (-4) into two numbers such that their sum is (-4) and their product is 3.
Clearly, -3+-1=-4 and -3×-1=3.

y2-4y+3=y2-3y-y+3                     =y(y-3)-1(y-3)                     =(y-3)(y-1)

Page No 69:

Question 6:

Factorize:
x2 − 21x + 108

Answer:

We have:
x2-21x+108
We have to split (-21) into two numbers such that their sum is (-21) and their product is 108.
Clearly, -12+-9=-21 and -12×-9=108.

x2-21x+108=x2-12x-9x+108                             =x(x-12)-9(x-12)                             =(x-12)(x-9)

Page No 69:

Question 7:

Factorize:
x2 − 11x − 80

Answer:

We have:
x2-11x-80
We have to split (-11) into two numbers such that their sum is (-11) and their product is (-80).
Clearly, -16+5=-11 and -16×5=-80.

x2-11x-80 = x2-16x+5x-80                           =x(x-16)+5(x-16)                           =(x-16)(x+5)

Page No 69:

Question 8:

Factorize:
x2x − 156

Answer:

We have:
x2-x-156
We have to split (-1) into two numbers such that their sum is (-1) and their product is (-156).
Clearly, -13+12=-1 and -13×12=-156.

x2-x-156=x2-13x+12x-156                        =x(x-13)+12(x-13)                        =(x-13)(x+12)

Page No 69:

Question 9:

Factorize:
z2 − 32z − 105

Answer:

We have:
z2-32z-105
We have to split (-32) into two numbers such that their sum is (-32) and their product is (-105).
Clearly, -35+3=-32 and -35×3=-105.

z2-32z-105=z2-35z+3z-105                            =z(z-35)+3(z-35)                            =(z-35)(z+3)

Page No 69:

Question 10:

Factorize:
40 + 3xx2

Answer:

We have:40+3x-x2=-x2-3x-40

We have to split (-3) into two numbers such that their sum is (-3) and their product is (-40).
Clearly, -8+5=-3 and -8×5=-40.

40 + 3x - x2 =-x2-3x-40                             =-x2-8x+5x-40                             =-xx-8+5x-8                             =-x-8x+5                             =8-x5+x

Page No 69:

Question 11:

Factorize:
6 − xx2

Answer:

We have:6-x-x2=-(x2+x-6)

We have to split 1 into two numbers such that their sum is 1 and their product is (-6).
Clearly, 3+-2=1 and 3×-2=-6.

6-x-x2=-x2+x-6                    =-x2+3x-2x-6                    =-xx+3-2x+3                    =-x+3x-2                    =(2-x)(3+x)

Page No 69:

Question 12:

Factorize:
7x2 + 49x + 84

Answer:

We have:7x2+49x+84=7x2+7x+12

We now have to split 7 into two numbers such that their sum is 7 and their product is 12.
Clearly, 3+4=7 and 3×4=12.

7x2+49x+84=7x2+7x+12                             =7x2+4x+3x+12                              =7xx+4+3x+4                             =7x+4x+3

Page No 69:

Question 13:

Factorize:
m2 + 17mn − 84n2

Answer:

We have:
m2+17mn-84n2
We have to split 17mn into two terms such that their sum is 17mn and their product is (-84 m2n2).
Clearly, 21mn+-4mn=17mn and 21mn×-4mn=-84m2n2.

m2+17mn-84n2=m2+21mn-4mn-84n2                                   =mm+21n-4nm+21n                                   =m+21nm-4n

Page No 69:

Question 14:

Factorize:
5x2 + 16x + 3

Answer:

We have:
5x2+16x+3
We have to split 16 into two numbers such that their sum is 16 and their product is 15, i.e., 5×3.
Clearly, 15+1=16 and 15×1=15.

5x2+16x+3=5x2+15x+x+3                           =5xx+3+1x+3                           =x+35x+1

Page No 69:

Question 15:

Factorize:
6x2 + 17x + 12

Answer:

We have:
6x2+17x+12
We have to split 17 into two numbers such that their sum is 17 and their product is 72, i.e., 6×12.
Clearly, 8+9=17 and 8×9=72.

6x2+17x+12=6x2+8x+9x+12                             =2x3x+4+33x+4                             =3x+42x+3

Page No 69:

Question 16:

Factorize:
9x2 + 18x + 8

Answer:

We have:
9x2+18x+8
We have to split 18 into two numbers such that their sum is 18 and their product is 72, i.e., 9×8.
Clearly, 12+6=18 and 12×6=72.

9x2+18x+8=9x2+12x+6x+8                           =3x3x+4+23x+4                           =3x+43x+2

Page No 69:

Question 17:

Factorize:
14x2 + 9x + 1

Answer:

We have:
14x2+9x+1

We have to split 9 into two numbers such that their sum is 9 and their product is 14, i.e., 14×1.
Clearly, 7+2=9 and 7×2=14.

14x2+9x+1=14x2+2x+7x+1                           =2x7x+1+17x+1                           =7x+12x+1

Page No 69:

Question 18:

Factorize:
2x2 + 3x − 90

Answer:

We have:
2x2+3x-90
We have to split 3 into two numbers such that their sum is 3 and their product is (-180), i.e., 2×-90.
Clearly, -12 + 15 = 3 and -12×15 = -180.

2x2+3x-90=2x2-12x+15x-90                          =2xx-6+15x-6                          =x-62x+15

Page No 69:

Question 19:

Factorize:
2x2 + 11x − 21

Answer:

We have:
2x2+11x-21
We have to split 11 into two numbers such that their sum is 11 and their product is (-42), i.e., 2×-21.
Clearly, 14+-3=11 and 14×-3=-42.

2x2+11x-21=2x2+14x-3x-21                             =2xx+7-3x+7                             =x+72x-3

Page No 69:

Question 20:

Factorize:
3x2 − 14x + 8

Answer:

We have:
3x2-14x+8
We have to split (-14) into two numbers such that their sum is (-14) and their product is 24, i.e., 3×8.
Clearly, -12+-2=-14 and -12×-2=24.

3x2-14x+8=3x2-12x-2x+8                           =3xx-4-2x-4                           =x-43x-2

Page No 69:

Question 21:

Factorize:
18x2 + 3x − 10

Answer:

We have:
18x2+3x-10
We have to split 3 into two numbers such that their sum is 3 and their product is (-180), i.e., 18×-10.
Clearly, 15+-12=3 and 15×-12=-180.

18x2+3x-10=18x2+15x-12x-10                             =3x6x+5-26x+5                             =6x+53x-2

Page No 69:

Question 22:

Factorize:
15x2 + 2x − 8

Answer:

We have:
15x2+2x-8
We have to split 2 into two numbers such that their sum is 2 and their product is (-120), i.e., 15×-8.
Clearly, 12+-10=2 and 12×-10=-120.

15x2+2x-8=15x2+12x-10x-8                          =3x5x+4-25x+4                          =5x+43x-2

Page No 69:

Question 23:

Factorize:
6x2 + 11x − 10

Answer:

We have:
6x2+11x-10
We have to split 11 into two numbers such that their sum is 11 and their product is (-60), i.e., 6×-10.
Clearly, 15+-4=11 and 15×-4=-60.

6x2+11x-10=6x2+15x-4x-10                             =3x2x+5-22x+5                             =2x+53x-2

Page No 69:

Question 24:

Factorize:
30x2 + 7x − 15

Answer:

We have:
30x2+7x-15

We have to split 7 into two numbers such that their sum is 7 and their product is (-450), i.e., 30×-15.
Clearly, 25+-18=7 and 25×-18=-450.

30x2+7x-15=30x2+25x-18x-15                             =5x6x+5-36x+5                             =6x+55x-3

Page No 69:

Question 25:

Factorize:
24x2 − 41x + 12

Answer:

We have:
24x2-41x+12
We have to split (-41) into two numbers such that their sum is (-41) and their product is 288, i.e., 24×12.
Clearly, -32+-9=-41 and -32×-9=288.

24x2-41x+12=24x2-32x-9x+12                               =8x3x-4-33x-4                               =3x-48x-3

Page No 69:

Question 26:

Factorize:
2x2 − 7x − 15

Answer:

We have:
2x2-7x-15
We have to split (-7) into two numbers such that their sum is (-7) and their product is (-30), i.e., 2×-15.
Clearly, -10+3=-7 and -10×3=-30.

2x2-7x-15=2x2-10x+3x-15                          =2xx-5+3x-5                          =x-52x+3

Page No 69:

Question 27:

Factorize:
6x2 − 5x − 21

Answer:

We have:
6x2-5x-21
We have to split (-5) into two numbers such that their sum is (-5) and their product is (-126), i.e., 6×-21.
Clearly, 9+-14=-5 and 9×-14=-126.

