Class 9 Maths Ch-2 Polynomials

NCERT Solutions of Class 9 Maths Ch-2 Polynomials बहुपद

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NCERT Solutions of Class 9 Maths Ch-2 Polynomials बहुपद

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Exercise 2.1

Question 1:
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
निम्नलिखित व्यंजकों में कौन-कौन एक चर में बहुपद हैं और कौन-कौन नहीं हैं ? कारण के साथ उत्तर दीजिए :
\((i) 4x^2 – 3x + 7 \quad (ii) y^2 + \sqrt{2} \quad (iii) 3\sqrt{t} + t\sqrt{2}\quad (iv) y + \frac{2}{y}\quad(v) x^{10} + y^{3} +t^{50}\)
Answer
(i) 4x² – 3x + 7
It is a polynomial in one variable i.e., x
because each exponent of x is a whole number.
(ii) y² + √2
It is a polynomial in one variable i.e., y
because each exponent of y is a whole number.
(iii) 3 √t + t√2
It is not a polynomial, because one of the exponents of t is ,\(\frac{1}{2}\)
which is not a whole number.
(iv) y + \(\frac{2}{y}\)
It is not a polynomial, because one of the exponents of y is -1,
which is not a whole number.
(v) x¹⁰+  y³ + t⁵⁰
Here, exponent of every variable is a whole number, but x¹⁰ + y³ + t⁵⁰ is a polynomial in x, y and t, i.e., in three variables.
So, it is not a polynomial in one variable.

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NCERT Solutions of Class 9 Maths Ch-2 Polynomials बहुपद

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Question 2:
Write the coefficients of x² in each of the following:
 निम्नलिखित में से प्रत्येक में x² का गुणांक लिखिए |
\((i) 2 + x^2 + x\quad (ii) 2 – x^2 + x^3 \quad (iii) \frac{\pi}{2}x^2 + x \quad (iv) \sqrt{2}x – 1 \)
Answer
\((i) 2 + x^2 + x\)
coefficient of x² = 1
\((ii) 2 – x^2 + x^3 \)
coefficient of x² = -1
\((iii) \frac{\pi}{2}x^2 + x\)
coefficient of x² = \(\frac{\pi}{2}\)
\((iv) \sqrt{2}x – 1 \)
coefficient of x² = 0

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NCERT Solutions of Class 9 Maths Ch-2 Polynomials बहुपद

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Question 3:
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
35 घात के द्विपद का और 100 घात के एकपदी का एक-एक उदाहरण दीजिए |
Answer
A binomial of degree 35 = 3\(x^35\) – 4;
A monomial of degree 100 = \(\sqrt{2} y^{100}\)

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NCERT Solutions of Class 9 Maths Ch-2 Polynomials

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Question 4:
Write the degree of each of the following polynomials:
निम्नलिखित बहुपदों में से प्रत्येक के घात लिखिए:
\((i)\;5x^3 +4x^2 + 7x \quad (ii)\;4 – y^2 \quad (iii)\;5t – \sqrt{7}\quad (iv)\;3\)
Answer
(i) 3; (ii) 2; (iii) 1; (iv) 0

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NCERT Solutions of Class 9 Maths Ch-2 Polynomials बहुपद

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Question 5:
Classify the following as linear, quadratic and cubic polynomials:
निम्नलिखित को रैखिक, द्विघात और त्रिघात बहुपद में वर्गीकृत कीजिए:
(i) x² + x (ii) x – x³ (iii) y + y² + 4 (iv) 1 + x (v) 3t (vi) r² (vii) 7x³
Answer
(i) Quadratic; (ii) Cubic; (iii) Quadratic; (iv) Linear; (v) Linear; (vi) Quadratic; (vii) Cubic

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NCERT Solutions of Class 9 Maths Ch-2 Polynomials बहुपद

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Exercise 2.2

Question 1:
Find the value of the polynomial 5x – 4x² + 3 at
(i) x = 0 (ii) x = –1 (iii) x = 2
Answer
Let P(x) = 5x−4x² + 3
(i) When x = 0
P(0) = 5(0)-4(0)²+3 = 3
(ii) When x = -1
P(−1) = 5(−1) − 4(−1)² + 3 = −5–4+3 = −6
(iii) When x = 2
P(2) = 5(2) − 4(2)² + 3 = 10–16 + 3 = −3

