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#### Question 1:

Write the correct answer in each of the following :
Which one of the following is a polynomial?
(A) $\frac{{x}^{2}}{2}-\frac{2}{{x}^{2}}$

(B) $\sqrt{2x}-1$

(C) ${x}^{2}+\frac{3{x}^{\frac{3}{2}}}{\sqrt{x}}$

(D) $\frac{x-1}{x+1}$

For an expression to be a polynomial, any variable in the expression must have the power of a whole number such as ,etc.
a variable term with negative power or fractional power such as , etc. are not a polynomial.
Option A, is not a polynomial because of $-\frac{2}{{x}^{2}}$.
Option B, is not a polynomial because of $\sqrt{2x}=\sqrt{2}{x}^{\frac{1}{2}}$
Option C, ${x}^{2}+\frac{3{x}^{\frac{3}{2}}}{\sqrt{x}}$
$={x}^{2}+\frac{3{x}^{\frac{3}{2}}}{{x}^{\frac{1}{2}}}={x}^{2}+3{x}^{\frac{3}{2}-\frac{1}{2}}\phantom{\rule{0ex}{0ex}}={x}^{2}+3{x}^{1}$
Hence it is a polynomial.
In option D, $\frac{x-1}{x+1}$
$=\frac{x-1+1-1}{x+1}\phantom{\rule{0ex}{0ex}}=\frac{x+1}{x+1}-\frac{2}{x+1}$
$=1-\frac{2}{x+1}$
Hence, it is not a polynomial.

We can conclude that option C is correct.

#### Question 2:

Write the correct answer in each of the following :
$\sqrt{2}$ is a polynomial of degree
(A) 2
(B) 0
(C) 1
(D) $\frac{1}{2}$

$\sqrt{2}$ is a constant polynomial.
It can be represented as $\sqrt{2}{x}^{0}$.
Thus, the degree of the non-zero constant polynomial is 0.

Hence, the correct answer is option B.

#### Question 3:

Write the correct answer in each of the following :
Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is
(A) 4
(B) 5
(C) 3
(D) 7

Given: 4x4 + 0x3 + 0x5 + 5x + 7.
The highest power of the variable in a polynomial is called degree of the polynomial.

Thus, the term with the highest power of x is 4x4. So, the degree of polynomial is 4.
Hence, the correct answer is option A.

#### Question 4:

Write the correct answer in each of the following :
Degree of the zero polynomial is
(A) 0
(B) 1
(C) Any natural number
(D) Not defined

Degree of the zero polynomial (0) is not defined.
Hence, the correct answer is option D.

#### Question 5:

Write the correct answer in each of the following :
If is equal to
(A) 0
(B) 1
(C) $4\sqrt{2}$
(D) $8\sqrt{2}+1$

Given: $p\left(x\right)={x}^{2}-2\sqrt{2}x+1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$\begin{array}{rcl}p\left(2\sqrt{2}\right)& =& {\left(2\sqrt{2}\right)}^{2}-2\sqrt{2}\left(2\sqrt{2}\right)+1\\ & =& \left(8\right)-\left(8\right)+1\\ & =& 1\end{array}$

Hence, the correct answer is option B.

#### Question 6:

Write the correct answer in each of the following :
The value of the polynomial 5x – 4x2 + 3, when x = –1 is
(A) – 6
(B) 6
(C) 2
(D) –2

Given: p(x) = 5– 4x+ 3
p(–1) – 5(–1) – 4(–1)2 + 3
= –5 – 4 + 3
= – 6
So, value of polynomial at x = –1 is –6.
Hence, the correct answer is option A.

#### Question 7:

Write the correct answer in each of the following :
If p(x) = x + 3, then p(x) + p(–x) is equal to
(A) 3
(B) 2x
(C) 0
(D) 6

Given: p(x) = + 3
∴ p(–x) = –x + 3
Now,
p(x) + p(–x) = (x + 3) + (–x + 3)
= 6
Hence, the correct answer is option D.

#### Question 8:

Write the correct answer in each of the following :
Zero of the zero polynomial is
(A) 0
(B) 1
(C) Any real number
(D) Not defined

Zero of a polynomial is the value of a variable for which polynomial becomes zero.
In zero polynomial, all coefficients are equal to zero.
Therefore, zero of a zero polynomial is not defined.

Hence, the correct answer is option D.

#### Question 9:

Write the correct answer in each of the following :
Zero of the polynomial p(x) = 2x + 5 is
(A) $-\frac{2}{5}$

(B) $-\frac{5}{2}$

(C) $\frac{2}{5}$

(D) $\frac{5}{2}$

Given: p(x) = 2+ 5
For zero of polynomial, p(x) = 0

Hence, zero of polynomial p(x) is $-\frac{5}{2}.$
Hence, the correct answer is option B.

#### Question 10:

Write the correct answer in each of the following :
One of the zeroes of the polynomial 2x2 + 7x – 4 is
(A) 2

(B) $\frac{1}{2}$

(C) $-\frac{1}{2}$

(D) –2

Given: p(x) = 2x+ 7– 4
For zero of polynomial, p(x) = 0

Hence, the correct answer is option B.

#### Question 11:

Write the correct answer in each of the following :
If x51 + 51 is divided by x + 1, the remainder is
(A) 0
(B) 1
(C) 49
(D) 50

Let p(x) = x51 + 51
By Remainder Theorem, we have
Remainder = p(–1), since divisor is (x + 1)
p(–1) = (1)51 + 51

​= –1 + 51

= 50

Hence, the correct answer is option D.

#### Question 12:

Write the correct answer in each of the following :
If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is
(A) –3
(B) 4
(C) 2
(D) –2

Let p(x) = 2xkx
By Remainder Theorem
Remainder = p(–1) = 0    ((x + 1) is a factor of p(x))
∴ p(–1) = 2(–1)2 + k(–1) = 0
⇒ 2 – k = 0
k = 2
Hence, the correct answer is option C.

#### Question 13:

Write the correct answer in each of the following :
x + 1 is a factor of the polynomial
(A) x3 + x2 x + 1
(B) x3 + x2 + x + 1
(C) x4 + x3 + x2 + 1
(D) x4 + 3x3 + 3x2 + x + 1

Apply Remainder Theorem.
+ 1 = 0
x = –1
Putting the value of x = –1 in all equations
(A) xx– + 1
= (–1)3 + (–1)2 – (–1) + 1
= –1 + 1 + 1 + 1
= 2
Therefore, (x + 1) is not a factor of equation.