6x2-5x-21=6x2+9x-14x-21                          =3x2x+3-72x+3                          =2x+33x-7

Page No 69:

Question 28:

Factorize:
10x2 − 9x − 7

Answer:

We have:
10x2-9x-7

We have to split (-9) into two numbers such that their sum is (-9) and their product is (-70), i.e., 10×-7.
Clearly, -14+5=-9 and -14×5=-70.

10x2-9x-7=10x2+5x-14x-7                          =5x2x+1-72x+1                          =2x+15x-7

Page No 69:

Question 29:

Factorize:
5x2 − 16x − 21

Answer:

We have:
5x2-16x-21
We have to split (-16) into two numbers such that their sum is (-16) and their product is (-105), i.e., 5×-21.
Clearly, -21+5=-16 and -21×5=-105.

5x2-16x-21=5x2+5x-21x-21                             =5xx+1-21x+1                             =x+15x-21

Page No 69:

Question 30:

Factorize:
2x2x − 21

Answer:

We have:
2x2-x-21
We have to split (-1) into two numbers such that their sum is (-1) and their product is (-42), i.e., 2×-21.
Clearly, -7+6=-1 and -7×6=-42.

2x2-x-21=2x2+6x-7x-21                        =2x(x+3)-7(x+3)                        =(x+3)(2x-7)

Page No 69:

Question 31:

Factorize:
15x2x − 128

Answer:

We have:
15x2-x-28
We have to split (-1) into two numbers such that their sum is (-1) and their product is (-420), i.e., 15×-28.
Clearly, -21+20=-1 and -21×20=-420.

15x2-x-28=15x2-21x+20x-28                          =3x(5x-7)+4(5x-7)                          =(5x-7)(3x+4)

Page No 69:

Question 32:

Factorize:
8a2 − 27ab + 9b2

Answer:

We have:
8a2-27ab+9b2

We have to split (-27) into two numbers such that their sum is (-27) and their product is 72, i.e., 8×9.
Clearly, -24+-3=-27 and -24×-3=72.

8a2-27ab+9b2=8a2-24ab-3ab+9b2                                 =8a(a-3b)-3b(a-3b)                                 =(a-3b)(8a-3b)

Page No 69:

Question 33:

Factorize:
5x2 + 33xy − 14y2

Answer:

We have:
5x2+33xy-14y2
We have to split 33 into two numbers such that their sum is 33 and their product is (-70), i.e., 5×-14.
Clearly, 35+-2=33 and 35×-2=-70.

5x2+33xy-14y2=5x2+35xy-2xy-14y2                                   =5x(x+7y)-2y(x+7y)                                   =(x+7y)(5x-2y)

Page No 69:

Question 34:

Factorize:
3x3x2 − 10x

Answer:

We have:3x3-x2-10x=x(3x2-x-10)
We have to split (-1) into two numbers such that their sum is (-1) and their product is (-30), i.e., 3×-10.
Clearly, -6+5=-1 and -6×5=-30.

3x3 - x2 - 10x =x3x2-x-10                                 = x3x2-6x+5x-10                                 =x3x(x-2)+5(x-2)                                 =x(x-2)(3x+5)

Page No 69:

Question 35:

Factorize:
13x2-2x-9

Answer:

We have:13x2-2x-9=x2-6x-273=13x2-6x-27
Now, we have to split (-6) into two numbers such that their sum is (-6) and their product is (-27).
Clearly, -9+3=-6 and -9×3=-27.

13x2 - 2x - 9 = 13(x2+-6x-27)                                =13(x2+3x-9x-27)                                =13x(x+3)-9(x+3)                                =13(x+3)(x-9)

Page No 69:

Question 36:

Factorize:
x2-2x+716

Answer:

We have:x2-2x+716=16x2-32x+716=11616x2-32x+7

Now, we have to split (-32) into two numbers such that their sum is (-32) and their product is 112, i.e., 16×7.
Clearly, -4+-28=-32 and -4×-28=112.

x2 - 2x + 716 =116(16x2-32x+7)                                =116(16x2-4x-28x+7)                                =1164x(4x-1)-7(4x-1)                                =116(4x-1)(4x-7)

Page No 69:

Question 37:

Factorize:
2x2+3x+2

Answer:

We have:
2x2+3x+2
We have to split 3 into two numbers such that their sum is 3 and their product is 2, i.e., 2×2.
Clearly, 2+1=3 and 2×1=2.

2x2+3x+2=2x2+2x+x+2                                =2xx+2+1x+2                                =x+22x+1

Page No 69:

Question 38:

Factorize:
5x2+2x-35

Answer:

We have:
5x2+2x-35
We have to split 2 into two numbers such that their sum is 2 and product is (-15), i.e.,5×-35.
Clearly, 5+-3=2 and 5×-3=-15.

5x2+2x-35=5x2+5x-3x-35                                  =5xx+5-3x+5                                  =x+55x-3

Page No 69:

Question 39:

Factorize:
2x2+33x+3

Answer:

We have:
2x2+33x+3
We have to split 33 into two numbers such that their sum is 33 and their product is 6, i.e.,2×3.
Clearly, 23+3=33 and 23×3=6.

2x2+33x+3=2x2+23x+3x+3                              =2xx+3+3x+3                              =x+32x+3

Page No 69:

Question 40:

Factorize:
23x2+x-53

Answer:

We have:
23x2+x-53
We have to split 1 into two numbers such that their sum is 1 and product is 30, i.e.,23×-53.
Clearly, 6+-5=1 and 6×-5=-30.

23x2+x-53=23x2+6x-5x-53                                  =23xx+3-5x+3                                  =x+323x-5

Page No 69:

Question 41:

Factorize:
55x2+20x+35

Answer:

We have:
55x2+20x+35
We have to split 20 into two numbers such that their sum is 20 and their product is 75.
Clearly, 
15+5=20 and 15×5=75

55x2+20x+35=55x2+15x+5x+35                                       =5x5x+3+5(5x+3)                                       =5x+35x+5

Page No 69:

Question 42:

Factorize:
72x3-10x-42

Answer:

We have:
72x2-10x-42
We have to split (-10) into two numbers such that their sum is (-10) and their product is (-56), i.e., 72×-42=-28×2.
Clearly, -14+4=-10 and -14×4=-56.

72x2-10x-42=72x2-14x+4x-42                                      =72xx-2+4x-2                                      =x-272x+4

Page No 69:

Question 43:

Factorize:
63x2-47x+53

Answer:

We have:
63x2-47x+53
Now, we have to split (-47) into two numbers such that their sum is (-47) and their product is 90.
Clearly, -45+-2=-47 and -45×-2=90.

63x2-47x+53 =63x2-2x-45x+53                                        =2x33x-1-5333x-1                                        =33x-12x-53

Page No 69:

Question 44:

Factorize:
7x2+214x+2

Answer:

We have:
7x2+214x+2
We have to split 214 into two numbers such that their sum is 214 and product is 14.
Clearly, 14+14=214 and 14×14=14.
7x2+214x+2=7x2+14x+14x+2                                 =7x7x+2+27x+2                                 =7x+27x+2                                 =7x+22

Page No 69:

Question 45:

Factorize:
2(x + y)2 − 9(x + y) − 5

Answer:

We have:
2x+y2-9x+y-5Let:(x+y)=u
Thus, the given expression becomes
2u2-9u-5
Now, we have to split (-9) into two numbers such that their sum is (-9) and their product is (-10).
Clearly, -10+1=-9 and -10×1=-10.

2u2-9u-5=2u2-10u+u-5                         =2u(u-5)+1(u-5)                         =(u-5)(2u+1)
Putting u=(x+y), we get:
2x+y2 - 9x+y - 5 = x+y-52x+y+1                                          = x+y-52x+2y+1

Page No 69:

Question 46:

Factorize:
9(2ab)2 − 4(2ab) − 13

Answer:

We have:
9(2a-b)2-4(2a-b)-13Let:(2a-b)=p
Thus, the given expression becomes
9p2-4p-13
Now, we must split (-4) into two numbers such that their sum is (-4) and their product is (-117).
Clearly, -13+9=-4 and -13×9=-117.
9p2-4p-13=9p2+9p-13p-13                           =9p(p+1)-13(p+1)                           =(p+1)(9p-13)
Putting p=(2a-b), we get:
92a-b2-42a-b-13=2a-b+192a-b-13                                           =2a-b+118a-9b-13

Page No 69:

Question 47:

Factorize:
7(x − 2y)2 − 25(x − 2y) + 12

Answer:

We have:
7(x-2y)2-25(x-2y)+12Let:(x-2y)=a
7x-2y2-25x-2y+12=7a2-25a+12
Now, we have to split (-25) into two numbers such that their sum is (-25) and their product is 84.
Clearly, -21+-4=-25 and -21×-4=84.

