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Page No 74:

Question 1:

Which of the following expressions are polynomials? In case of a polynomial, write its degree.
(i) x5-2x3+x+3
(ii) y3+3y
(iii) t2-25t+5
(iv) x100-1
(v) 12x2-2x+2
(vi) x-2+2x-1+3
(vii) 1
(viii) -35
(ix) x22-2x2
(x) 23x2-8
(xi) 12x2
(xii) 15x12+1
(xiii) 35x2-73x+9
(xiv) x4-x32+x-3
(xv) 2x3+3x2+x-1

Answer:

(i) x5-2x3+x+3 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+5 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) x100-1 is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 100, so, it is a polynomial of degree 100.

(v) 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.

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

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

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

(ix) x22-2x2=x22-2x-2 
This is an expression having negative integral power of x i.e. −2. So, it is not a polynomial.

(x) 23x2-8 is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.

(xi) 12x2=12x-2 is an expression having negative integral power of x. So, it is not a polynomial.

(xii) 15x12+1 
In this expression, the power of x is 12 which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.

(xiii) 35x2-73x+9 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.

(xiv) x4-x32+x-3
In this expression, one of the powers of x is 32 which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.

(xv) 2x3+3x2+x-1=2x3+3x2+x12-1
In this expression, one of the powers of x is 12 which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.



Page No 75:

Question 2:

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.
(i) –7 + x
(ii) 6y
(iii) –z3
(iv) 1 – yy3
(v) xx3 + x4
(vi) 1 + x + x2
(vii) – 6x2
(viii) – 13
(ix) – p

Answer:


(i) –7 + x is a polynomial with degree 1. So, it is a linear polynomial.

(ii) 6is a polynomial with degree 1. So, it is a linear polynomial.

(iii) –z3 is a polynomial with degree 3. So, it is a cubic polynomial.

(iv) 1 – y – y3 is a polynomial with degree 3. So, it is a cubic polynomial.

(v) x – x3 + x4 is a polynomial with degree 4. So, it is a quartic polynomial.

(vi) 1 + x + x2 is a polynomial with degree 2. So, it is a quadratic polynomial.

(vii) – 6x2 is a polynomial with degree 2. So, it is a quadratic polynomial.

(viii) –13 is a polynomial with degree 0. So, it is a constant polynomial.

(ix) – p is a polynomial with degree 1. So, it is a linear polynomial.

Page No 75:

Question 3:

Write
(i) the coefficient of x3 in x+3x2-5x3+x4.
(ii) the coefficient of x in 3-22x+6x2.
(iii) the coefficient of x2 in 2x – 3 + x3.
(iv) the coefficient of x in 38x2-27x+16.
(v) the constant term in π2x2+7x-25π.

Answer:

(i) The coefficient of x3 in x+3x2-5x3+x4 is −5.

(ii) The coefficient of x in 3-22x+6x2 is -22.

(iii) 2x – 3 + x= – 3 + 2x + 0xx3
The coefficient of x2 in 2x – 3 + x3 is 0.

(iv) The coefficient of x in 38x2-27x+16 is -27.

(v) The constant term in π2x2+7x-25π is -25π.

Page No 75:

Question 4:

Determine the degree of each of the following polynomials.
(i) 4x-5x2+6x32x
(ii) y2(y y3)
(iii) (3x – 2) (2x3 + 3x2)
(iv) -12x+3
(v) – 8
(vi) x–2(x4+ x2)

Answer:


(i) 4x-5x2+6x32x=4x2x-5x22x+6x32x=2-52x+3x2
Here, the highest power of x is 2. So, the degree of the polynomial is 2. 

(ii) y2(– y3) = y3 – y5
Here, the highest power of y is 5. So, the degree of the polynomial is 5. 

(iii) (3x – 2)(2x3 + 3x2) = 6x4 + 9x3 – 4x3 – 6x2 = 6x4 + 5x3 – 6x2
Here, the highest power of x is 4. So, the degree of the polynomial is 4. 