Question 2:
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y² – y + 1
(ii) p(t) = 2 + t + 2t² – t³
(iii) p(x) = x³
(iv) p(x) = (x – 1)(x + 1)
Answer
p(y) = y²–y+1
∴p(0) = (0)²−(0) + 1 = 1
p(1) = (1)²–(1) + 1 = 1
p(2) = (2)²–(2) + 1 = 3

(ii) p(t)=2+t+2t²−t³
∴p(0) = 2+0+2(0)²–(0)³=2
p(1) = 2+1+2(1)²–(1)³=2+1+2–1=4
p(2) = 2+2+2(2)²–(2)³=2+2+8–8=4

(iii) p(x)=x³
∴p(0) = (0)³ = 0
p(1) = (1)³ = 1
p(2) = (2)³ = 8

(iv) P(x) = (x−1)(x+1)
∴p(0) = (0–1)(0+1) = (−1)(1) = –1
p(1) = (1–1)(1+1) = 0(2) = 0
p(2) = (2–1)(2+1) = 1(3) = 3

Question 3:
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1; x = \(\frac{-1}{3}\)
(ii) p(x) = 5x – \(\pi\), x = \(\frac{4}{5}\)
(iii) p(x) = \(x^2 – 1\); x = 1, -1
(iv) p(x) = (x + 1) (x – 2); x = -1, 2
(v) p(x) = \(x^2\); x = 0
(vi) p(x) = \(lx + m; x = \frac{-m}{l}\)
(vii) p(x) = 3\(x^2\) – 1; x = -\(\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
(viii) p(x) = 2x + 1; x = \(\frac{1}{2}\)

Question 4:
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2 (v) p(x) = 3x
(vi) p(x) = ax, a\(\ne\)0 (vii) p(x) = cx + d, a\(\ne\)0, c, d are real numbers.
Answer
(i) p(x) = x + 5
\(\quad\) p(x) = 0
\(\Rightarrow\) x + 5 = 0
\(\Rightarrow\) x = – 5

(ii) p(x) = x – 5
\(\quad\) p(x) = 0
\(\Rightarrow\) x – 5 = 0
\(\Rightarrow\) x = 5

(iii) p(x) = 2x + 5
\(\quad\) p(x) = 0
\(\quad\) 2x + 5 = 0
\(\Rightarrow\) 2x = – 5
\(\Rightarrow\) x = \(\frac{-5}{2}\)

(iv) p(x) = 3x – 2
\(\quad\) p(x) = 0
\(\Rightarrow\) 3x – 2 = 0
\(\Rightarrow\) 3x = 2
\(\Rightarrow\) x = \(\frac{2}{3}\)

(v) p(x) = 3x
\(\quad\) p(x) = 0
\(\Rightarrow\) 3x = 0
\(\Rightarrow\) x = \(\frac{0}{3}\) = 0

(vi) p(x) = ax, a\(\ne\)0
\(\quad\) p(x) = 0
\(\Rightarrow\) ax = 0
\(\Rightarrow\) x = \(\frac{0}{a}\) = 0

(vii) p(x) = cx + d, a\(\ne\)0, c, d are real numbers.
\(\quad\) p(x) = 0
\(\Rightarrow\) cx + d = 0
\(\Rightarrow\) cx = -d
\(\Rightarrow\) x = \(\frac{-d}{c}\)

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Exercise 2.3

Question 1:
Determine which of the following polynomials has (x + 1) a factor :
(i) x³ + x² + x + 1 (ii) x⁴ + x³ + x² + x + 1 (iii) x⁴ + 3x³ + 3x² + x + 1 (iv) x³ – x² – (2 + \(\sqrt{2}\))x + \(\sqrt{2}\)
Answer
(i) Let p(x) = x³ + x² + x + 1
\(\quad\) x + 1 = 0
\(\Rightarrow\) x = – 1
p(-1) = (-1)³ + (-1)² + (-1) + 1
\(\quad\)= – 1 + 1 – 1 + 1 = 0
So, (x + 1) is a factor of p(x).