7a2-25a+12=7a2-21a-4a+12                             =7a(a-3)-4(a-3)                             =(a-3)(7a-4)
Putting a=x-2y, we get:
7(x-2y)2-25(x-2y)+12 = x-2y-37x-2y-4                                               =x-2y-37x-14y-4

Page No 69:

Question 48:

Factorize:
4x4 + 7x2 − 2

Answer:

We have:4x4+7x2-2=4x22+7x2-2Let: x2=a4x4+7x2-2=4a2+7a-2Splitting 7 such that the sum is 7 and the product is 4×-2=-8.Clearly, 8+-1=7 and 8×-1=-8.Thus, we have:4a2+7a-2=4a2+8a-a-2                        =4a(a+2)-1(a+2)                        =(a+2)(4a-1)
Substituting a=x2, we get:
4x4 + 7x2 - 2 =x2+24x2-1                            = x2+22x2-12                           = x2+22x-12x+1



Page No 72:

Question 1:

Expand:
(i) (a + 2b + 5c)2
(ii) (2bb + c)2
(iii) (a − 2b − 3c)2

Answer:

i a+2b+5c2=a2 + 2b2 +5c2+2a2b+22b5c+25ca                           =a2+4b2+25c2+4ab+20bc+10ac

ii 2a-b+c2=2a+-b+c2                          =2a2+-b2+c2+22a-b+2-bc+4ac                          =4a2+b2+c2-4ab-2bc+4ac

iii a-2b-3c2=a+-2b+-3c2                             =a2+-2b2+-3c2+2a-2b+2-2b-3c+2a-3c                             =a2+4b2+9c2-4ab+12bc-6ac

Page No 72:

Question 2:

Expand:
(i) (2a − 5b − 7c)2
(ii) (−3a + 4b − 5c)2
(iii) 12a-14a+22

Answer:

i 2a-5b-7c2=2a+-5b+-7c2                             =2a2+-5b2+-7c2+22a-5b+2-5b-7c+22a-7c                             =4a2+25b2+49c2-20ab+70bc-28ac

ii -3a+4b-5c2=-3a+4b+-5c2                                 =-3a2+4b2+-5c2+2-3a4b+24b-5c+2-3a-5c                                 =9a2+16b2+25c2-24ab-40bc+30ac

iii 12a-14b+22=a2+-b4+22                                  =a22+-b42+22+2a2-b4+2-b42+2a22                                  =a24+b216+4-ab4-b+2a

Page No 72:

Question 3:

Factorize: 4x2 + 9y2 + 16z2 + 12xy − 24yz − 16xz.

Answer:


We have: 4x2+9y2+16z2+12xy-24yz-16xz=2x2+3y2+-4z2+2(2x)(3y)+2(3y)(-4z)+2(-4z)(2x)=2x+3y+-4z2=2x+3y-4z2

Page No 72:

Question 4:

Factorize: 9x2 + 16y2 + 4z2 − 24xy + 16yz − 12xz

Answer:

We have:9x2+16y2+4z2-24xy+16yz-12xz=-3x2+4y2+2z2+2(-3x)(4y)+2(4y)(2z)+2(2z)(-3x)=-3x+4y+2z2=-3x+4y+2z2

Page No 72:

Question 5:

Factorize: 25x2 + 4y2 + 9z2 − 20xy − 12yz + 30xz.

Answer:

We have:25x2+4y2+9z2-20xy-12yz+30xz=5x2+-2y2+3z2+2(5x)(-2y)+2(-2y)(3z)+2(3z)(5x)=5x+-2y+3z2=5x-2y+3z2

Page No 72:

Question 6:

Evaluate:
(i) (99)2
(ii) (998)2

Answer:

i 992=100-12             =100+-12             = 1002+2×100×-1+-12             =10000-200+1             =9801         

ii 9982=1000-22                =1000+-22                =10002+2×1000×-2+-22                =1000000-4000+4                =996004



Page No 73:

Question 1:

Expand:
(i) (3x + 2)3
(ii) (3a − 2b)3
(iii) 23x+13

Answer:

i 3x+23=3x3+3×3x2x2+3×3x×22+23                   =27x3+54x2+36x+8

ii 3a-2b3=3a3-3×3a2×2b+3×3a×2b2-2b3                      = 27a3-54a2b+36ab2-8b3

iii 23x+13=23x3+3×23x2×1+3x23x×12+13                         =827x3+43x2+2x+1

Page No 73:

Question 2:

Expand:
(i) 2x-2x3
(ii) 3a+14b3
(iii) 45x-23

Answer:


 i 2x-2x3=2x3-2x3+32x2-2x+32x-2x2                      =8x3-8x3+34x2-2x+32x4x2                      =8x3-8x3+-24x+24x


ii 3a+14b3=3a3+14b3+33a214b+33a14b2                          =27a3+164b3+27a24b+9a16b2


iii 45x-23=45x3-23-345x22+345x22                        =64125x3-8-9625x2+485x

Page No 73:

Question 3:

Evaluate:
(i) (95)3
(ii) (999)3

Answer:

i 953=90+53             =903+3×902×5+3×90×52+53             =729000+121500+6750+125             =857375

ii 9993=1000-13                =10003-3×10002×1+3×1000×12-13                =1000000000-3000000+3000-1                =997002999



Page No 76:

Question 1:

Factorize:
x3 + 27

Answer:

x3+27=x3+33            =x+3x2-3x+32            =x+3x2-3x+9

Page No 76:

Question 2:

Factorize:
8x3 + 27y3

Answer:

8x3+27y3=2x3+3y3                  =2x+3y2x2-2x×3y+3y2                  =2x+3y4x2-6xy+9y2

Page No 76:

Question 3:

Factorize:
343 + 125b3

Answer:

343+125b3=73+5b3                     =7+5b72-7×5b+5b2                     =7+5b49-35b+25b2

Page No 76:

Question 4:

Factorize:
1 + 64x3

Answer:


1+64x3=13+4x3               =1+4x12-1×4x+4x2               =1+4x1-4x+16x2

Page No 76:

Question 5:

Factorize:
125a3+18

Answer:

125a3+18=5a3+123                   =5a+125a2-5a×12+122                   =5a+1225a2-5a2+14

Page No 76:

Question 6:

Factorize:
216x3+1125

Answer:

216x3+1125=6x3+153                        =6x+156x2-6x×15+152                        =6x+1536x2-6x5+125

Page No 76:

Question 7:

Factorize:
16x4 + 54x

Answer:


16x4+54x=2x8x3+27                    =2x2x3+33                    =2x2x+32x2-2x×3+32                    =2x2x+34x2-6x+9

Page No 76:

Question 8:

Factorize:
7a3 + 56b3

Answer:

7a3+56b3=7a3+8b3                   =7a3+2b3                   =7a+2ba2-a×2b+2b2                   =7a+2ba2-2ab+4b2

Page No 76:

Question 9:

Factorize:
x5 + x2

Answer:

x5+x2=x2x3+1            =x2x3+13            =x2x+1x2-x×1+12            =x2x+1x2-x+1

Page No 76:

Question 10:

Factorize:
a3 + 0.008

Answer:

a3+0.008=a3+0.23                  =a+0.2a2-a×0.2+0.22                  =a+0.2a2-0.2a+0.04



Page No 77:

Question 11:

Factorize:
x6 + y6

Answer:

x6+y6=x23+y23            =x2+y2x22-x2y2+y22            =x2+y2x4-x2y2+y4

Page No 77:

Question 12:

Factorize:
2a3 + 16b3 − 5a − 10b

Answer:

2a3+16b3-5a-10b=2a3+8b3-5a+2b                                        =2a3+2b3-5a+2b                                        =2a+2ba2-a×2b+2b2-5a+2b                                        =2a+2ba2-2ab+4b2-5a+2b                                        =a+2b2a2-2ab+4b2-5

Page No 77:

Question 13:

Factorize:
x3 − 512

Answer:

x3-512 =x3-83               =x-8x2+8x+82               =x-8x2+8x+64

Page No 77:

Question 14:

Factorize:
64x3 − 343

Answer:

64x3-343=4x3-73                   =4x-716x2+4x×7+72                   =4x-716x2+28x+49

Page No 77:

Question 15:

Factorize:
1 − 27x3

Answer:


1-27x3=13-3x3              =1-3x12+1×3x+3x2              =1-3x1+3x+9x2

Page No 77:

Question 16:

Factorize:
x3 − 125y3

Answer:

x3-125y3=x3-5y3                  =x-5yx2+x×5y+5y2                  =x-5yx2+5xy+25y2