(iv) -12x+3
Here, the highest power of x is 1. So, the degree of the polynomial is 1. 

(v) – 8
–8 is a constant polynomial. So, the degree of the polynomial is 0. 

(vi) x–2(x4 x2) = x2 xx2 + 1 
Here, the highest power of x is 2. So, the degree of the polynomial is 2. 

Page No 75:

Question 5:

(i) Give an example of a monomial of degree 5.
(ii) Give an example of a binomial of degree 8.
(iii) Give an example of a trinomial of degree 4.
(iv) Give an example of a monomial of degree 0.

Answer:


(i) A polynomial having one term is called a monomial. Since the degree of required monomial is 5, so the highest power of x in the monomial should be 5.
An example of a monomial of degree 5 is 2x5.    

(ii) A polynomial having two terms is called a binomial. Since the degree of required binomial is 8, so the highest power of x in the binomial should be 8.
An example of a binomial of degree 8 is 2x8 − 3x.

(iii) A polynomial having three terms is called a trinomial. Since the degree of required trinomial is 4, so the highest power of x in the trinomial should be 4.
An example of a trinomial of degree 4 is 2x4 − 3x + 5.

(iv) A polynomial having one term is called a monomial. Since the degree of required monomial is 0, so the highest power of x in the monomial should be 0.
An example of a monomial of degree 0 is 5. 

Page No 75:

Question 6:

Rewrite each of the following polynomials in standard form.
(i) x-2x2+8+5x3
(ii) 23+4y2-3y+2y3
(iii) 6x3+2x-x5-3x2
(iv) 2+t-3t3+t4-t2

Answer:

A polynomial written either in ascending or descending powers of a variable is called the standard form of a polynomial.

(i) 8+x-2x2+5x3 is a polynomial in standard form as the powers of x are in ascending order.
(ii) 23-3y+4y2+2y3 is a polynomial in standard form as the powers of y are in ascending order.
(iii) 2x-3x2+6x3-x5 is a polynomial in standard form as the powers of x are in ascending order.
(iv) 2+t-t2-3t3+t4 is a polynomial in standard form as the powers of t are in ascending order.



Page No 78:

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 78:

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 78:

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 78:

Question 4:

If px=x3-3x2+2x, find p(0), p(1), p(2). What do you conclude?

Answer:


px=x3-3x2+2x      .....(1)

Putting x = 0 in (1), we get

p0=03-3×02+2×0=0

Thus, x = 0 is a zero of p(x).

Putting x = 1 in (1), we get

p1=13-3×12+2×1=1-3+2=0

Thus, x = 1 is a zero of p(x).

Putting x = 2 in (1), we get

p2=23-3×22+2×2=8-3×4+4=8-12+4=0

Thus, x = 2 is a zero of p(x).

Page No 78:

Question 5:

If p(x) = x3 + x2 – 9x – 9, find p(0), p(3), p(–3) and p(–1). What do you conclude about the zero of p(x)? Is 0 a zero of p(x)?

Answer:


p(x) = x3 + x2 – 9x – 9         .....(1)

Putting x = 0 in (1), we get

p(0) = 03 + 02 – 9 × 0 – 9 = 0 + 0 – 0 – 9 = –9 ≠ 0

Thus, x = 0 is not a zero of p(x).

Putting x = 3 in (1), we get

p(3) = 33 + 32 – 9 × 3 – 9 = 27 + 9 – 27 – 9 = 0

Thus, x = 3 is a zero of p(x).

Putting x = –3 in (1), we get

p(–3) = (–3)3 + (–3)2 – 9 × (–3) – 9 = –27 + 9 + 27 – 9 = 0

Thus, x = –3 is a zero of p(x).

Putting x = –1 in (1), we get

p(–1) = (–1)3 + (–1)2 – 9 × (–1) – 9 = –1 + 1 + 9 – 9 = 0

Thus, x = –1 is a zero of p(x).