(B) xx+ 1
= (–1)3 + (–1)2 + (–1) + 1
= –1 + 1 – 1 + 1
= 0
Therefore, (x + 1) is a factor of equation.

(C) xxx+ 1
= (–1)4 + (–1)3 + (–1)2 + 1
= 1 – 1 + 1 + 1
= 2
Therefore, (x + 1) is not a factor of equation.

(D) x+ 3x+ 3x+ 1
= (–1)4 + 3(–1)3 + 3(–1)2 + (–1) + 1
= 1 – 3 + 3 – 1 +  1
Therefore, (x + 1) is not a factor of equation.

Hence, the correct answer is option B.

#### Question 14:

Write the correct answer in each of the following :
One of the factors of (25x2 – 1) + (1 + 5x)2 is
(A) 5 + x
(B) 5 – x
(C) 5x – 1
(D) 10x

Let Q(x) = (25x– 1) + (1 + 5x)2
= (25x– 1) + (25x2 + 10x + 1)
= 50x2 + 10x
= 10x (5x + 1)
Thus, factors of Q(x) are 10x and (5x + 1).

Hence, the correct answer is option D.

#### Question 15:

Write the correct answer in each of the following :
The value of 2492 – 2482 is
(A) 12
(B) 477
(C) 487
(D) 497

Given: (249)– (248)2
we know,
a2b2 = (a – b) (a + b)
∴ (249)– (248)2
= (249 – 248)(249 + 248)
= (1)(497)
= 497

Hence, the correct answer is option D.

#### Question 16:

Write the correct answer in each of the following :
The factorisation of 4x2 + 8x + 3 is
(A) (x + 1) (x + 3)
(B) (2x + 1) (2x + 3)
(C) (2x + 2) (2x + 5)
(D) (2x – 1) (2x – 3)

Let p(x) = 4x+ 8+ 3
= 4x+ 6x + 2+ 3
= 2x(2x + 3) + 1(2x + 3)
= (2x + 1) (2x + 3)
Hence, the correct answer is option B.

#### Question 17:

Write the correct answer in each of the following :
Which of the following is a factor of (x + y)3 – (x3 + y3)?
(A) x2 + y2 + 2xy
(B) x2 + y2 xy
(C) xy2
(D) 3xy

(y)– (xy3)
= x3 + y3 + 3xy (x + y) – x3  – y3
= 3xy (x + y)
Thus, factors are 3xy and (x + y).

Hence, the correct answer is option D.

#### Question 18:

Write the correct answer in each of the following :
The coefficient of x in the expansion of (x + 3)3 is
(A) 1
(B) 9
(C) 18
(D) 27

We know,
(+ y)= x3 + y3 + 3x2y + 3xy2

∴ (+ 3)x3 + 27 + 9x2 + 27x
So, coefficient of x is 27.

Hence, the correct answer is option D.

#### Question 19:

Write the correct answer in each of the following :
If , the value of x3 y3 is
(A) 1
(B) –1
(C) 0
(D) $\frac{1}{2}$

Given:
⇒ xy2 = –xy
⇒ xy2 + xy = 0      ...(1)
Now,
x– y3 = (x – y) (x2 + y2 + xy)
= (x – y) (0)        (from (1))
= 0
Hence, the correct answer is option C.

#### Question 20:

Write the correct answer in each of the following :
If , then the value of b is
(A) 0

(B) $\frac{1}{\sqrt{2}}$

(C) $\frac{1}{4}$

(D) $\frac{1}{2}$

From RHS, we have

Now, in LHS, we have
⇒ 49x2b
Comparing LHS with RHS, we get
$b=\frac{1}{4}$

Hence, the correct answer is option C.

#### Question 21:

Write the correct answer in each of the following :
If a + b + c = 0, then a3 + b3 + c3 is equal to
(A) 0
(B) abc
(C) 3abc
(D) 2abc

Given: = 0
we know,
abc3 – 3abc = () (a2 b2 c2 ab – bc – ca)
∴ abc3 – 3abc = 0
⇒ abc3 = 3abc
Hence, the correct answer is option C.

#### Question 1:

Which of the following expressions are polynomials? Justify your answer:
(i) 8

(ii) $\sqrt{3}{x}^{2}-2x$

(iii) $1-\sqrt{5x}$

(iv) $\frac{1}{5{x}^{-2}}+5x+7$

(v) $\frac{\left(x-2\right)\left(x-4\right)}{x}$

(vi) $\frac{1}{x+1}$

(vii) $\frac{1}{7}{a}^{3}-\frac{2}{\sqrt{3}}{a}^{2}+4a-7$

(viii) $\frac{1}{2x}$

(i) Polynomial, because the exponent of the variable of 8 or $8{x}^{0}$ is 0, which is a whole number.
(ii) Polynomial, because the exponent of  variable $\sqrt{3}{x}^{2}-2x$ is a whole number.
(iii) Not polynomial, because the exponent of the variable of $\left(1-\sqrt{5x}\right)$ is $\frac{1}{2},$ whole number.
(iv) Polynomial, because the exponent of the variable $\frac{1}{5{x}^{-2}}+5x+7=\frac{1}{5}{x}^{2}+5x+7$ is a whole number.
(v) Not polynomial, because the exponent of $\frac{\left(x-2\right)\left(x-4\right)}{x}=\frac{{x}^{2}-6x+8}{x}=x-6+8{x}^{-1}$ is –1, which is not a whole number.
(vi) Not polynomial, $\frac{1}{x+1}$ is a rational expression
(vii) Polynomial. because exponent of  $\frac{1}{7}{a}^{3}-\frac{2}{\sqrt{3}}{a}^{2}+4a-7$ is whole number.
(viii) Not polynomial, because exponent of  $\frac{1}{2x}=\frac{1}{2}{x}^{-1}$ is –1, which is not a whole number.

#### Question 2:

Write whether the following statements are True or False. Justify your answer.
(i) A binomial can have at most two terms
(ii) Every polynomial is a binomial
(iii) A binomial may have degree 5
(iv) Zero of a polynomial is always 0
(v) A polynomial cannot have more than one zero
(vi) The degree of the sum of two polynomials each of degree 5 is always 5.