Page No 77:

Question 17:

Factorize:
8x3-127y3

Answer:

8x3-127y3=2x3-13y3                    =2x-13y2x2+2x×13y+13y2                    =2x-13y4x2+2x3y+19y2

Page No 77:

Question 18:

Factorize:
a3 − 0.064

Answer:

a3-0.064=a3-0.43                  =a-0.4a2+a×0.4+0.42                  =a-0.4a2+0.4a+0.16

Page No 77:

Question 19:

Factorize:
(a + b)3 − 8

Answer:

a+b3-8=a+b3-23                  =a+b-2a+b2+a+b×2+22                  =a+b-2a2+2ab+b2+2a+b+4                     

Page No 77:

Question 20:

Factorize:
x6 − 729

Answer:

x6-729=x23-93             =x2-9x22+x2×9+92             =x2-32x4+9x2+81             =x+3x-3x4+18x2+81-9x2             =x+3x-3x22+2×x2×9+92-9x2             =x+3x-3x2+92-3x2             =x+3x-3x2+9+3xx2+9-3x             =x+3x-3x2+3x+9x2-3x+9

Page No 77:

Question 21:

Factorize:
(a + b)3 − (ab)3

Answer:

a + b3-a-b3=a+b-a-ba+b2+a+ba-b+a-b2                            =a+b-a+ba2+2ab+b2+a2-b2+a2-2ab+b2                            =2b3a2+b2

Page No 77:

Question 22:

Factorize:
x − 8xy3

Answer:

x-8xy3=x1-8y3              =x13-2y3              =x1-2y12+1×2y+2y2              =x1-2y1+2y+4y2

Page No 77:

Question 23:

Factorize:
32x4 − 500x

Answer:

32x4-500x=4x8x3-125                      =4x2x3-53                      =4x2x-52x2+2x×5+52                      =4x2x-54x2+10x+25

Page No 77:

Question 24:

Factorize:
3a7b − 81a4b4

Answer:

3a7b-81a4b4=3a4ba3-27b3                         =3a4ba3-3b3                         =3a4ba-3ba2+a×3b+3b2                         =3a4ba-3ba2+3ab+9b2

Page No 77:

Question 25:

Factorize:
a3-1a3-2a+2a

Answer:

a3-1a3-2a+2a=a3-1a3-2a-1a                                 =a3-1a3-2a-1a                                 =a-1aa2+a×1a+1a2-2a-1a                                 =a-1aa2+1+1a2-2a-1a                                 =a-1aa2+1+1a2-2                                 =a-1aa2-1+1a2

Page No 77:

Question 26:

Factorize:
8a3b3 − 4ax + 2bx

Answer:

8a3-b3-4ax+2bx=2a3-b3-2x2a-b                                    =2a-b2a2+2ab+b2-2x2a-b                                    =2a-b4a2+2ab+b2-2x2a-b                                    =2a-b4a2+2ab+b2-2x

Page No 77:

Question 27:

Factorize:
a3 + 3a2b + 3ab2 + b3 − 8

Answer:

a3+3a2b+3ab2+b3-8=a3+b3+3a2b+3ab2-8                                            =a3+b3+3aba+b-8                                            =a+b3-23                                            =a+b-2a+b2+2a+b+22                                            =a+b-2a+b2+2a+b+4                                            



Page No 80:

Question 1:

Factorize:
125a3 + b3 + 64c3 − 60abc

Answer:

125a3+b3+64c3-60abc=5a3+b3+4c3-3×5a×b×4c                                              =5a+b+4c5a2+b2+4c2-5a×b-b×4c-5a×4c                                              =5a+b+4c25a2+b2+16c2-5ab-4bc-20ac                          

Page No 80:

Question 2:

Factorize:
a3 + 8b3 + 64c3 − 24abc

Answer:

a3+8b3+64c3-24abc=a3+2b3+4c3-3×a×2b×4c                                          =a+2b+4ca2+2b2+4c2-a×2b-2b×4c-4c×a                                          =a+2b+4ca2+4b2+16c2-2ab-8bc-4ca

Page No 80:

Question 3:

Factorize:
1 + b3 + 8c3 − 6bc

Answer:

1+b3+8c3-6bc=13+b3+2c3-3×1×b×2c                               =1+b+2c12+b2+2c2-1×b-b×2c-1×2c                               =1+b+2c1+b2+4c2-b-2bc-2c

Page No 80:

Question 4:

Factorize:
216 + 27b3 + 8c3 − 108abc

Answer:

216+27b3+8c3-108abc=63+3b3+2c3-3×6×3b×2c                                               =6+3b+2c62+3b2+2c2-6×3b-3b×2c-2c×6                                               =6+3b+2c36+9b2+4c2-18b-6bc-12c

Page No 80:

Question 5:

Factorize:
27a3b3 + 8c3 + 18abc

Answer:

27a3-b3+8c3+18abc=3a3+-b3+2c3-3×3a×-b×2c                                         =3a+-b+2c3a2+-b2+2c2-3a-b--b2c-3a×2c                                         =3a-b+2c9a2+b2+4c2+3ab+2bc-6ac

Page No 80:

Question 6:

Factorize:
8a3 + 125b3 − 64c3 + 120abc

Answer:

8a3+125b3-64c3+120abc=2a3+5b3+-4c3-3×2a×5b×-4c                                                   =2a+5b-4c2a2+5b2+-4c2-2a5b-5b-4c-2a×-4c                                                   =2a+5b-4c4a2+25b2+16c2-10ab+20bc+8ac

Page No 80:

Question 7:

Factorize:
8 − 27b3 − 343c3 − 126bc

Answer:

8-27b3-343c3-126bc=23+-3b3+-7c3-3×2×-3b×-7c                                             =2+-3b+-7c22+-3b2+-7c2-2-3b--3b-7c-2-7c                                             =2-3b-7c4+9b2+49c2+6b-21bc+14c

Page No 80:

Question 8:

Factorize:
125 − 8x3 − 27y3 − 90xy

Answer:

125-8x3-27y3-90xy=53+-2x3+-3y3-3×5×-2x×-3y                                          =5+-2x +-3y52+-2x2+-3y2-5×-2x--2x-3y-5×-3y                                          =5-2x-3y25+4x2+9y2+10x-6xy+15y

Page No 80:

Question 9:

Factorize:
22a3 + 162b3+c3-12abc

Answer:

22a3+162b3+c3-12abc=2a3+22b3+c3-3×2a×22b×c                                                    =2a+22b+c2a2+22b2+c2-2a×22b-22b×c-2a×c                                                    =2a+22b+c2a2+8b2+c2-4ab-22bc-2ac

Page No 80:

Question 10:

Factorize:
x3 + y3 − 12xy + 64

Answer:

x3+y3-12xy+64=x3+y3+64-12xy                                =x3+y3+43-3×x×y×4                                =x+y+4x2+y2+42-xy-4y-4x                                 =x+y+4x2+y2+16-xy-4y-4x 

Page No 80:

Question 11:

Factorize:
(ab)3 + (bc)3 + (ca)3

Answer:

a-b3+b-c3+c-a3
Putting a-b=x, b-c=y and c-a=z, we get:a-b3+b-c3+c-a3=x3+y3+z3        [Where x+y+z=a-b+b-c+c-a=0]=3xyz                 x+y+z=0 x3+y3+z3=3xyz=3a-bb-cc-a

Page No 80:

Question 12:

Factorize:
(3a − 2b)3 + (2b − 5c)3 + (5c − 3a)3

Answer:

Put 3a-2b=x, 2b-5c=y and 5c-3a=z.We have:x+y+z = 3a-2b+2b-5c+5c-3a=0Now,3a-2b3+2b-5c3+5c-3a3=x3+y3+z3                                                   =3xyz    Here, x+y+z=0. So, x3 + y3 +z3 = 3xyz                                                   =33a-2b2b-5c5c-3a

Page No 80:

Question 13:

Factorize:
a3(bc)3 + b3(ca)3 + c3(ab)3

Answer:

We have:a3b-c3+b3c-a3+c3a-b3 = ab-c3+bc-a3+ca-b3Put ab-c = x      bc-a = y      ca-b = z Here, x+y+z = ab-c+bc-a+ca-b              =ab - ac + bc - ab + ac - bc              =0Thus, we have:a3b-c3+b3c-a3+c3a-b3 =x3 + y3+ z3                                                   =3xyz      When x+y+z =0, x3 + y3+ z3 =3xyz.                                                   =3 ab-cbc-aca-b                                                   =3abca-bb-cc-a

Page No 80:

Question 14:

Factorize:
(5a − 7b)3 + (9c − 5a)3 + (7b − 9c)3

Answer:

Put 5a-7b=x, 9c-5a=z and 7b-9c=y.Here, x+y+z = 5a - 7b + 9c-5a+7b-9c=0Thus, we have:5a-7b3+9c-5a3+7b-9c3=x3+z3+y3                                                   =3xzy   When x+y+z=0, x3+y3+z3 = 3xyz.                                                   =3 5a-7b9c-5a7b-9c



Page No 81:

Question 15:

Find the product:
(x + yz) (x2 + y2 + z2xy + yz + zx)

Answer:

   x+y-zx2+y2+z2-xy+yz+zx=x+y+-zx2+y2+-z2-xy-y×-z--z×x=x3+y3+-z3-3x×y×-z=x3+y3-z3+3xyz

Page No 81:

Question 16:

Find the product:
(x − 2y + 3) (x2 + 4y2 + 2xy − 3x + 6y + 9)

Answer:

   x − 2y + 3x2+4y2+2xy-3x+6y+9=x − 2y + 3x2+4y2+9+2xy+6y-3x=x+-2y+3x2+-2y2+32-x×-2y--2y×3-3×x=x3+-2y3+33-3x-2y3=x3-8y3+27+18xy

Page No 81:

Question 17:

Find the product:
(x − 2y z) (x2 + 4y2 + z2 + 2xy + zx − 2yz)

Answer:

   x-2y-zx2+4y2+z2+2xy+zx-2yz=x-2y-zx2+4y2+z2+2xy-2yz+zx=x+-2y+-zx2+4y2+z2-x×-2y--2y-z--z×x=x3+-2y3+-z3-3×x-2y×-z=x3-8y3-z3-6xyz

Page No 81:

Question 18:

If x + y + 4 = 0, find the value of (x3 + y3 − 12xy + 64).

Answer:

  x3+y3-12xy+64=x3+y3+64-12xy=x3+y3+43-3×x×y×4=x+y+4x2+y2+42-xy-4y-4x=0×x2+y2+16-xy-4y-4x            Given: x + y + 4 = 0=0

Page No 81:

Question 19:

If x = 2y + 6, find the value of (x3 − 8y3 − 36xy − 216)

Answer:

    x3-8y3-216-36xy=x3+-2y3+-63-3×x×-2y×-6=x+-2y+-6x2+-2y2+-62-x-2y--2y×-6--6x=x-2y-6x2+4y2+36+2xy-12y+6x=2y+6-2y-6 x2+4y2+36+2xy-12y+6x     Given: x=2y+6=0×x2+4y2+36+2xy-12y+6x=0                                                               



Page No 82:

Question 1:

Which of the following expressions is a polynomial in one variable?
(a) x+2x+3
(b) 3x+2x+5
(c) 2x2 - 3x + 6
(d) x10 + y5 + 8

Answer:

(c) 2x2 - 3x + 6

Clearly, 2x2-3x+6 is a polynomial in one variable because it has only non-negative integral powers of x.

Page No 82:

Question 2:

Which of the following expression is a polynomial?
(a) x-1
(b) x-1x+1
(c) x2-2x2+5
(d) x2+2x3/2x+6

Answer:

(d) x2+2x3/2x+6

We have:

x2+2x32x+6=x2+2x32x-12+6
               =x2+2x+6
 
It a polynomial because it has only non-negative integral powers of x.

Page No 82:

Question 3:

(a) y3+4
(b) y-3
(c) y
(d) 1y+7

Answer:

(c) y

 y is a polynomial because it has a non-negative integral power 1.

Page No 82:

Question 4:

Which of the following is a polynimial?
(a) x-1x+2
(b) 1x+5
(c) x+3
(d) −4

Answer:

(d) −4

-4 is a constant polynomial of degree zero.

Page No 82:

Question 5:

Which of the following is a polynomial?
(a) x−2 + x−1 + 3
(b) x + x−1 + 2
(c) x−1
(d) 0

Answer:

(d) 0

0 is a polynomial whose degree is not defined.

Page No 82:

Question 6:

Which of the following is quadratic polynomial?
(a) x + 4
(b) x3 + x
(c) x3 + 2x + 6
(d) x2 + 5x + 4

Answer:

(d) x2 + 5x + 4

x2+5x+4 is a polynomial of degree 2. So, it is a quadratic polynomial.

Page No 82:

Question 7:

Which of the following is a linear polynomial?
(a) x + x2
(b) x + 1
(c) 5x2x + 3
(d) x+1x

Answer:

(b) x + 1

Clearly, x+1 is a polynomial of degree 1. So, it is a linear polynomial.

Page No 82:

Question 8:

Which of the following is a binomial?
(a) x2 + x + 3
(b) x2 + 4
(c) 2x2
(d) x+3+1x

Answer:

(b) x2 + 4

Clearly, x2+4 is an expression having two non-zero terms. So, it is a binomial.

Page No 82:

Question 9:

3is a polynomial of degree
(a) 12
(b) 2
(c) 1
(d) 0

Answer:

(d) 0

3 is a constant term, so it is a polynomial of degree 0.

Page No 82:

Question 10:

Degree of the zero polynomial is
(a) 1
(b) 0
(c) not defined
(d) none of these

Answer:

(c) not defined

Degree of the zero polynomial is not defined.

Page No 82:

Question 11:

Zero of the polynomial p(x) = 2x + 3 is
(a) 32
(b) -32
(c) -23
(d) 12

Answer:

(b) -32

Zero of the polynomial p(x) is given by p(x) = 0.
Thus, we have:
2x+3=02x=-3x=-32



Page No 83:

Question 12:

Zero is the polynomial p(x) = 2 − 5x is
(a) 25
(b) 52
(c) -25
(d) -52

Answer:

(a) 25

Zero of the polynomial p(x) is given by p(x) = 0.
Thus, we have:
2-5x=02=5xx=25

Page No 83:

Question 13:

Zero of the zero polynomial is
(a) 0
(b) 1
(c) every real number
(d) not defined

Answer:

(d) not defined

Zero of the zero polynomial is not defined.

Page No 83:

Question 14:

If p(x) = x = 4, then p(x) + p(−x) = ?
(a) 0
(b) 4
(c) 2x
(d) 8

Answer:

(d) 8

Let:
px=x+4

 p-x=-x+4             =-x+4
Thus, we have:
px+p-x=x+4+-x+4
               = 4 + 4
               =8

Page No 83:

Question 15:

If p(x)=x2-22x+1, then p22=?
(a) 0
(b) 1
(c) 42
(d) −1

Answer:

(b) 1

px=x2-22 x+1 p22=222-22×22+1
             = 8 - 8 + 1
             = 1

Page No 83:

Question 16:

The zeroes of the polynomial p(x) = x2 + x − 6 are
(a) 2, 3
(b) −2, 3
(c) 2, −3
(d) −2, −3

Answer:

(c) 2, −3

The zeroes of the polynomial p(x) is given by p(x) = 0.
We have:
x2+x-6=0x2+3x-2x-6=0xx+3-2x+3=0x+3x-2=0x=-3 or x=2

Hence, 2 and -3 are the zeroes of the given polynomial.

Page No 83:

Question 17:

The zeroes of the polynomial p(x) = 2x2 + 5x − 3 are
(a) 12, 3
(b) 12, -3
(c) -12, 3
(d) 1, -12

Answer:

(b) 12, -3

The zeroes of the polynomial p(x) is given by p(x) = 0.
We have:
2x2+5x-3=02x2+6x-x-3=02xx+3-1x+3=0x+32x-1=0x=-3 or x=12
Hence, 12 and -3 are the zeroes of the given polynomial.

Page No 83:

Question 18:

If (x2 + kx − 3) = (x − 3) (x + 1), then k = ?
(a) 2
(b) −2
(c) 3
(d) −1

Answer:

(b) −2

x-3x+1=x2-2x-3
Thus, we have:
x2+kx-3=x2-2x-3k=-2

Page No 83:

Question 19:

If (x + 1) is a factor of (2x2 + kx), then k = ?
(a) −3
(b) −2
(c) 2
(d) 4

Answer:

(c) 2

x+1 is a factor of 2x2+kx.So, -1 is a zero of 2x2+kx.Thus, we have:2×-12+k×-1=02-k=0k=2

Page No 83:

Question 20:

The coefficient of the highest power of x in the polynomial 2x3 − 4x4 + 5x2x5 + 3 is
(a) 2
(b) −4
(c) 3
(d) − 1

Answer:

(d) −1

The highest power of x in the given polynomial is 5.
The coefficient of x5 is −1.