Page No 78:

Question 6:

Verify that:
(i) 4 is a zero of the polynomial p(x) = x − 4.
(ii) −3 is a zero of the polynomial q(x) = x + 3.
(iii) 25is a zero of the polynomial, f(x) = 2 − 5x.
(iv) -12is a zero of the polynomial g(y) = 2y + 1.

Answer:

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

ii px=-3+3p3=0
Hence, 3 is the zero of the given polynomial.

iii px=2-5xp25=2-5×25
          =2-2=0

Hence, 25 is the zero of the given polynomial.

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

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



Page No 79:

Question 7:

Verify that
(i) 1 and 2 are the zeros of the polynomial p(x) = x2 − 3x + 2.
(ii) 2 and −3 are the zeros of the polynomial q(x) = x2 + x − 6.
(iii) 0 and 3 are the zeros of the polynomial r(x) = x2 − 3x.

Answer:

i px=x2-3x+2=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.

ii 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.

iii 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.

Page No 79:

Question 8:

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

Answer:

i px=0x-5=0x=5Hence, 5 is the zero of the polynomial px.ii qx=0x+4=0x=-4Hence, -4 is the zero of the polynomial qx.iii rx=02x+5=0t=-52Hence, -52 is the zero of the polynomial pt.

iv fx=03x+1=0x=-13Hence, -13 is the zero of the polynomial fx.v gx=05-4x=0x=54Hence, 54 is the zero of the polynomial gx.vi hx=06x-2=0x=26=13Hence, 13 is the zero of the polynomial hx.

vii px=0ax=0x=0Hence, 0 is the zero of the polynomial px.viii qx=04x=0x=0Hence, 0 is the zero of the polynomial qx.

Page No 79:

Question 9:

If 2 and 0 are the zeros of the polynomial fx=2x3-5x2+ax+b then find the values of a and b.
Hint f(2) = 0 and f(0) = 0.

Answer:


It is given that 2 and 0 are the zeroes of the polynomial fx=2x3-5x2+ax+b.
∴ f(2) = 0
2×23-5×22+a×2+b=016-20+2a+b=0-4+2a+b=02a+b=4            .....1
Also,
f(0) = 0
2×03-5×02+a×0+b=00-0+0+b=0b=0
Putting b = 0 in (1), we get
2a+0=42a=4a=2
Thus, the values of a and b are 2 and 0, respectively.



Page No 84:

Question 1:

By actual division, find the quotient and the remainder when (x4 + 1) is divided by (x – 1).
Verify that remainder = f(1).

Answer:

Let f(x) = x4 + 1 and g(x) = x – 1.



Quotient = x3x2x + 1

Remainder = 2

Verification:

Putting x = 1 in f(x), we get

f(1) = 14 + 1 = 1 + 1 = 2 = Remainder, when f(x) = x4 + 1 is divided by g(x) = x – 1

Page No 84:

Question 2:

Verify the division algorithm for the polynomials px=2x4-6x3+2x2-x+2 and gx=x+2.

Answer:


px=2x4-6x3+2x2-x+2 and gx=x+2


Quotient = 2x3-10x2+22x-45

Remainder = 92

Verification:

Divisor × Quotient + Remainder
=x+2×2x3-10x2+22x-45+92=x2x3-10x2+22x-45+22x3-10x2+22x-45+92=2x4-10x3+22x2-45x+4x3-20x2+44x-90+92=2x4-6x3+2x2-x+2
= Dividend

Hence verified.

Page No 84:

Question 3:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=x3-6x2+9x+3, gx=x-1.

Answer:


px=x3-6x2+9x+3

​gx=x-1

By remainder theorem, when p(x) is divided by (x − 1), then the remainder = p(1).

Putting x = 1 in p(x), we get

p1=13-6×12+9×1+3=1-6+9+3=7 

∴ Remainder = 7

Thus, the remainder when p(x) is divided by g(x) is 7.

Page No 84:

Question 4:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=2x3-7x2+9x-13, gx=x-3.

Answer:


px=2x3-7x2+9x-13

​gx=x-3

By remainder theorem, when p(x) is divided by (x − 3), then the remainder = p(3).