(i) False, because a binomial has exactly two terms.
(ii) False, because every polynomial is not binomial
e.g., 5x is a polynomial but not binomial.
(iii) True, because a binomial is a polynomial whose degree is a whole number greater than or equal to one.
(iv) False, because zero of polynomial can be any real number.
e.g., Q(x) = x – 6, then 6 is zero of polynomial Q(x).
(v) False, because a polynomial can have any number of zeros, it depends upon degree of polynomial.
(vi) False, because the sum of any two polynomial of same degree is not always the same.
e.g.,  f(x) = x5 + 3x2 ; g(x) = –x5 + 2
f(x) + g(x) = (x5 + 3x2) + (–x5 + 2)
= 3x2 + 2
Which is not a polynomial of degree 5.

#### Question 1:

Classify the following polynomials as polynomials in one variable, two variables etc.
(i) x2 + x + 1
(ii) y3 – 5y
(iii) xy + yz + zx
(iv) x2 – 2xy + y2 + 1

(i) x+ 1, Polynomial in one variable i.e. x.
(ii) y– 5y, Polynomial in one variable because it contains only one variable i.e., y.
(iii) xy yz zx , polynomial in three variable, i.e., x, y and z.
(iv)  x– 2xy y+ 1, polynomial in two variable, i.e., x and y.

#### Question 2:

Determine the degree of each of the following polynomials :
(i) 2x – 1
(ii) –10
(iii) x3 – 9x + 3x5
(iv) y3(1 – y4)

(i) 2– 1, degree is 1, because the maximum exponent of x is 1.
(ii) –10, degree is 0, because the exponent of x is 0.
(iii) x– 9+ 3x5, degree is 5, because the maximum exponent of x is 5.
(iv) y3(1 – y4) = y3 – y7, degree is 7, because the maximum exponent of y is 7.

#### Question 3:

For the polynomial $\frac{{x}^{3}+2x+1}{5}-\frac{7}{2}{x}^{2}-{x}^{6}$, write
(i) the degree of the polynomial
(ii) the coefficient of x3
(iii) the coefficient of x6
(iv) the constant term

Given: $\frac{{x}^{3}+2x+1}{5}-\frac{7}{2}{x}^{2}-{x}^{6}$

(i) Degree of polynomial is the highest degree of its terms.
Here, degree of   is 6.
(ii) Coefficient of x3 is $\frac{1}{5}$.
(iii) Coefficient of x6 is –1.
(iv) The constant term in polynomial is $\frac{1}{5}$.

#### Question 4:

Write the coefficient of x2 in each of the following :
(i) $\frac{\pi }{6}x+{x}^{2}-1$
(ii) 3x – 5
(iii) (x –1) (3x – 4)
(iv) (2x – 5) (2x2 – 3x + 1)

(i) Coefficient of x2 in $\frac{\pi }{6}x+{x}^{2}-1$ is 1.
(ii) Coefficient of  x2 in 3– 5 is 0.
(iii)  (x – 1) (3x – 4) = 3x2 – 7x + 4
Coefficient of x2 in 3x2 – 7x + 4 is 3.
(iv) (2– 5) (2x– 3+ 1) = 4x3 – 6x2 + 2x – 10x2 + 15x – 5
= 4x3 – 16x2 + 17x – 5
Coefficient of x2 in 4x3 – 16x2 + 17x – 5 is –16.

#### Question 5:

Classify the following as a constant, linear, quadratic and cubic polynomials :
(i) 2 – x2 + x3
(ii) 3x3
(iii) $5t-\sqrt{7}$
(iv) 4 – 5y2
(v) 3
(vi) 2 + x
(vii) y3 y
(viii) 1 + x + x2
(ix) t2
(x) $\sqrt{2}x-1$

(i) 2 – xx3 is a cubic polynomial.
(ii) 3x3 is a cubic polynomial because its highest degree is 3.
(iii) $5t-\sqrt{7}$ is a linear polynomial.
(iv) 4 – 5y2 is a quadratic polynomial.
(v) 3 is a constant polynomial.
(vi) 2 + is a linear polynomial.
(vii) y– y is a cubic polynomial.
(viii) 1 + x2 is a quadratic polynomial.
(ix) t2 is a quadratic polynomial.
(x) $\sqrt{2}x-1$ is a linear polynomial.

#### Question 6:

Give an example of a polynomial, which is :
(i) monomial of degree 1
(ii) binomial of degree 20
(iii) tri-nomial of degree 2

(i) Monomial of degree 1  :  5x
(ii) Binomial of degree 20  :  6y20 + 5
(iii) Tri-nomial of degree 2  :  x2 + x + 1

#### Question 7:

Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = –3.

Given: p(x) = 3x– 4x+ 7– 5
∴ p(x) = 3(3)3 – 4(3)2 + 7(3) – 5
= 81 – 36 + 21 – 5

= 61

Also,
p(x) = 3(–3)3 – 4(–3)2 + 7(–3) – 5
= –81 – 36 – 21 – 5
​       = –143
Hence, value of polynomial x = 3 is 61 and at x = –3 is –143.

#### Question 8:

If p(x) = x2 – 4x + 3, evaluate : p(2) – p(–1) + p$\left(\frac{1}{2}\right)$

Given: p(x) = x– 4+ 3
Now, p(2) = (2)2 – 4(2) + 3

= 4 – 8 + 3
= –1
p(–1) = (–1)2 – 4(–1) + 3
= 1 + 4 + 3
= 8
p$\left(\frac{1}{2}\right)$ = ${\left(\frac{1}{2}\right)}^{2}-4\left(\frac{1}{2}\right)+3$
​​$=\frac{5}{4}$

​p(2) – p(–1) + p$\left(\frac{1}{2}\right)$ = –1 – 8 + $\frac{5}{4}$
$=\frac{-31}{4}$
Hence, required value of ​p(2) – p(–1) + p$\left(\frac{1}{2}\right)$ is $\frac{-31}{4}.$

#### Question 9:

Find p(0), p(1), p(–2) for the following polynomials :
(i) p(x) = 10x – 4x2 – 3
(ii) p(y) = (y + 2) (y – 2)

(i)  Given p(x) = 10– 4x– 3
Now,
p(0) = 10(0) – 4(0)– 3

–3
p(1) = 10(1) – 4(1)– 3
= 3
p(–2) = 10(–2) – 4(–2)– 3
= –20 – 16 – 3
= –39

(ii) Given: p(y) = (+ 2) (– 2)
Now,
p(0) = (0 + 2) (0 – 2)
= –4
p(1) = (1 + 2) (1 – 2)
= –​3
​       p(– 2) = (–2 + 2) (–2 – 2)
​​        = 0