Page No 83:

Question 21:

When (x31 = 31) is divided by (x + 1), the remainder is
(a) 0
(b) 1
(c) 30
(d) 31

Answer:

(c) 30

Let:
px=x31+31
x+1=0x=-1
By the remainder theorem, we know that when p(x) is divided by (x + 1), the remainder is p(-1).
Thus, we have:
p-1=-131+31        =-1+31        =30

Page No 83:

Question 22:

When p(x) = x3ax2 + a is divided by (xa), the remainder is
(a) 0
(b) a
(c) 2a
(d) 3a

Answer:

(b) a

x-a=0x=a
By the remainder theorem, we know that when p(x) is divided by (x - a), the remainder is p(a).
Thus, we have:
pa=a3-a×a2+a      =a3-a3+a      =a

Page No 83:

Question 23:

When p(x) = x3 + ax2 + 2x + a is divided by (x + a), the remainder is
(a) 0
(b) −1
(c) −15
(d) 21

Answer:

(c) −a

x+a=0x=-a
By the remainder theorem, we know that when p (x) is divided by (x + a), the remainder is p (−a).
Thus, we have:
p-a=-a3+a×-a2+2×-a+a         =-a3+a3-2a+a         =-a

Page No 83:

Question 24:

When p(x) = x4 + 2x3 − 3x2 + x − 1 is divided by (x − 2), the remainder is
(a) 0
(b) −1
(c) −15
(d) 21

Answer:

(d) 21

x-2=0x=2
By the remainder theorem, we know that when p(x) is divided by (x - 2), the remainder is p(2).
Thus, we have:
p2=24+2×23-3×22+2-1       =16+16-12+1       =21

Page No 83:

Question 25:

When p(x) = x3 − 3x2 + 4x + 32 is divided by (x + 2), the remainder is
(a) 0
(b) 32
(c) 36
(d) 4

Answer:

(d) 4

x+2=0x=-2
By the remainder theorem, we know that when p(x) is divided by (x + 2), the remainder is p(-2).
Now, we have:
p-2=-23-3×-22+4×-2+32         =-8-12-8+32         =4

Page No 83:

Question 26:

When p(x) = 4x3 − 12x2 + 11x − 5 is divided by (2x − 1), the remainder is
(a) 0
(b) −5
(c) −2
(d) 2

Answer:

(c) −2

2x-1=0x=12
By the remainder theorem, we know that when p(x) is divided by (2x - 1), the remainder is p12.
Now, we have:
 p12=4×123-12×122+11×12-5        =12-3+112-5        =-2



Page No 84:

Question 27:

(x + 1) is a factor of the polynomial
(a) x3 − 2x2 + x + 2
(b) x3 + 2x2 + x − 2
(c) x3 − 2x2 − x − 2
(d) x3 − 2x2 − x + 2

Answer:

(c) x3 − 2x2 − x − 2

Let:
fx=x3-2x2+x+2
By the factor theorem, (x + 1) will be a factor of f (x) if f (−1) = 0.
We have:
f-1=-13-2×-12+-1+2        =-1-2-1+2        =-20
Hence, (x + 1) is not a factor of fx=x3-2x2+x+2.

Now,
Let:
fx=x3+2x2+x-2

By the factor theorem, (x + 1) will be a factor of f (x) if f (-1) = 0.
We have:
f-1=-13+2×-12+-1-2        =-1+2-1-2        =-20
Hence, (x + 1) is not a factor of fx=x3+2x2+x-2.

Now,
Let:
fx=x3+2x2-x-2

By the factor theorem, (x + 1) will be a factor of f (x) if f (-1) = 0.
We have:
f-1=-13+2×-12--1-2        =-1+2+1-2        =0
Hence, (x + 1) is a factor of fx=x3+2x2-x-2.

Page No 84:

Question 28:

(4x2 + 4x − 3) = ?
(a) (2x − 1) (2x − 3)
(b) (2x + 1) (2x − 3)
(c) (2x + 3) (2x − 1)
(d) none of these

Answer:

(c) (2x + 3) (2x − 1)

4x2+4x-3=4x2+6x-2x-3
             =2x2x+3-12x+3=2x+32x-1

Page No 84:

Question 29:

6x2 + 17x + 5 = ?
(a) (2x + 1)(3x + 5)
(b) (2x + 5)(3x + 1)
(c) (6x + 5)(x + 1)
(d) none of these

Answer:

(b) (2x + 5)(3x + 1)

6x2+17x+5=6x2+15x+2x+5
              =3x2x+5+12x+5=2x+53x+1

Page No 84:

Question 30:

(x2 − 4x − 21) = ?
(a) (x − 7)(x − 3)
(b) (x + 7)(x − 3)
(c) (x − 7)(x + 3)
(d) none of these

Answer:

(c) (x − 7)(x + 3)

x2-4x-21=x2-7x+3x-21
             =xx-7+3x-7=x-7x+3

Page No 84:

Question 31:

If (x + 5) is a factor of p(x) = x3 − 20x + 5k, then k = ?
(a) −5
(b) 5
(c) 3
(d) −3

Answer:

(b) 5

x+5 is a factor of px=x3-20x+5k. p-5=0-53-20×-5+5k=0-125+100+5k=05k=25k=5

Page No 84:

Question 32:

3x3 + 2x2 + 3x + 2 = ?
(a) (3x − 2)(x2 − 1)
(b) (3x − 2)(x2 + 1)
(c) (3x + 2)(x2 − 1)
(d) (3x + 2)(x2 + 1)

Answer:

(d) (3x + 2)(x2 + 1)

3x3+2x2+3x+2=x23x+2+13x+2
                   =3x+2x2+1

Page No 84:

Question 33:

If xy+yx=-1, where x ≠ 0 and y ≠ 0, then the value of (x3y3) is
(a) 1
(b) −1
(c) 0
(d) 12

Answer:

 (c) 0

   xy+yx=-1x2+y2xy=-1
x2 + y2 = -xy
x2 + y2 + xy = 0

Thus, we have:
x3-y3=x-yx2+y2+xy
         =x-y×0=0

Page No 84:

Question 34:

If a + b + c = 0, then a3 + b3 + c3 = ?
(a) 0
(b) abc
(c) 2abc
(d) 3abc

Answer:

(d) 3abc

  a+b+c=0a+b=-c

a+b3=-c3a3+b3+3aba+b=-c3a3+b3+3ab-c=-c3a3+b3+c3=3abc

Page No 84:

Question 35:

If (x + 2) and (x − 1) are factors of (x3 + 10x2 + mx + n), then
(a) m = 5, n = −3
(b) m = 7, n = −18
(c) m = 17, n = −8
(d) m = 23, n = −19

Answer:

(b) m = 7, n = −18

Let:
px=x3+10x2+mx+n
Now,
x+2=0x=-2
(x + 2) is a factor of p(x).
So, we have p(-2)=0
-23+10×-22+m×-2+n=0-8+40-2m+n=032-2m+n=02m-n=32   ...i
Now,
x-1=0x=1
Also, 
(x - 1) is a factor of p(x).
We have:
p(1) = 0
13+10×12+m×1+n=01+10+m+n=011+m+n=0m+n=-11   ...iiFrom i and ii, we get:3m=21m=7
By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18

Page No 84:

Question 36:

The value of (369)2 − (368)2 = ?
(a) 12
(b) 81
(c) 37
(d) 737

Answer:

(d) 737

3692-3682=369+368369-368
                =737×1=737

Page No 84:

Question 37:

104 × 96 = ?
(a) 9864
(b) 9984
(c) 9684
(d) 9884

Answer:

(b) 9984

104×96=100+4100-4
         =1002-42=10000-16=9984

Page No 84:

Question 38:

4a2 + b2 + 4ab + 8a + 4b + 4 = ?
(a) (2a + b + 2)2
(b) (2ab + 2)2
(c) (a + 2b + 2)2
(d) none of these

Answer:

(a) (2a + b + 2)2

    4a2+b2+4ab+8a+4b+4=4a2+b2+4+4ab+4b+8a=2a2+b2+22+2×2a×b+2×b×2+2×2a×2=2a+b+22

Page No 84:

Question 39:

The coefficient of x in the expansion of (x + 3)3 is
(a) 1
(b) 9
(c) 18
(d) 27

Answer:

(d) 27

x+33=x3+33+3×x×3x+3
        =x3+27+9x×x+3=x3+27+9x2+27x

Hence, the coefficient of x is 27.



Page No 85:

Question 40:

If a + b + c = 0, then a2bc+b2ca+c2ab=?