Putting x = 3 in p(x), we get

p3=2×33-7×32+9×3-13=54-63+27-13=5 

∴ Remainder = 5

Thus, the remainder when p(x) is divided by g(x) is 5.

Page No 84:

Question 5:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=3x4-6x2-8x-2, gx=x-2.

Answer:


px=3x4-6x2-8x-2

​gx=x-2

By remainder theorem, when p(x) is divided by (x − 2), then the remainder = p(2).

Putting x = 2 in p(x), we get

p2=3×24-6×22-8×2-2=48-24-16-2=6 

∴ Remainder = 6

Thus, the remainder when p(x) is divided by g(x) is 6.

Page No 84:

Question 6:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=2x3-9x2+x+15, gx=2x-3.

Answer:


px=2x3-9x2+x+15

​gx=2x-3=2x-32

By remainder theorem, when p(x) is divided by (2x − 3), then the remainder = p32.

Putting x=32 in p(x), we get

p32=2×323-9×322+32+15=274-814+32+15 =27-81+6+604=124=3

∴ Remainder = 3

Thus, the remainder when p(x) is divided by g(x) is 3.

Page No 84:

Question 7:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=x3-2x2-8x-1, gx=x+1.

Answer:


px=x3-2x2-8x-1

​gx=x+1

By remainder theorem, when p(x) is divided by (x + 1), then the remainder = p(−1).

Putting x = −1 in p(x), we get

p-1=-13-2×-12-8×-1-1=-1-2+8-1=4 

∴ Remainder = 4

Thus, the remainder when p(x) is divided by g(x) is 4.

Page No 84:

Question 8:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=2x3+x2-15x-12, gx=x+2.

Answer:


px=2x3+x2-15x-12

​gx=x+2

By remainder theorem, when p(x) is divided by (x + 2), then the remainder = p(−2).

Putting x = −2 in p(x), we get

p-2=2×-23+-22-15×-2-12=-16+4+30-12=6 

∴ Remainder = 6

Thus, the remainder when p(x) is divided by g(x) is 6.

Page No 84:

Question 9:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=6x3+13x2+3, gx=3x+2.

Answer:


px=6x3+13x2+3

​gx=3x+2=3x+23=3x--23

By remainder theorem, when p(x) is divided by (3x + 2), then the remainder = p-23.

Putting x=-23 in p(x), we get

p-23=6×-233+13×-232+3=-169+529+3 =-16+52+279=639=7

∴ Remainder = 7

Thus, the remainder when p(x) is divided by g(x) is 7.

Page No 84:

Question 10:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=x3-6x2+2x-4, gx=1-32x.

Answer:


px=x3-6x2+2x-4

​gx=1-32x=-32x-23

By remainder theorem, when p(x) is divided by 1-32x, then the remainder = p23.

Putting x=23 in p(x), we get

p23=233-6×232+2×23-4=827-83+43-4 =8-72+36-10827=-13627

∴ Remainder = -13627

Thus, the remainder when p(x) is divided by g(x) is -13627.

Page No 84:

Question 11:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=2x3+3x2-11x-3, gx=x+12.

Answer:


px=2x3+3x2-11x-3

​gx=x+12=x--12

By remainder theorem, when p(x) is divided by x+12, then the remainder = p-12.

Putting x=-12 in p(x), we get

p-12=2×-123+3×-122-11×-12-3=-14+34+112-3 =-1+3+22-124=124=3

∴ Remainder = 3

Thus, the remainder when p(x) is divided by g(x) is 3.

Page No 84:

Question 12:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=x3-ax2+6x-a, gx=x-a.

Answer:


px=x3-ax2+6x-a

​gx=x-a

By remainder theorem, when p(x) is divided by (x − a), then the remainder = p(a).

Putting x = a in p(x), we get

pa=a3-a×a2+6×a-a=a3-a3+6a-a=5a 

∴ Remainder = 5a

Thus, the remainder when p(x) is divided by g(x) is 5a.