#### Question 10:

Verify whether the following are True or False :
(i) –3 is a zero of x – 3
(ii) $-\frac{1}{3}$ is a zero of 3x + 1
(iii) $\frac{-4}{5}$ is a zero of 4 – 5y
(iv) 0 and 2 are the zeroes of t2 – 2t
(v) –3 is a zero of y2 + y – 6

(i) False, because
– 3 = 0
x = 3
Thus, zero of x – 3 is 3 not –3.
(ii) True, because
3x + 1 =0
x$\frac{-1}{3}$
Thus, $\frac{-1}{3}$ is zero of 3x + 1.
(iii) False, because
4 – 5y = 0
y = $\frac{4}{5}$
Thus, $\frac{4}{5}$ is zero of 4 – 5y.
(iv) True, because
t– 2t = 0
t(– 2) = 0
t = 0, 2
Thus, 0 and 2 are zeros of t2 – 2t.
(v) True, because
y2 + y – 6 = 0
y2 + 3y – 2y – 6 = 0
⇒ (y + 3)(– 2) = 0
y = –3, 2
Thus, –3 is a zero of y2 + – 6.

#### Question 11:

Find the zeroes of the polynomial in each of the following :
(i) p(x) = x – 4
(ii) g(x) = 3 – 6x
(iii) q(x) = 2x – 7
(iv) h(y) = 2y

Given: (i) p(x) = – 4
For zero of polynomial, p(x) = 0
x – 4 = 0
⇒ x = 4
Hence, zero of polynomial, p(x) is 4.
(ii) Given: g(x) = 3 – 6x
For zero of polynomial, g(x) = 0
⇒ 3 – 6x = 0
x$\frac{1}{2}$
Hence, zero of polynomial, g(x) is $\frac{1}{2}.$
(iii) Given: q(x) = 2– 7
For zero of polynomial, q(x) = 0
⇒ 2– 7 = 0
⇒ x = $\frac{7}{2}.$
Hence, zero of polynomial q(x) is $\frac{7}{2}.$
(iv) Given: h(y) = 2y
For zero of polynomial, h(y) = 0
⇒ 2y = 0
y = 0
Hence, zero of polynomial h(y) is 0.

#### Question 12:

Find the zeroes of the polynomial :
p(x) = (x – 2)2 – (x + 2)2

Given: p(x) = (– 2)– (+ 2)2
For zero of polynomial,
p(x) = 0
⇒ (– 2)– (+ 2)2 = 0
⇒ (x2 – 4x + 4) – (x2 + 4x + 4) = 0
⇒ –8x = 0
x = 0
Hence, zero of polynomial p(x) is 0.

#### Question 13:

By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial : x4 + 1; x –1

From Long Division Method, we have Hence, Quotient : x3 + x2 + x + 1 and Remainder : 2

#### Question 14:

By Remainder Theorem find the remainder, when p(x) is divided by g(x), where
(i) p(x) = x3 – 2x2 – 4x – 1, g(x) = x + 1
(ii) p(x) = x3 – 3x2 + 4x + 50, g(x) = x – 3
(iii) p(x) = 4x3 – 12x2 + 14x – 3, g(x) = 2x – 1
(iv) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – $\frac{3}{2}x$

(i) Given: p(x) = x– 2x– 4– 1, g(x) = + 1
Now, g(x), we get
x + 1 = 0
x = –1
By remainder theorem, when p(x) divided by g(x), remainder is p(–1).
p(–1) = (–1)3 – 2(–1)2 – 4(–1) – 1
= –1 – 2 + 4 – 1
= 0
Hence, value of remainder is 0.

(ii) Given: p(x) = x– 3x+ 4+ 50, g(x) = – 3
Now, p(x) = 0, we get
x – 3 = 0
x = 3
By remainder theorem, when p(x) is divided by g(x), remainder is p(3).
p(3) = (3)– 3(3)+ 4(3) + 50
= 27 – 27 + 12 + 50
= 62
Hence, value of remainder is 62.
(iii)  Given:  p(x) = 4x– 12x+ 14– 3, g(x) = 2– 1
Now, g(x) = 0 we get
2– 1 = 0
x$\frac{1}{2}$
By Remainder Theorem, when p(x) is divided by g(x) remainder is $p\left(\frac{1}{2}\right)$.

Hence, value of remainder is $\frac{3}{2}$.
(iv) Given: p(x) = x– 6x+ 2– 4, g(x) = $1-\frac{3}{2}x$
Now, g(x) = 0, we get
$1-\frac{3x}{2}=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{2}{3}$
By remainder theorem, when p(x) is divided by g(x) remainder is $p\left(\frac{2}{3}\right)$.

Hence, value of remainder is $-\frac{136}{27}$.

#### Question 15:

Check whether p(x) is a multiple of g(x) or not :
(i) p(x) = x3 – 5x2 + 4x – 3, g(x) = x – 2
(ii) p(x) = 2x3 – 11x2 – 4x + 5, g(x) = 2x + 1

(i) Given: p(x) = x– 5x+ 4– 3, g(x) = – 2
zero of polynomial g(x)
⇒ g(x) = 0
⇒ – 2 = 0
⇒ – 2
For p(x) to be a multiple of g(x), remainder must be equal to zero
​From remainder theorem, we have
Remainder = p(2) = (2)3 – 5(2)2 + 4(2) – 3
= 8 – 20 + 8 – 3
= –7
Remainder ≠ 0
Here, p(x) is not a multiple of g(x).
(ii) Given: p(x) = 2x– 11x– 4+ 5, g(x) = 2+ 1
zero of polynomial g(x),
⇒ g(x) = 0
⇒ 2+ 1 = 0
x$\frac{-1}{2}$
For p(x) to be a multiple of g(x), remainder must be equal to zero.
From remainder theorem, we have
Remainder = $\begin{array}{rcl}p\left(\frac{-1}{2}\right)& =& 2{\left(\frac{-1}{2}\right)}^{3}-11{\left(\frac{-1}{2}\right)}^{2}-4\left(\frac{-1}{2}\right)+5\end{array}$

$\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & \frac{1}{4}\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & \frac{11}{4}\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & 7\end{array}\begin{array}{rcl}& & \end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 4\end{array}$
Remainder ≠ 0
Hence, p(x) is not a multiple of g(x).