(ii) Let p(x) = x⁴ + x³ + x² + x + 1
\(\quad\) x + 1 = 0
\(\Rightarrow\) x = – 1
p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
\(\quad\)= 1 – 1 + 1 – 1 + 1 = 1
So, (x + 1) is not a factor of p(x).

(iii) Let p(x) = x⁴ + 3x³ + 3x² + x + 1
\(\quad\) x + 1 = 0
\(\Rightarrow\) x = – 1
p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1
\(\quad\)= 1 – 3 + 3 – 1 + 1 = 1
So, (x + 1) is not a factor of p(x).

(iv) Let p(x) = x³ – x² – (2 + \(\sqrt{2}\))x + \(\sqrt{2}\)
\(\quad\) x + 1 = 0
\(\Rightarrow\) x = – 1
p(-1) = (-1)³ – (-1)² – (2 + \(\sqrt{2}\))(-1) + \(\sqrt{2}\)
\(\quad\)= – 1 – 1 + 2 + \(\sqrt{2}\) + \(\sqrt{2}\)
\(\quad\) = – 2 + 2 + 2\(\sqrt{2}\) = 2\(\sqrt{2}\)
So, (x + 1) is not a factor of p(x).

Question 2:
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x³ + x² -2x – 1 , g(x) = x + 1
(ii) p(x) = x³ + 3x² + 3x + 1 , g(x) = x + 2
(iii) p(x) = x³ – 4x² + x + 6 , g(x) = x – 3
Answer
(i) p(x) = 2x³ + x² -2x – 1
\(\quad\) g(x) = x + 1 = 0
\(\Rightarrow\) x = -1
p(-1) = 2(-1)³ + (-1)² -2(-1) – 1
\(\quad\) = -2 + 1 + 2 – 1 = 0
So, g(x) is a factor of p(x).

(ii) p(x) = x³ + 3x² + 3x + 1
\(\quad\) g(x) = x + 2 = 0
\(\Rightarrow\) x = – 2
\(\quad\) p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1
\(\quad\) = – 8 + 12 – 6 + 1 = – 1
So, g(x) is not a factor of p(x).

(iii) p(x) = x³ – 4x² + x + 6
\(\quad\) g(x) = x – 3 = 0
\(\Rightarrow\) x = 3
\(\quad\) p(3) = (3)³ – 4(3)² + 3 + 6
\(\quad\) = 27 – 36 + 3 + 6= 0
So, g(x) is a factor of p(x).

Question 3:
Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:
(i) p(x) = x² + x + k
(ii) p(x) = 2x² + kx + \(\sqrt{2}\)
(iii) p(x) = kx² – \(\sqrt{2}\)x + 1
(iv) p(x) = kx² – 3x + k
Answer
(i) p(x) = x² + x + k
\(\quad\) x – 1 is a factor of p (x).
\(\Rightarrow\) x – 1 = 0 \(\Rightarrow\) x = 1
\(\therefore\) p(1) = 0
\(\Rightarrow\) (1)² + 1 + k = 0
\(\Rightarrow\) 1 + 1 + k = 0
\(\Rightarrow\) 2 + k = 0
\(\Rightarrow\) k = – 2

(ii) p(x) = 2x² + kx + \(\sqrt{2}\)
\(\quad\) x – 1 is a factor of p (x).
\(\Rightarrow\) x – 1 = 0 \(\Rightarrow\) x = 1
\(\therefore\) p(1) = 0
\(\Rightarrow\) 2(1)² + k(1) + \(\sqrt{2}\) = 0
\(\Rightarrow\) 2 + k + \(\sqrt{2}\) = 0
\(\Rightarrow\) k = – 2 – \(\sqrt{2}\)

(iii) p(x) = kx² – \(\sqrt{2}\)x + 1
\(\quad\) x – 1 is a factor of p (x).
\(\Rightarrow\) x – 1 = 0 \(\Rightarrow\) x = 1
\(\therefore\) p(1) = 0
\(\Rightarrow\) k(1)² – \(\sqrt{2}\)(1) + 1 = 0
\(\Rightarrow\) k – \(\sqrt{2}\) + 1 = 0
\(\Rightarrow\) k = \(\sqrt{2}\) – 1