Page No 84:

Question 13:

The polynomials 2x3+x2-ax+2 and 2x3-3x2-3x+a when divided by (x – 2) leave the same remainder. Find the value of a.

Answer:


Let fx=2x3+x2-ax+2 and gx=2x3-3x2-3x+a.

By remainder theorem, when f(x) is divided by (x – 2), then the remainder = f(2).

Putting x = 2 in f(x), we get

f2=2×23+22-a×2+2=16+4-2a+2=-2a+22 

By remainder theorem, when g(x) is divided by (x – 2), then the remainder = g(2).

Putting x = 2 in g(x), we get

g2=2×23-3×22-3×2+a=16-12-6+a=-2+a

It is given that,

f2=g2-2a+22=-2+a-3a=-24a=8
Thus, the value of a is 8.

Page No 84:

Question 14:

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

Answer:

 Let:px=x4-2x3+3x2-ax+b

Now,When px is divided by x-1, the remainder is p1.When px is divided by x+1, the remainder is p-1.
Thus, we have:
p1=14-2×13+3×12-a×1+b      =1-2+3-a+b      =2-a+b
And,
p-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 px is divided by x-2, the remainder is p2.
Thus, we have:
p2=24-2×23+3×22-5×2+8   a=5 and b=8      =16-16+12-10+8      =10



Page No 85:

Question 15:

If px=x3-5x2+4x-3 and gx=x-2, show that p(x) is not a multiple of g(x).

Answer:


px=x3-5x2+4x-3

gx=x-2

Putting x = 2 in p(x), we get

p2=23-5×22+4×2-3=8-20+8-3=-70

Therefore, by factor theorem, (x − 2) is not a factor of p(x).

Hence, p(x) is not a multiple of g(x).

Page No 85:

Question 16:

If px=2x3-11x2-4x+5 and gx=2x+1, show that g(x) is not a factor of p(x).

Answer:


px=2x3-11x2-4x+5

gx=2x+1=2x+12=2x--12

Putting x=-12 in p(x), we get

p-12=2×-123-11×-122-4×-12+5=-14-114+2+5 =-124+7=-3+7=40

Therefore, by factor theorem, (2x + 1) is not a factor of p(x).

Hence, g(x) is not a factor of p(x).



Page No 90:

Question 1:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = x3 – 8, g(x) = x – 2

Answer:

Let:
p(x) = x3 – 8
Now,
gx=0x-2=0x=2
By the factor theorem, (x – 2) is a factor of the given polynomial if p(2) = 0.
Thus, we have:
p2=23-8=0
Hence, (x - 2) is a factor of the given polynomial.

Page No 90:

Question 2:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x3 + 7x2 – 24x – 45, g(x) = x – 3

Answer:

Let:
p(x) = 2x3 + 7x2 – 24x – 45
Now,
x-3=0x=3
By the factor theorem, (x - 3) is a factor of the given polynomial if p(3) = 0.
Thus, we have:
p3=2×33-7×32-24×3-45      =54+63-72-45      =0
Hence, (x - 3) is a factor of the given polynomial.

Page No 90:

Question 3:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x4 + 9x3 + 6x2 – 11x – 6, g(x) = x – 1

Answer:

Let:
p(x) = 2x4 + 9x3 + 6x2 – 11x – 6
Here, 
x-1=0x=1​
By the factor theorem, (x - 1) is a factor of the given polynomial if p(1) = 0.
Thus, we have:
p1=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 91:

Question 4:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = x4x2 – 12, g(x) = x + 2

Answer:

Let:
p(x) = x4x2 – 12
Here, 
 x+2=0x=-2
By the factor theorem, (x + 2) is a factor of the given polynomial if p (-2) = 0.
Thus, we have:
p-2=-24--22-12        =16-4-12        =0
Hence, (x + 2) is a factor of the given polynomial.

Page No 91:

Question 5:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 69 + 11xx2 + x3, g(x) = x + 3

Answer:

px=69+11x-x2+x3

gx=x+3

Putting x = −3 in p(x), we get

p-3=69+11×-3--32+-33=69-33-9-27=0 

Therefore, by factor theorem, (x + 3) is a factor of p(x).