Answer:

(d) 3

a+b+c=0a3+b3+c3=3abc

Thus, we have:
a2bc+b2ca+c2ab=a3+b3+c3abc
                     =3abcabc=3

Page No 85:

Question 41:

If x + y + z = 9 and xy + yz + zx = 23, then the value of (x3 + y3 + z3 − 3xyz) = ?
(a) 108
(b) 207
(c) 669
(d) 729

Answer:

(a) 108

x3+y3+z3-3xyz=x+y+zx2+y2+z2-xy-yz-zx
                   =x+y+zx+y+z2-3xy+yz+zx=9×81-3×23=9×12=108

Page No 85:

Question 42:

If (x100 + 2x99 + k) is divisible by (x + 1), then the value of k is
(a) 1
(b) 2
(c) −2
(d) −3

Answer:

(a) 1

Let:
px=x100+2x99+k
Now,
x+1=0x=-1
x+1 is divisible by px.p-1=0-1100+2×-199+k=01-2+k=0-1+k=0k=1

Page No 85:

Question 43:

In a polynomial in x, the indices of x must be
(a) integers
(b) positive integers
(c) non-negative integers
(d) real numbers

Answer:

(c) non-negative integers

Page No 85:

Question 44:

For what value of k is the polynomial p(x) = 2x3kx + 3x + 10 exactly divisible by (x + 2)?
(a) -13
(b) 13
(c) 3
(d) −3

Answer:

(d) −3

Let:
px=2x3-kx2+3x+10
Now,
x+2=0x=-2
px is completely divisible by x+2. p-2=02×-23-k×-22+3×-2+10=0-16-4k-6+10=0-12-4k=04k=-12k=-124k=-3

Page No 85:

Question 45:

207 × 193 = ?
(a) 39851
(b) 39951
(c) 39961
(d) 38951

Answer:

(b) 39951

  207×193=200+7200-7=2002-72=40000-49=39951

Page No 85:

Question 46:

305 × 308 = ?
(a) 94940
(b) 93840
(c) 93940
(d) 94840

Answer:

(c) 93940

305×308=300+5×300+8             =3002+300×5+8+5×8             =90000+3900+40             =93940

Page No 85:

Question 47:

The zeroes of the polynomial p(x) = x2 − 3x are
(a) 0, 0
(b) 0, 3
(c) 0, −3
(d) 3, −3

Answer:

(b) 0, 3

Let:
px=x2-3x
Now, we have:
px=0x2-3x=0
         xx-3=0x=0 and x-3=0x=0 and x=3

Page No 85:

Question 48:

The zeroes of the polynomial p(x) = 3x2 − 1 are
(a) 13
(b) 13
(c) -13
(d) 13and-13

Answer:

(d) 13 and -13

Let:
px=3x2-1

To find the zeroes of px, we have:px=03x2-1=0
         3x2=1x2=13x=±13x=13 and x=-13



Page No 86:

Question 49:

Assertion: If (x − 1) is a factor of p(x) = x2 + kx + 1, then k = −2.
Reason: If (xa) is a factor of p(x), then p(a) = 0.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and but Reason is not a correct explanation of assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Assertion:
Let:
px=x2+kx+1
Now,
x-1=0x=1
x-1 is a factor of px.p1=012+k×1+1=01+k+1=0k+2=0k=-2
Hence, Assertion is true.
Reason: If (xa) is a factor of p(x), then p(a) = 0.
The given statement is true.
Therefore, both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Page No 86:

Question 50:

Assertion: If p(x) = x3ax2 + 6xa is divided by (xa), then the remainder is 5a.
Reason: If p(x) is divided by (xa),then the remainder is p(a).

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Assertion:
Let:
px=x3-ax2+6x-a
Now,
x-a=0x=a
By the remainder theorem, we know that when px is divided by x-a, the remainder is pa.
Now, we have:
pa=a3-a×a2+6×a-a     =a3-a3+6a-a     =5a
Hence, Assertion is true.
Reason: If p(x) is divided by (xa), then the remainder is p(a).
The given statement is true.
Therefore, both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Page No 86:

Question 51:

Assertion: If (x − 2) is a factor p(x) = x3 − 2x + 3k, then k=-43.
Reason: If p(x) is divided by (xa), then the remainder is p(a).

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

Assertion:
Let:
px=x3-2x+3k
Now,
x-2=0x=2
x-2 is a factor of px.p2=023-2×2+3k=08-4+3k=04+3k=0k=-43
Hence, Assertion is true.
Reason: If p(x) is divided by (xa), then the remainder is p(a).
The given statement is true, but it is not correct explanation of Assertion.
Therefore, both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

Page No 86:

Question 52:

Assertion: The value of (25)3 + (−16)3 + (−9)3 is 10800.
Reason: If a + b + c = 0, then a3 + b3 + c3 = 3abc.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true, and Reason is a correct explanation of Assertion.

Assertion:
Let:
a=25, b=-16 and c=-9
Now,
a+b+c=25+-16+-9
          = 0
a+b+c=0a3+b3+c3=3abc

Thus, we have:
3abc=3×25×-16×-9
     = 10800
Hence, Assertion is true.
Reason: If a + b + c = 0, then a3 + b3 + c3 = 3abc.
The given statement is true.
Therefore,  both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Page No 86:

Question 53:

Match of following columns:

Column I Column II (a) When p(x) = 81x4 + 54x3 − 9x2 − 3x + 2 is divided by (3x + 2), then reaminder ....... . (p) 1 (b) p(x) = ax3 + 3x2 − 3 and q(x) = 2x3 − 5x + a divided by (x − 4) leave the same remainder. Then, a ...... . (q) −7 (c) If p(x)=7x2-42x+cis completely divisible by x-2, then c = ...... . (r) 0 (d) If q(x) = 2x3 + bx2 + 11x + b + 3 is divisibl by (2x + 1), then b = ....... . (s) −6 Column II
(a) If p(x) = x3 + 2x2 + 3x + 1 is divided by (x + 1), then remainder = ...... . (p) 78
(b) If (2x − 1) is a factor of q(x) = x3 + 2x2 − 3x + k, then k = ..... . (q) 50
(c) The degree of the constant polynomial (−5) is = ....... . (r) −1
(d) When x51 + 51 is divided by (x + 1), the remainder is = ...... . (s) 0

(a) ...... .
(b) ...... .
(c) ...... .
(d) ...... .

Answer:

(a)
Let:
px=x3+2x2+3x+1
x+1=0x=-1
By the remainder theorem, we know that when px is divided by x+1, the remainder is p-1.
Thus, we have:
p-1=-13+2×-12+3×-1+1        =-1+2-3+1        =-1
∴ ar

(b)
Let:
qx=x3+2x2-3x+k
Now, 
2x-1=0x=12
2x-1 is a factor of qx.q12=0123+2×122-3×12+k=018+12-32+k=0-78+k=0k=78
∴ bp

(c)
The degree of the constant polynomial (-5) is 0.
 ∴ cs

(d)
Let:
px=x51+51
Now,
x+1=0x=-1
By the remainder theorem, we know that when px is divided by x+1, the remainder is p-1.
Thus, we have:
p-1=-151+51
       = -1+51
       = 50
∴ dq



Page No 87:

Question 54:

Match of the following columns:

Column I Column II
(a) When p(x) = 81x4 + 54x3 − 9x2 − 3x + 2 is divided by (3x + 2), then reaminder ....... . (p) 1
(b) p(x) = ax3 + 3x2 − 3 and q(x) = 2x3 − 5x + a divided by (x − 4) leave the same remainder. Then, a ...... . (q) −7
(c) If p(x)=7x2-42x+cis completely divisible by x-2, then c = ...... . (r) 0
(d) If q(x) = 2x3 + bx2 + 11x + b + 3 is divisibl by (2x + 1), then b = ....... . (s) −6

(a) ....... .
(b) ....... .
(c) ....... .
(d) ....... .

Answer:

(a)
Let:
px=81x4+54x3-9x2-3x+2
3x+2=0x=-23
By the remainder theorem, we know that when px is divided by 3x+2, the remainder is p-23.
Thus, we have:
p-23=81×-234+54×-233-9×-232-3×-23+2
         =81×1681-54×827-9×49+2+2=16-16-4+2+2=0
∴ ar

(b)
Let:
px=ax3+3x2-3
x-4=0x=4
By the remainder theorem, we know that when px is divided by x-4, the remainder is p4.
Thus, we have:
p4=a×43+3×42-3     =64a+48-3     =64a+45

Also,
Let:
qx=2x3-5x+a
By the remainder theorem, we know that when​ qx is divided by x-4, the remainder is q4.
Thus, we have:
q4=2×43-5×4+a     =128-20+a     =108+a
Now,
64a+45=108+a63a=63a=1
∴ bp

(c)
Let:
px=7x2+-42x+c
Now, 
x-2=0x=2
px is completely divisible by x-2.p2=07×22+-42×2+c=014-8+c=06+c=0c=-6
cs

(d)
Let:
qx=2x3+bx2+11x+b+3
Now,
2x-1=0x=12
qx is completely divisible by 2x-1.q12=02×123+b×122+11×12+b+3=014+b4+112+b+3=0

1+b+22+4b+124=035+5b=0b=-355b=-7
∴ dq



Page No 91:

Question 1:

Let p(x) = 3x3 + 4x2 − 5x + 8. Find p(−2).