(iv) p(x) = kx² – 3x + k
\(\quad\) x – 1 is a factor of p (x).
\(\Rightarrow\) x – 1 = 0 \(\Rightarrow\) x = 1
\(\quad\) p(1) = 0
\(\Rightarrow\) k(1)² – 3(1) + k = 0
\(\Rightarrow\) k – 3 + k = 0
\(\Rightarrow\) 2k – 3 = 0
\(\Rightarrow\) k = \(\frac{3}{2}\)

Question 4:
Factorise :
(i) 12x² – 7x + 1 (ii) 2x² + 7x + 3 (iii) 6x² + 5x – 6 (iv) 3x² – x – 4
Answer
(i) 12x² – 7x + 1
\(\;\) = 12x² – 4x – 3x + 1
\(\;\) = 4x(3x – 1 ) – 1(3x – 1)
\(\;\) = (3x -1) (4x – 1)

(ii) 2x² + 7x + 3
\(\;\) = 2x² + 6x + x + 3
\(\;\) = 2x(x + 3 ) + 1(x + 3)
\(\;\) = (x + 3) (2x + 1)

(iii) 6x² + 5x – 6
\(\;\) = 6x² + 9x – 4x – 6
\(\;\) = 3x(2x + 3 ) – 2(2x + 3)
\(\;\) = (2x + 3) (3x – 2)

(iv) 3x² – x – 4
\(\;\) = 3x² – 4x + 3x – 4
\(\;\) = x(3x – 4 ) + 1(3x – 4)
\(\;\) = (3x – 4) (x + 1)

Question 5:
Factorise :
(i) x³ – 2x² – x + 2 (ii) x³ – 3x² – 9x – 5 (iii) x³ + 13x² + 32x + 20 (iv) 2y³ + y² – 2y – 1
Answer
(i) x³ – 2x² – x + 2
\(\;\) = x³ – 2x² – x + 2 = x²(x − 2) −1(x − 2) = (x²−1)(x−2)= (x + 1)(x – 1)(x – 2)

(ii) x³ – 3x² – 9x – 5
\(\;\) Putting x = – 1
\(\;\) = (-1)³ – 3(-1)² – 9(-1) – 5 = -1 – 3 + 9 – 5 = 0
\(\therefore\) x + 1 is a factor of given polynomial.
use logn division, \(\frac{x^3 – 3x^2 – 9x – 5}{(x + 1)} = x^2 – 4x – 5\) = (x + 1)(x – 5)
\(\therefore\) x³ – 2x² – x + 2 = (x + 1) (x + 1)(x – 5)

(iii) x³ + 13x² + 32x + 20
\(\;\) Putting x = – 1
\(\;\) = (-1)³ + 13(-1)² + 32(-1) + 20 = -1 + 13 – 32 + 20 = 0
\(\therefore\) x + 1 is a factor of given polynomial.
use logn division, \(\frac{x^3 + 13x^2 + 32x + 20}{(x + 1)} = x^2 + 12x + 20\) = (x + 2 )(x + 10)
\(\therefore\) x³ + 13x² + 32x + 20 = (x + 1) (x + 2)(x + 10)

(iv) 2y³ + y² – 2y – 1
\(\;\) = y²(2y + 1) – 1(2y +1 ) = ( y² – 1) (2y + 1) = (y + 1) (y -1) (2y + 1)

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Exercise 2.4

Question 1:
Use suitable identities to find the following products:
(i) (x + 4) (x + 10) (ii) (x + 8) (x – 10) (iii) (3x + 4) (3x – 5) (iv) (\(y^2 + \frac{3}{2})\) (\(y^2 – \frac{3}{2})\) (v) (3 – 2x) (3 + 2x)
Answer
(i) (x + 4) (x + 10)
[(x+a)(x+b) = x² + (a+b)x + ab]
(x + 4) (x + 10) = x² + (4 + 10)x + \(4\times 10\) = x² + 14x + 40

(ii) (x + 8) (x – 10)
[(x+a)(x+b) = x² + (a+b)x + ab]
(x + 8) (x – 10) = x² + (8 – 10)x + \(8\times (-10)\) = x² – 2x – 80