Hence, g(x) is a factor of p(x).

Page No 91:

Question 6:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x3 + 9x2 – 11x – 30, g(x) = x + 5

Answer:

Let:
px=2x3+9x2-11x-30
Here, 
x+5=0x=-5
By the factor theorem, (x + 5) is a factor of the given polynomial if p (-5) = 0.
Thus, we have:
p-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 91:

Question 7:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x4 + x3 – 8x2 x + 6, g(x) = 2x – 3

Answer:

Let:
px=2x4+x3-8x2-x+6
Here, 
2x-3=0x=32
By the factor theorem, (2x - 3) is a factor of the given polynomial if p32=0.
Thus, we have:

p32=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 91:

Question 8:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 3x3 + x2 – 20x+ 12, g(x) = 3x – 2

Answer:


px=3x3+x2-20x+12

gx=3x-2=3x-23

Putting x=23 in p(x), we get

p23=3×233+232-20×23+12=89+49-403+12 =8+4-120+1089=120-1209=0

Therefore, by factor theorem, (3x − 2) is a factor of p(x).

Hence, g(x) is a factor of p(x).

Page No 91:

Question 9:

Using factor theorem, show that g(x) is a factor of p(x), when
px=7x2-42x-6, gx=x-2

Answer:

Let:
px=7x2-42x-6
Here, 
x-2=0x=2
By the factor theorem, x-2 is a factor of the given polynomial if p2=0
Thus, we have:
p2=7×22-42×2-6         =14-8-6         =0
Hence, x-2 is a factor of the given polynomial.

Page No 91:

Question 10:

Using factor theorem, show that g(x) is a factor of p(x), when
px=22x2+5x+2, gx=x+2

Answer:

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

Page No 91:

Question 11:

Show that (p – 1) is a factor of (p10 – 1) and also of (p11 – 1).

Answer:


Let f(p) = p10 – 1 and g(p) = p11 – 1.

Putting p = 1 in f(p), we get

f(1) = 110 − 1 = 1 − 1 = 0

Therefore, by factor theorem, (p – 1) is a factor of (p10 – 1).

Now, putting p = 1 in g(p), we get

g(1) = 111 − 1 = 1 − 1 = 0

Therefore, by factor theorem, (p – 1) is a factor of (p11 – 1).

Page No 91:

Question 12:

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 91:

Question 13:

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 91:

Question 14:

Find the value of a for which (x + 1) is a factor of (ax3+ x2 – 2x + 4a – 9).

Answer:


Let f(x) = ax3 x2 – 2x + 4a – 9

It is given that (+ 1) is a factor of f(x).

Using factor theorem,  we have

f(−1) = 0
a×-13+-12-2×-1+4a-9=0-a+1+2+4a-9=03a-6=03a=6a=2
Thus, the value of a is 2.

Page No 91:

Question 15:

Find the value of a for which (x + 2a) is a factor of (x5– 4a2 x3 + 2x + 2a +3).

Answer:

Let f(x) = x5 – 4a2 x3 + 2x + 2a +3

It is given that (+ 2a) is a factor of f(x).

Using factor theorem,  we have

f(−2a) = 0
-2a5-4a2×-2a3+2×-2a+2a+3=0-32a5-4a2×-8a3+2×-2a+2a+3=0-32a5+32a5-4a+2a+3=0-2a+3=0
2a=3a=32
Thus, the value of a is 32.

Page No 91:

Question 16:

Find the value of m for which (2x – 1) is a factor of (8x4+ 4x3 – 16x2 + 10x + m).

Answer:


Let f(x) = 8x4 + 4x3 – 16x2 + 10x + m

It is given that 2x-1=2x-12 is a factor of f(x).

Using factor theorem,  we have

f12=08×124+4×123-16×122+10×12+m=012+12-4+5+m=0
2+m=0m=-2
Thus, the value of m is −2.