#### Question 16:

Show that :
(i) x + 3 is a factor of 69 + 11x x2 + x3.
(ii) 2x – 3 is a factor of x + 2x3 – 9x2 + 12.

(i) Let p(x) = 69 + 11– xx3 and g(x) = x + 3
Now, g(x) = 0
x + 3 = 0
⇒ x = –3
Now, for g(x) to be a factor of p(x).
p(–3) = 0
p(–3) = 69 + 11(–3) – (–3)2 + (–3)3
= 69 – 33 – 9 – 27
= 0
Hence, (x + 3) is a factor of 69 + 11x – x2 + x3.

(ii) Let p(x) = + 2x– 9x+ 12 and g(x) = 2– 3
Now, g(x) = 0
⇒ 2– 3 = 0
⇒ x$\frac{3}{2}$
Now, for g(x) to be a factor of g(x).
$p\left(\frac{3}{2}\right)=0$

Hence, (2– 3) is a factor of x + 2x3 – 9x2 + 12.

#### Question 17:

Determine which of the following polynomials has x – 2 a factor :
(i) 3x2 + 6x – 24
(ii) 4x2 + x – 2

(i) Let p(x) = 3x+ 6– 24 and g(x) = x – 2
Now, g(x) = 0
x – 2 = 0
x = 0
For g(x) to be a factor of p(x).
p(2) = 0
p(2) = 3(2)2 + 6(2) – 24
= 12 + 12 – 24
= 0
Hence, (– 2) is a factor of 3x2 + 6x – 24.

(ii) Let p(x) =  4x– 2 and g(x) = x – 2
Now, g(x) = 0
x – 2 = 0
x = 2
For g(x) to be a factor of p(x).
p(2) = 0
∴ p(2) = 4(2)2 + 2 – 2
= 16
Hence, (– 2) is not a factor of 4x2 + x – 2.

#### Question 18:

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

(i) Let p(x) = p10 – 1;  q(x) = p11 – 1 and g(x) = p – 1
⇒ p – 1 = 0
⇒ ​p = 1
For (p – 1) to be a factor of p(x) and q(x),
p(1) = 0 and q(1) = 0
p(1) = (1)10 – 1 = 0
q(1) = (1)11 – 1 = 0

Hence, (p – 1) is a factor of both (p10 – 1) and (p11 – 1).

#### Question 19:

For what value of m is x3 – 2mx2 + 16 divisible by x + 2?

Let p(x) = x– 2mx+ 16
Since, p(x) is divisible by (x + 2).
p(–2) = 0 (By remainder theorem)
⇒ (–2)– 2m(–2)2 + 16 = 0
⇒ –8 – 8m + 16 = 0
⇒  8m = 8
m = 1
Hence, value of m is 1.

#### Question 20:

If x + 2a is a factor of x5 – 4a2x3 + 2x + 2a + 3, find a.

Let p(x) = x– 4a2x+ 2+ 2+ 3 and g(x) = x + 2a
Since, g(x) is a factor of p(x).
∴ By remainder theorem.
p(–2a) = 0
⇒ (–2a)5 – 4a2(–2a)3 + 2(–2a) + 2a + 3 = 0
⇒ –32a5 + 32a5 – 4a + 2a + 3 = 0
⇒ –2a + 3 = 0
a $\frac{3}{2}$
Hence, value of a is $\frac{3}{2}$.

#### Question 21:

Find the value of m so that 2x – 1 be a factor of 8x4 + 4x3 – 16x2 + 10x + m.

Let p(x) =  8x+ 4x– 16x+ 10m and g(x) =  2– 1
g(x) = 0
x$\frac{1}{2}$
For g(x) to be a factor of p(x),
$p\left(\frac{1}{2}\right)=0$
$⇒8{\left(\frac{1}{2}\right)}^{4}+4{\left(\frac{1}{2}\right)}^{3}-16{\left(\frac{1}{2}\right)}^{2}+10\left(\frac{1}{2}\right)+m=0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}+\frac{1}{2}-4+5+m=0$
$⇒1+1+m=0\phantom{\rule{0ex}{0ex}}⇒m=-2$

Hence, value of m is –2.

#### Question 22:

If x + 1 is a factor of ax3 + x2 – 2x + 4a – 9, find the value of a.

Given: p(x) = axx– 2+ 4– 9
g(x) = x + 1
⇒ x + 1 = 0
⇒ x = –1
If g(x) is a factor of p(x)
p(–1) = 0
a(–1)3 + (–1)2 – 2(–1) + 4a – 9 = 0
⇒ –a + 1 + 2 + 4a – 9 = 0
⇒ 3a – 6 = 0
a = 2
Hence, value of a is 2.

#### Question 23:

23. Factorise :
(i) x2 + 9x + 18
(ii) 6x2 + 7x – 3
(iii) 2x2 – 7x – 15
(iv) 84 – 2r – 2r2

(i) x+ 9+ 18
⇒ x+ 3x +  6+ 18
x(x + 3) + 6(x + 3)
⇒ (x + 3) (x + 6)

(ii) 6x+ 7– 3
⇒ 6x+ 9x – 2– 3
⇒ 3x(2x + 3) – 1(2x + 3)
⇒ (2x + 3) (3x – 1)

(iii) 2x– 7– 15
⇒ 2x– 10x + 3– 15
⇒ 2x(x – 5) + 3(x – 5)
⇒ (2x + 3) (x – 5)

(iv) 84 – 2– 2r2
⇒ –2(r2 + r – 42)
⇒ –2(r2 + 7r – 6r – 42)
⇒ –2(r(r + 7) – 6(r + 7))
⇒ –2(r + 7) (r – 6)

#### Question 24:

Factorise :
(i) 2x3 – 3x2 – 17x + 30
(ii) x3 – 6x2 + 11x – 6
(iii) x3 + x2 – 4x – 4
(iv) 3x3 x2 – 3x + 1

(i) 2x– 3x– 17+ 30
⇒  2x3 + 2x2 – 12x – 5x– 5+ 30
⇒ 2x(x2 + x – 6) – 5(x2 + x – 6)
⇒ (2x – 5) (x2 + x –  6)
⇒  (2x – 5) (x2 + 3x – 2x – 6)
⇒  (2x – 5) (x(x + 3) – 2(x + 3))
⇒  (2x – 5) (–  2) (x + 3)