(iii) (3x + 4) (3x – 5)
[(x+a)(x+b) = x² + (a+b)x + ab]
(3x + 4) (3x – 5) = (3x)² + (4 – 5)3x + \(4\times (-5)\) =9 x² + (-1)3x – 20 = 9 x² -3x – 20

(iv) (\(y^2 + \frac{3}{2})\) (\(y^2 – \frac{3}{2})\)
[(x+a)(x+b) = x² + (a+b)x + ab]
(\(y^2 + \frac{3}{2})\) (\(y^2 – \frac{3}{2})\) = (y²)² + \((\frac{3}{2} – \frac{3}{2})\dot(y^2) + \frac{3}{2}\times(\frac{-3}{2}\)) = \(y^4 – \frac{9}{4}\)

(v) (3 – 2x) (3 + 2x)
[(x+a)(x+b) = x² + (a+b)x + ab]
(3 – 2x) (3 + 2x) = (3)² + (-2x + 2x)3 + \((-2x)\times 2x\) = 9 – 4x².

Question 2:
Evaluate the following products without multiplying directly:
(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96
Answer
(i) 103 × 107
[(x+a)(x+b) = x² + (a+b)x + ab]
= (100 + 3) (100 + 7)
= \(100^2\) +(3 + 7)100 + 3\(\times\)7
= 10000 + \(10\times100\) + 21
= 10000 + 1000 + 21
= 11021

(ii) 95 × 96
[(x+a)(x+b) = x² + (a+b)x + ab]
= (100 – 5) (100 – 4)
= \(100^2 +(-5 – 4)100 + (-5)\times(-4)\)
= 10000 + \((-9)\times100\) + 20
= 10000 – 900 + 20
= 9120

(iii) 104 × 96
[(x+a)(x+b) = x² + (a+b)x + ab]
= (100 + 4) (100 – 4)
= \(100^2 +(4 – 4)100 + 4\times(-4)\)
= 10000 + \(0\times100\) – 16
= 10000 – 16
= 9984

Question 3:
Factorise the following using appropriate identities:
(i) 9x² + 6xy +y² (ii) 4y² – 4y + 1 (iii) \(x^2 – \frac{y^2}{100}\)
Answer
(i) 9x² + 6xy +y²
= (3x)² + 2(3x)(y) + y²
[A² + 2AB + B² = (A + B)²]
= (3x + y)²

(ii) 4y² – 4y + 1
= (2y)² + 2(2x)(1) + (1)²
[A² + 2AB + B² = (A + B)²]
= (2x + 1)²

(iii) \(x^2 – \frac{y^2}{100}\)
= \(x^2 – \left( \frac{y}{10} \right)^2 \)
[A² – B² = (A+B) (A-B)]
= (\(x + \frac{y}{10})\) (\(x – \frac{y}{10})\)

Question 4:
Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)² (ii) (2x – y + z)² (iii) (-2x + 3y + 2z)² (iv) (3a – 7b – c)² (v) (-2x + 5y – 3z)² (vi) \(\left[\frac{1}{4}a-\frac{1}{2}b+1\right]^2\)
Answer
(i) (x + 2y + 4z)²
[(A + B + C)² = A² + B² + C² + 2AB + 2BC + 2CA]
= (x)² + (2y)²+ (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
= x² + 4y² + 16z² + 4xy + 16yz + 4zx

(ii) (2x – y + z)²
[(A + B + C)² = A² + B² + C² + 2AB + 2BC + 2CA]
= (2x)² + (-y)²+ (z)² + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
= 4x² + y² + z² – 4xy – 2yz + 4zx

(iii) (-2x + 3y + 2z)²
[(A + B + C)² = A² + B² + C² + 2AB + 2BC + 2CA]
= (-2x)² + (3y)²+ (2z)² + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
= 4x² + 9y² + 4z² – 12xy + 12yz – 8zx

(iv) (3a – 7b – c)²
[(A + B + C)² = A² + B² + C² + 2AB + 2BC + 2CA]
= (3a)² + (-7b)²+ (-c)² + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)
= 9a² + 49b² + c² – 42ab + 14bc – 6ca

(v) (-2x + 5y – 3z)²
[(A + B + C)² = A² + B² + C² + 2AB + 2BC + 2CA]
= (-2x)² + (5y)²+ (-3z)² + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