Page No 91:

Question 17:

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 91:

Question 18:

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 91:

Question 19:

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 91:

Question 20:

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 91:

Question 21:

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 91:

Question 22:

If both (x – 2) and x-12 are factors of px2 + 5x + r, prove that p = r.

Answer:


Let f(x) = px2 + 5x + r

It is given that (x – 2) is a factor of f(x).

Using factor theorem,  we have

f2=0p×22+5×2+r=04p+r=-10           .....1
Also,  x-12 is a factor of f(x).

Using factor theorem,  we have

f12=0p×122+5×12+r=0p4+r=-52p+4r=-10         .....2
From (1) and (2), we have

4p+r=p+4r4p-p=4r-r3p=3rp=r

 

Page No 91:

Question 23:

Without actual division, prove that 2x4 – 5x3 + 2x2x + 2 is divisible by x2 – 3x + 2.

Answer:


Let f(x) = 2x4 – 5x3 + 2x2 – x + 2 and g(x) = x2 – 3x + 2.

x2-3x+2=x2-2x-x+2=xx-2-1x-2=x-1x-2
Now, f(x) will be divisible by g(x) if f(x) is exactly divisible by both (x − 1) and (x − 2).

Putting x = 1 in f(x), we get

f(1) = 2 × 14 – 5 × 13 + 2 × 12 – 1 + 2 = 2 – 5 + 2 – 1 + 2 = 0

By factor theorem, (x − 1) is a factor of f(x). So, f(x) is exactly divisible by (x − 1).

Putting x = 2 in f(x), we get

f(2) = 2 × 24 – 5 × 23 + 2 × 22 – 2 + 2 = 32 – 40 + 8 – 2 + 2 = 0

By factor theorem, (x − 2) is a factor of f(x). So, f(x) is exactly divisible by (x − 2).

Thus, f(x) is exactly divisible by both (x − 1) and (x − 2).

Hence, f(x) = 2x4 – 5x3 + 2x2 – x + 2 is exactly divisible by (x − 1)(x − 2) = x2 – 3x + 2.

Page No 91:

Question 24:

What must be added to 2x4– 5x3+ 2x2x – 3 so that the result is exactly divisible by (x – 2)?

Answer:


Let k be added to 2x4 – 5x3 + 2x2 – – 3 so that the result is exactly divisible by (– 2). Here, k is a constant.

∴ f(x) = 2x4 – 5x3 + 2x2 – – 3 + k is exactly divisible by (– 2).

Using factor theorem, we have

f2=02×24-5×23+2×22-2-3+k=032-40+8-5+k=0-5+k=0k=5
Thus, 5 must be added to 2x4 – 5x3 + 2x2 – – 3 so that the result is exactly divisible by (– 2).

Page No 91:

Question 25:

What must be subtracted from (x4+ 2x3– 2x2 + 4x + 6) so that the result is exactly divisible by (x2 + 2x – 3)?

Answer:


Dividing (x4 + 2x3 – 2x2 + 4x + 6) by (x+ 2x – 3) using long division method, we have


Here, the remainder obtained is (2x + 9).

Thus, the remainder (2x + 9) must be subtracted from (x4 + 2x3 – 2x2 + 4x + 6) so that the result is exactly divisible by (x+ 2x – 3).

Page No 91:

Question 26:

Use factor theorem to prove that (x + a) is a factor of (xn+ an) for any odd positive integer.

Answer:


Let f(x) = xn an

Putting x = −a in f(x), we get

f(−a) = (−a)n an

If n is any odd positive integer, then

f(−a) = (−a)n an = −an an = 0

Therefore, by factor theorem, (x + a) is a factor of (xn an) for any odd positive integer.



Page No 92:

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 92:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

Question 11:

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 93:

Question 12:

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 93:

Question 13:

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 93:

Question 14:

If p(x) = 5x – 4x2 + 3 then p(–1) =?
(a) 2
(b) –2
(c) 6
(d) –6

Answer:


p(x) = 5x – 4x2 + 3

Putting x = –1 in p(x), we get

p(–1) = 5 × (–1) – 4 × (–1)2 + 3 = –5 – 4 + 3 = –6

Hence, the correct answer is option (d).