(ii) x– 6x+ 11– 6
⇒ x3 – 5x2 – x2 + 5+ 6x + 30
⇒ x3 – 5x2 + 6x – x2 + 5– 6
xx2 – 5x + 6) – 1(x2 – 5x + 6)
⇒  (x – 1) (x2 – 5x + 6)
⇒   (x – 1) (x(x – 2) – 3 (x – 2))
⇒   (x – 1) (x – 2) (x – 3)

(iii) xx– 4– 4
⇒  x2(x + 1) – 4(x + 1)
⇒ (x + 1) (x2 – 4)
⇒ (x + 1) (x – 2) (x + 2)

(iv) 3x– x– 3+ 1
⇒  3x– 3x – x2 + 1
⇒  3x(x2 – 1) – 1(x2 – 1)
⇒  (3x – 1) (x2 – 1)
⇒  (3x – 1) (x – 1) ​(x + 1)

#### Question 25:

Using suitable identity, evaluate the following:
(i) 1033
(ii) 101 × 102
(iii) 9992

(i) (103)3
⇒ (100 + 3)3
⇒ (100)3 + (3)3 + 3 × 100 × 3 (100 + 3)
⇒ 1000000 + 27 + 92700
⇒ 1092727

(ii) 101 × 102
⇒(100 + 1) (100 + 2)
⇒ (100)2 + 100(1 + 2) + 1 × 2
⇒ 10000 + 300 + 2
⇒ 10302

(iii) 9992
⇒ (1000 – 1)2
⇒ (1000)2 + (1)2 – 2 × 1000 × 1
⇒ 1000000 + 1 – 2000
⇒ 998001

#### Question 26:

Factorise the following:
(i) 4x2 + 20x + 25
(ii) 9y2 – 66yz + 121z2
(iii) ${\left(2x+\frac{1}{3}\right)}^{2}-{\left(x-\frac{1}{2}\right)}^{2}$

(i) 4x+ 20+ 25
= (2x)2 + 2 × (2x) × (5) + (5)2
= (2x + 5)2                (using (a + b)2 = a2 + 2ab + b2)

(ii) 9y– 66yz + 121z2
= (3y)2 – 2 × (3y) × (11z) + (11z)2
=(3y – 11z)2            (using (b)2 = a2 – 2ab + b2)

(iii) ${\left(2x+\frac{1}{3}\right)}^{2}-{\left(x-\frac{1}{2}\right)}^{2}$

#### Question 27:

Factorise the following :
(i) 9x2 – 12x + 3
(ii) 9x2 – 12x + 4

(i) 9x– 12+ 3
= 9x2 – 9x – 3x + 3
= 9x(x – 1) – 3(x – 1)
= (9x – 3) (x – 1)
= 3(3x – 1) (x – 1)

(ii) 9x– 12+ 4
= (3x)2 – 2 × (3x) × (2) + (2)2
=  (3– 2)2                           (∵ (ab)2 = a2 – 2ab + b2)

#### Question 28:

Expand the following :
(i) (4a b + 2c)2
(ii) (3a – 5b c)2
(iii) (– x + 2y – 3z)2

(i) (4– + 2c)2

Using Identity.
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (4– + 2c)2 = (4a)2 + (–b)2 + (2c)2 + 2(4a) (–b) + 2(–b) (2c) + 2(2c) (4a)
= 16a2 + b2 + 4c2 – 8ab – 4bc + 16ac

(ii) (3– 5– c)
Using Identity,
(a + b + c)2a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (3– 5– c)2 = (3a)2 + (–5b)2 + (–c)2 + 2(3a) (–5b) + 2(–5b) (–c) + 2(3a) (–c)
= 9a2 + 25b2 + c2 – 30ab + 10bc – 6ac

(iii) (– + 2– 3z)2
Using Identity,
(a + b + c)2a2 + b2 + c2 + 2ab + 2bc + 2ca
∴  (– + 2– 3z)2 = (–x)2 + (2y)2 + (–3z)2 + 2(–x) (2y) + 2(2y) (–3z) + 2(–3z) (–x)
= x2 + 4y2 + 9z2 – 4xy – 12yz + 6zx

#### Question 29:

Factorise the following :
(i) 9x2 + 4y2 + 16z2 + 12xy – 16yz – 24xz
(ii) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz
(iii) 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz

(i) 9x+ 4y+ 16z+ 12xy – 16yz – 24xz
=
(3x)+ (2y)+ (–4z)+ 2(3x)(2y) + 2(2y)(–4z) + 2(–4z)(3x)
= (3x + 2y – 4z)2
= (3x + 2y – 4z) (3x + 2y – 4z

(ii) 25x+ 16y+ 4z– 40xy + 16yz – 20xz
(–5x)+ (4y)+ (2z)+ 2(–5x)(4y) + 2(4y)(2z) + 2(2z)(–5x)
= (–5x + 4y + 2z)2
= (–5x + 4y + 2z) (–5x + 4y + 2z)

(iii) 16x+ 4y+ 9z– 16xy – 12yz + 24xz
(–4x)+ (2y)+ (–3z)+ 2(–4x)(2y) + 2(2y)(–3z) + 2(–3z)(–4x)
= (–4x + 2y – 3z)2
= (–4x + 2y – 3z)  (–4x + 2y – 3z)​

#### Question 30:

If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 + c2.

Given: = 9 and ab bc ca = 26
we know,
(c)2 = (abc2) + 2(ab bc ca)
⇒ (9)2 = (abc2) + 2(26)
⇒ 81 = (abc2) + 52
⇒ (abc2) = 29

Hence, value of (abc2) is 29.