Answer:

Let:
px=3x3+4x2-5x+8
Thus, we have:
p-2=3×-23+4×-22-5×-2+8
       =3×-8 + 4×4 +10 +8=-24+16+10+8=10

Page No 91:

Question 2:

Find the remainder when p(x) = 4x3 + 8x2 − 17x + 10 is divided by (2x − 1).

Answer:

Let:
px=4x3+8x2-17x+10
And,2x-1=0x=12
By the remainder theorem, we know that when px is divided by 2x-1, the remainder is p12.
Thus, we have:
p12=4×123+8×122-17×12+10
       =12+2-172+10=4

∴ Remainder = 4

Page No 91:

Question 3:

If (x − 2) is a factor of 2x3 − 7x2 + 11x + 5a, find the value of a.

Answer:

Let:
px=2x3-7x2+11x+5a
Now, we have:
x-2=0x=2
x-2 is a factor of px. p2=0 2×23-7×22+11×2+5a=0 16-28+22+5a=0 10+5a=0 a=-105 a=-2

Hence, the value of a is -2.

Page No 91:

Question 4:

For what value of m, p(x) = (x3 − 2mx2 + 16) is divisible by (x + 2)?

Answer:

Let:px=x3-2mx2+16
Now, we have:
x+2=0x=-2
px is divisible by x+2. p-2=0 -23-2m-22+16=0 -8-8m+16=0 8-8m=0 m=88 m=1

Hence, the value of m is 1.

Page No 91:

Question 5:

If (a + b + c) = 8 and (ab + bc + ca) = 19, find (a2 + b2 + c2).

Answer:

a+b+c2 =a2+b2+c2 + 2ab+bc+caa2+b2+c2=a+b+c2-2ab+bc+ca
                   
               =82-2×19=64-38=26

Page No 91:

Question 6:

Expand: (3a + 4b + 5c)2.

Answer:

3a+4b+5c2=3a2+4b2+5c2+23a4b+24b5c+23a5c
                =9a2+16b2+25c2+24ab+40bc+30ac

Page No 91:

Question 7:

Expand: (3x + 2)3.

Answer:

3x+23=3x3+23+3×3x×2×3x+2
         =27x3+8+18x×3x+2=27x3+8+54x2+36x=27x3+54x2+36x+8

Page No 91:

Question 8:

Evaluate: {(28)3 + (−15)3 + (−13)3}.

Answer:

Let:
a=28, b=-15 and c=-13
Now, we have:
a+b+c=28+-15+-13
          = 0
 If a+b+c=0, then a3+b3+c3=3abc.

 283 + -153 + -133=a3 + b3 + c3                                             =3abc                                            =3×28×-15×-13                                            =16380



Page No 92:

Question 9:

If (x60 + 60) is divided by (x + 1), the remainder is

(a) 0
(b) 59
(c) 61
(d) 2

Answer:

(c) 61

Let:
px=x60+60
x+1=0x=-1
By the remainder theorem, we know that when px is divided by x+1, the remainder is p-1.
Now, we have:
p-1=-160+60         =1+60         =61

Page No 92:

Question 10:

One of the factors of (36x2 − 1) + (1 + 6x)2 is

(a) (6x − 1)
(b) (6x + 1)
(c) 6x
(d) 6 − x

Answer:

(b) (6x + 1)

36x2-1+1+6x2=6x2-12+1+6x2
                      =6x+16x-1+6x+12=6x+16x-1+1=6x+16x

Hence, one of the factors of the given polynomial is (6x + 1).

Page No 92:

Question 11:

If ab+ba=-1, then (a3b3) = ?

(a) −1
(b) −3
(c) −2
(d) 0

Answer:

(d) 0

We have:ab+ba=-1a2+b2ab=-1

a2+b2+ab=0

Thus, we have:
a3-b3=a-ba2+ab+b2
       =a-b×0=0

Page No 92:

Question 12:

The coefficient of x in the expansion of (x + 5)3 is

(a) 1
(b) 15
(c) 45
(d) 75

Answer:

(d) 75

x+53=x3+53+3×x×5x+5
        =x3+125+15x×x+5=x3+125+15x2+75x

Hence, the coefficient of x is 75.

Page No 92:

Question 13:

5is a polynomial of degree

(a) 12
(b) 2
(c) 0
(d) 1

Answer:

(c) 0

5 is a non-zero constant term and a non-zero constant is a polynomial of degree 0.

Page No 92:

Question 14:

One of the zeroes of the polynomial 2x2 + 7x − 4 is

(a) 2
(b) 12
(c) −2
(d) -12

Answer:

(b) 12

Let:
px=2x2+7x-4
px=02x2+7x-4=0
         2x2+8x-x-4=02xx+4-1x+4=0x+42x-1=0x+4=0 or 2x-1=0x=-4 or x=12
Hence, one of the zeroes of the given polynomial is 12.

Page No 92:

Question 15:

Zero of the zero polynomial is

(a) 0
(b) 1
(c) every real number
(d) not defined

Answer:

(d) not defined

Zero of the zero polynomial is not defined.

Page No 92:

Question 16:

If (x + 1) and (x − 1) are factors of p(x) = ax3 + x2 − 2x + b, find the values of a and b.

Answer:

Let:
px=ax3+x2-2x+b
Now, 
x=1=0x=-1
x+1 is a factor of px. p-1=0 a×-13+-12-2×-1+b=0 -a+1+2+b=0 -a+3+b=0 a-b=3   ...i
Also,
x-1=0x=1
x-1 is a factor of px. p1=0 a×13+12-2×1+b a+1-2+b=0 a+b-1=0 a+b=1   ...ii
 From i and ii, we get:2a=4a=2
By substituting the value of a, we get the value of b, i.e., −1.
∴ a = 2 and b = −1

Page No 92:

Question 17:

If (x + 2) is a factor of p(x) = ax3 + bx2 + x − 6 and p(x) when divided by (x − 2) leaves a remainder 4, prove that a = 0 and b = 2.

Answer:

Let:
px=ax3+bx2+x-6
Now,
x+2=0  x=-2

x+2 is a factor px. p-2=0 a×-23+b×-22+-2-6=0 -8a+4b-8=0 8a-4b=-8   ...i

Also,
x-2=0x=2
When p(x) is divided by (x - 2), the remainder is 4.
p(2) = 4

Here,p2=a×23+b×22+2-6      =8a+4b-4

Thus, we have:
8a+4b-4=48a+4b=8      ...ii

From i and ii, we get:16a=0a=0

By substituting the value of a, we get the value of b, i.e., 2.
∴ a = 0 and b = 2

Page No 92:

Question 18:

The expanded form of (3x − 5)3 is

(a) 27x3 + 135x2 + 225x − 125
(b) 27x3 + 135x2 − 225x − 125
(c) 27x3 − 135x2 + 225x − 125
(d) none of these

Answer:

(c) 27x3 − 135x2 + 225x − 125

3x-53=3x3-53-3×3x×53x-5
         =27x3-125-45x3x-5=27x3-125-135x2+225x=27x3-135x2+225x-125

Page No 92:

Question 19:

If a + b + c = 5 and ab + bc + ca = 10, prove that a3 + b3 + c3 − 3abc = −25.

Answer:

We have:a2+b2+c2=a+b+c2-2ab+bc+ca
                 =52 - 2×10=25-20=5
Now,
a3+b3+c3-3abc=a+b+ca2+b2+c2-ab-bc-ca                         =a+b+ca2+b2+c2-ab+bc+ca
                  =5×5-10=5×-5=-25

Page No 92:

Question 20:

If p(x) = 2x3 + ax2 + 3x − 5 and q(x) = 2x3 + ax2 + 3x − 5 leave the same remainder when divided by (x − 2), show that a=-133.

Answer:

Let:
px=2x3+ax2+3x-5
Now, 
x-2=0x=2
When p(x) is divided by (x - 2), then the remainder is p(2).
Here,p2=2×23+a×22+3×2-5     =16+4a+6-5     =4a+17

Let:
qx=x3+x2-4x+a
When q(x) is divided by (x - 2), then the remainder is q(2).
q2=23+22-4×2+a     =8+4-8+a     =a+4
According to the question, p(2) and q(2) are the same.
Thus, we have:
4a+17=a+43a=-13a=-133



View NCERT Solutions for all chapters of Class 9