 

Page No 93:

Question 15:

If (x51 + 51) is divided by (x + 1) then the remainder is
(a) 0
(b) 1
(c) 49
(d) 50

Answer:


Let f(x) = x51 + 51

By remainder theorem, when f(x) is divided by (x + 1), then the remainder = f(−1).

Putting x = −1 in f(x), we get

 f(−1) = (−1)51 + 51 = −1 + 51 = 50

∴ Remainder = 50

Thus, the remainder when (x51 + 51) is divided by (x + 1) is 50.

Hence, the correct answer is option (d).

Page No 93:

Question 16:

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

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

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Question 17:

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



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Question 18:

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

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Question 19:

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

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Question 20:

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

Answer:


By remainder theorem, when p(x) = x3 – ax2 + x is divided by (x – a), then the remainder = p(a).

Putting xa in p(x), we get

p(a) = a3 – a × a2 + aa3 – a3 + aa

∴ Remainder = a

Hence, the correct answer is option (b).

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Question 21:

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

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Question 22:

(x + 1) is a factor of the polynomial
(a) x3+ x2x + 1
(b) x3 + 2x2x – 2
(c) x3 + 2x2 − x + 2
(d) x4 + x3 + x2 + 1

Answer:

(b) x3 − 2x2 − x − 2

Let:
f(x) = 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.

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Question 23:

Zero of the polynomial p(x) = 2x + 5 is

(a) -25

(b) -52

(c) 25

(d) 52

Answer:


The zero of the polynomial p(x) can be obtained by putting p(x) = 0.

px=02x+5=02x=-5x=-52
Hence, the correct answer is option (b).

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Question 24:

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

Answer:


The given polynomial is p(x) = x2 + x – 6.

Putting x = 2 in p(x), we get

p(2) = 22 + 2 – 6 = 4 + 2 – 6 = 0

Therefore, x = 2 is a zero of the polynomial p(x).

Putting x = –3 in p(x), we get

p(–3) = (–3)2 – 3 – 6 = 9 – 9 = 0

Therefore, x = –3 is a zero of the polynomial p(x).

Thus, 2 and –3 are the zeroes of the given polynomial p(x).

Hence, the correct answer is option (c).

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Question 25:

The zeros of the polynomial p(x) = 2x2 + 5x – 3 are

(a) 12, 3

(b) 12, -3

(c) -12, 3

(d) 1,-12

Answer:

The given polynomial is p(x) = 2x2 + 5x – 3.

Putting x=12 in p(x), we get

p12=2×122+5×12-3=12+52-3=3-3=0

Therefore, x=12 is a zero of the polynomial p(x).

Putting x = –3 in p(x), we get

p-3=2×-32+5×-3-3=18-15-3=0

Therefore, x = –3 is a zero of the polynomial p(x).

Thus, 12 and –3 are the zeroes of the given polynomial p(x).

Hence, the correct answer is option (b).

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Question 26:

The zeros of the polynomial p(x) = 2x2 + 7x – 4 are

(a) 4, -12

(b) 4, 12 

(c) -4, 12

(d) -4, -12

Answer:


The given polynomial is p(x) = 2x2 + 7x – 4.

Putting x=12 in p(x), we get

p12=2×122+7×12-4=12+72-4=4-4=0

Therefore, x=12 is a zero of the polynomial p(x).

Putting x = –4 in p(x), we get

p-4=2×-42+7×-4-4=32-28-4=32-32=0

Therefore, x = –4 is a zero of the polynomial p(x).

Thus, 12 and –4 are the zeroes of the given polynomial p(x).

Hence, the correct answer is option (c).

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Question 27:

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

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Question 28:

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

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Question 29:

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

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Question 30:

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

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Question 31:

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



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Question 32:

The zeroes of the polynomial p(x) = 3x2 − 1 are
(a) 13 and 3
(b) 13 and 3
(c) -13 and 3
(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
Hence, the correct answer is option D.



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