#### Question 31:

Expand the following :
(i) (3a – 2b)3

(ii) ${\left(\frac{1}{x}+\frac{y}{3}\right)}^{3}$

(iii) ${\left(4-\frac{1}{3x}\right)}^{3}$

(i) (3– 2b)3
we know,
(b)3 = a3 – b3 – 3ab(a – b)
∴ (3– 2b)3 = (3a)3 – (2b)3 – 3(3a)(2b)(3a – 2b)
= 27a3 – 8b3 – 18ab (3– 2b)
= 27a3 – 8b3 – 54a2b + 36ab2

(ii) ${\left(\frac{1}{x}+\frac{y}{3}\right)}^{3}$
we know,
(a + b)3 = a3 + b3 + 3a2b + 3ab2
$\begin{array}{rcl}{\left(\frac{1}{x}+\frac{y}{3}\right)}^{3}& =& {\left(\frac{1}{x}\right)}^{3}+{\left(\frac{y}{3}\right)}^{3}+3{\left(\frac{1}{x}\right)}^{2}\left(\frac{y}{3}\right)+3\left(\frac{1}{x}\right){\left(\frac{y}{3}\right)}^{2}\\ & =& \frac{1}{{x}^{3}}+\frac{{y}^{3}}{27}+\frac{y}{{x}^{2}}+\frac{{y}^{2}}{3x}\end{array}$

(iii) ${\left(4-\frac{1}{3x}\right)}^{3}$
we know,
(a – b)3 = a3 – b3 – 3a2b + 3ab2

#### Question 32:

Factorise the following :
(i) 1 – 64a3 – 12a + 48a2
(ii) $8{p}^{3}+\frac{12}{5}{p}^{2}+\frac{6}{25}p+\frac{1}{125}$

(i) 1 – 64a– 12+ 48a2
= (1)3 – (4a)3 – 3(1)2(4a) + 3(1)(4a)2
= (1 – 4a)3   (∵ Using (a – b)3 = a3 – b3 – 3a2b + 3ab2)

(ii) $8{p}^{3}+\frac{12}{5}{p}^{2}+\frac{6}{25}p+\frac{1}{125}$

#### Question 33:

Find the following products :
(i)
(ii) (x2 – 1) (x4 + x2 + 1)

(i)

Using Identity,
a3 + b3 = (a + b) (a2 – ab + b2)
we get,

(ii) (x– 1) (xx+ 1)
= (x– 1) ((x2)2 + (x)2(1) + (1)2)
Using Identity,
a3 – b3 = (a – b) (a2 + ab + b2)
we get,
= (x2)– (1)3
= x6 – 1

#### Question 34:

Factorise :
(i) 1 + 64x3
(ii) ${a}^{3}-2\sqrt{2}{b}^{3}$

(i) 1 + 64x3
we know,
a3 + b3 = (a + b) (a2 ab + b2)
Now,
1 + 64x3 = (1)3 + (4x)3
= (1 + 4x) (1 – 4x + 16x2)

(ii) ${a}^{3}-2\sqrt{2}{b}^{3}$
we know,
a3 – b3 = (a – b) (a2 + ab b2)
Now,

#### Question 35:

Find the following product :
(2x y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz)

we know,
(a + b + c) (a2 + b2 + c2 – ab – bc – ca) = a3 + b3 + c3 – 3abc
Now,
(2– + 3z) (4xy+ 9z+ 2xy + 3yz – 6xz)
= (2x + (–y) + 3z)((2x)2 + (–y)2 + (3z)2 – (2x)(–y) – (–y)(3z) – (3z)(2x))
= (2x)3 + (–y)3 + (3z)3 – 3(2x) (–y) (3z)
= 8x3 – y3 + 27z3 + 18xyz

#### Question 36:

Factorise :
(i) a3 – 8b3 – 64c3 – 24abc
(ii) $2\sqrt{2}{a}^{3}+8{b}^{3}-27{c}^{3}+18\sqrt{2}abc$.

(i) a– 8b– 64c– 24abc
we know,
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2xyyz – zx)
Now,
(a)3 + (–2b)3 + (–4c)3 – 3(a)(–2b)(–4c)
= {(a – 2b – 4c) ((a)2 + (–2b)2 + (–4c)2 –(a)(–2b) – (–2b)(–4c) – (–4c)(a)}
= {(a – 2b –4c) (a2 + 4b2 + 16c2 + 2ab – 8bc + 4ca)}

(ii) $2\sqrt{2}{a}^{3}+8{b}^{3}-27{c}^{3}+18\sqrt{2}abc$.
we know,
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
Now,

#### Question 37:

Without actually calculating the cubes, find the value of :
(i) ${\left(\frac{1}{2}\right)}^{3}+{\left(\frac{1}{3}\right)}^{3}-{\left(\frac{5}{6}\right)}^{3}$

(ii) (0.2)3 – (0.3)3 + (0.1)3

(i) ${\left(\frac{1}{2}\right)}^{3}+{\left(\frac{1}{3}\right)}^{3}-{\left(\frac{5}{6}\right)}^{3}$
Let
$\therefore x+y+z=0$
We know,
if  x + y + z = 0
then, x3 + y3 + z3 = 3xyz

(ii) (0.2)– (0.3)+ (0.1)3
Let x = 0.2, y = –0.3, z = 0.1
$\therefore x+y+z=0$
we know,
if  x + y + z = 0
then, x3 + y3 + z3 = 3xyz

#### Question 38:

Without finding the cubes, factorise (x – 2y)3 + (2y – 3z)3 + (3z x)3

(– 2y)+ (2– 3z)+ (3– x)
Let a = x – 2y , b = 2y – 3z, c = 3z – x
If a + b + c = 0  (∴ x – 2y + 2y + 2y – 3z + 3z + 3z + 3z – x = 0)
then,  a3 + b3 + c3 = 3abc
∴ (x – 2y)3 +  (2y – 3z)3 + (3z – x)3
= 3(– 2y) (2– 3z) (3z x)

#### Question 39:

Find the value of
(i) x3 + y3 – 12xy + 64, when x + y = – 4
(ii) x3 – 8y3 – 36xy – 216, when x = 2y + 6

(i) we know
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2c2 – ab – bc – ca)
x3 + y3 – 12xy + 64
= (x)3 + (y)3 + (4)– 3(x)(y)(4)
= 0(x2 + y2 + 16 – xy – 4y – 4x)                        (∵ x + y = –4)
= 0

(ii) we know,
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
x3 – 8y3 – 36xy – 216
= (x)3 + (–2y)3 + (–6)– 3(x)(–2y)(–6)
= (x – 2y – 6) (x2 + 4y2 + 36 + 2xy – 12y + 6x)
Now,  x – 2y = 6   (given)
= (6 – 6) (x2 + 4y2 + 36 + 2xy – 2xy – 12y + 6x)
= 0

#### Question 40:

Give possible expressions for the length and breadth of the rectangle whose area is given by 4a2 + 4a – 3.

Given: Area = 4a2 + 4a – 3
we know,
Area of rectangle = length × breadth
Using the Method of splitting middle term,
4a2 + 4a – 3
= 4a2 + 6a – 2a – 3
= 2a(2a + 3) – 1(2a + 3)
= (2a – 1) (2a + 3)
∴ length × breadth = (2a – 1) (2a + 3)
Hence, Possible expression for length and breadth
length = (2a – 1), breadth = 2a + 3 or
length = (2a + 3), breadth = 2a – 1

#### Question 1:

If the polynomials az3 + 4z2 + 3z – 4 and z3 – 4z + a leave the same remainder when divided by z – 3, find the value of a.

Let p(z) = az3 + 4z2 + 3z – 4, q(z) = z3 – 4z + a, r(z) = z – 3
Now, zero of r(z),
r(z) = 0
z – 3 = 0
z = 3

By remainder theorem, we have
p(3) = q(3)
p(3) = q(3)
a(3)3 + 4(3)2 + 3(3) – 4 = (3)3 – 4(3) + a
⇒ 27a + 36 + 9 – 4 = 27 – 12 + a
⇒ 27aa = 15 – 41
⇒ 26a = –26
a = –1

#### Question 2:

The polynomial p(x) = x4 – 2x3 + 3x2 ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also find the remainder when p(x) is divided by x + 2.

Given:  p(x) = x– 2x+ 3x– ax + 3– 7
when p(x) is divided by (x + 1), remainder is 19.
By Remainder Theorem, we have
p(–1) = 19
⇒ (–1)4 – 2(–1)3 + 3(–1)2a(–1) + 3a – 7 = 19
⇒ 1 + 2 + 3 + a + 3a – 7 = 19
⇒ 4a – 1 = 19
⇒4a = 20
a = 5
Thus, the value of a is 5.

Therefore,  p(x) = x– 2x+ 3x– 5+ 8
Now, when p(x) is divided by (x + 2), remainder will be p(–2)
p(–2) = (–2)4 – 2(–2)3 + 3(–2)2 –5(–2) + 8
= 16 + 16 + 12 + 10 + 8
= 62

Hence, the remainder of polynomial p(x) when divided by (x + 2) is 62.

#### Question 3:

If both x – 2 and $\left(x-\frac{1}{2}\right)$ are factors of px2 + 5x + r, show that p = r.

Let Q(x) = px+ 5r
(x – 2) and $\left(x-\frac{1}{2}\right)$ are factors of Q(x).
Q(2) = 0
p(2)2 + 5(2) + r = 0
​⇒ 4p + r + 10 = 0          .....(1)

Also,
$Q\left(\frac{1}{2}\right)=0$

Now, subtracting the equations (1) and (2),
(4pp) + (r – 4r) + (10 – 10) = 0
⇒ 3p – 3r = 0
p = r

Hence, proved.

#### Question 4:

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

Let p(x) = 2x– 5x+ 2x– + 2
Now,
x
2 –3x + 2
= x2 – 2xx + 2        (By splitting Middle term)
= x(x – 2) – 1(x – 2)
= (x – 1) (x – 2)

Therefore, p(x) is divisible by x2 –3x + 2, if p(1) = 0 and p(2) = 0
p(1) = 2(1)4 – 5(1)3 + 2(1)2 – (1) + 2
= 2 – 5 + 2 – 1 + 2
= 0
Also,
p(2) = 2(1)4 – 5(2)3 + 2(2)2 – (2) + 2
= 32 – 40 + 8 – 2 + 2
= 0
Hence, p(x) is divisible by x2 – 3x + 2.

#### Question 5:

Simplify (2x – 5y)3 – (2x + 5y)3.

(2– 5y)– (2+ 5y)3
we have,
(ab)3 = a3b3 – 3ab(ab)
(a + b)3 = a3 + b3 + 3ab(a + b)

∴ (2– 5y)– (2+ 5y)3
= (8x3 – 125y3 – 30xy(2x – 5y)) – (8x3 + 125y3 + 30xy(2x + 5y))
= 8x3 – 125y3 – 60x2y + 150xy2 – 8x3 – 125y3 – 60x2y – 150xy
= – 250y3 – 120x2y

#### Question 6:

Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (–z + x – 2y).

Now, (x+ 4yz+ 2xy xz – 2yz) (–– 2y)
= x2(–z + x – 2y) + 4y2(–z + x – 2y) + z2(–z + x – 2y) + 2xy(–z + x – 2y) + xz(–z + x – 2y) – 2yz(–z + x – 2y)
= –x2z + x3 – 2x2y – 4y2z + 4xy2 – 8y3 z3 + xz2 – 2yz2 – 2xyz + 2x2y – 4xy2 – xz2 + x2z – 2xyz + 2yz2 – 2xyz + 4y2z
= (–x2z + x2z) + x3 + (–2yz2 –2x2y) + (–4y2z + 4y2z) + (4xy2 – 4xy2) – 3y3z3 + (xz2xz2) + (2yz2 – 2yz2) + (–2xyz – 2xyz – 2xyz)
= x3 – 3y3z3 – 6xyz

#### Question 7:

If a, b, c are all non-zero and a + b + c = 0, prove that $\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}=3$.

Given: = 0
Now,
$\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}=3$

Hence, the required value is 3.

#### Question 8:

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

Given: = 5, ab bc ca = 10
Now,
(c)2 = 52
abc2 + 2(ab + bc + ca) = 25
abc2 + 2 × 10 = 25
abc2 = 5

We know,
abc– 3abc = (c) (abc2 – (ab + bc ca))

= 5(5 –10)
= 5 × (–5)
= –25
Hence proved.

#### Question 9:

Prove that (a + b + c)3 a3 b3 c3 = 3(a + b) (b + c) (c + a).

[(b) c]= (a + b)3 + 3(a + b)2c + 3(a + b)c2 + c3
= (a3 + 3a2b + 3ab2 + b3) + 3(a2 + 2ab + b2)c +3(a + b)c2 + c3
= a3 + 3a2b + 3ab2 + b3 + 3a2c + 6abc + b2c + 3ac2 + 3bc2 + c3
= a3 + b3 + c3 + 3a2b + 3a2c + 3ab2 + 3b2c + 3ac2 + 3bc2 + 6abc
= a3 + b3 + c3 + 3a(ab + ac + b2 + bc) + 3c(ab + ac + b2 + bc)
= a3 + b3 + c3 + 3(a + c) (ab + ac + b2 + bc)
= a3 + b3 + c3 + 3(a + c) (a + b) (b + c)
(a + b + c)3a3b3 – c3 = 3(a + c) (b + c) (a + b)

Hence, proved.

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