(vi) \(\left[\frac{1}{4}a-\frac{1}{2}b+1\right]^2\)
[(A + B + C)² = A² + B² + C² + 2AB + 2BC + 2CA]
= \(\left(\frac{1}{4}a\right)^2 + \left(\frac{-1}{2}b\right)^2 + (1)^2 + 2\left(\frac{1}{4}a\right)\times \left(\frac{-1}{2}b\right) + 2\times \left(\frac{-1}{2}b\right) \times (1) + 2\times(1) \times \left(\frac{1}{4}a\right)\)
= \(\frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 – \frac{1}{4}ab – b + \frac{1}{2}a\)

Question 5:
Factorise:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz – 8xz
Answer
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
[A² + B² + C² + 2AB + 2BC + 2CA = (A + B + C)²]
= (2x)² + (3y)² + (-4z)² + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)
= (2x + 3y – 4z)² = (2x + 3y – 4z)(2x + 3y – 4z)

(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz – 8xz
[A² + B² + C² + 2AB + 2BC + 2CA = (A + B + C)²]
= \((-\sqrt{2}x)^2\) + y² + \((2\sqrt{2}z)^2\) + 2\((-\sqrt{2}x)\)y + 2(y)\((2\sqrt{2}z)\) + 2\((2\sqrt{2}z)(-\sqrt{2}x)\)
= \(\left(-\sqrt{2}x + y + 2\sqrt{2}z\right)\)² = \(\left(-\sqrt{2}x + y + 2\sqrt{2}z\right)\)\(\left(-\sqrt{2}x + y + 2\sqrt{2}z\right)\)

Question 6:
Write the following cubes in expanded form:
(i) (2x + 1)³ (ii) (2a – 3b)³ (iii) \([\frac{3}{2}x+1]^3 (iv) [x – \frac{2}{3}]^3\)
Answer
(i) (2x + 1)³
\(\;\) [(A + B)³ = A³ + 3A²B + 3AB² + B³]
\(\;\) = (2x)³+ 3(2x)²(1) + 3(2x)(1)² + (1)³
\(\;\) = 8x³ + 12x² + 6x + 1

(ii) (2a – 3b)³
\(\;\) [(A – B)³ = A³ – 3A²B + 3AB² – B³]
\(\;\) = (2a)³– 3(2a)²(3b) + 3(2a)(3b)² + (3b)³
\(\;\) = 8x³ – 36a²b + 54ab² + 27b³

(iii) \([\frac{3}{2}x+1]^3\)
\(\;\) [(A + B)³ = A³ + 3A²B + 3AB² + B³]
\(\;\) = \(\left(\frac{3}{2}x\right)^3 \)+ 3\(\left(\frac{3}{2}x\right)^2\)(1) + 3\((\frac{3}{2}x)\)(1)² + (1)³
\(\;\) = \(\frac{27}{8}x^3 + \frac{27}{2}x^2 + \frac{9}{2}x + 1 \)

Question 7:
Evaluate the following using suitable identities:
\((i) (99)^3 (ii) (102)^3 (iii) (998)^3\)
Answer
(i) (99)³ = (100 – 1)³
\(\;\) [(A – B)³ = A³ – 3A²B + 3AB² – B³]
\(\;\) = (100)³ – 3(100)²(1) + 3(100)(1)² – (1)³
\(\;\) = 1000000 – 30000 + 300 – 1
\(\;\) = 970299

(ii) (102)³ = (100 + 2)³
\(\;\) [(A + B)³ = A³ + 3A²B + 3AB² + B³]
\(\;\) = (100)³ + 3(100)²(2) + 3(100)(2)² + (2)³
\(\;\) = 1000000 + 60000 + 1200 + 8
\(\;\) = 1061208

(iii) (998)³ = (1000 – 2)³
\(\;\) [(A – B)³ = A³ – 3A²B + 3AB² – B³]
\(\;\) = (1000)³ – 3(1000)²(2) + 3(1000)(2)² – (2)³
\(\;\) = 1000000000 – 6000000 + 12000 – 8
\(\;\) = 994011992

Question 8:
Factorise each of the following:
(i) 8a³ + b³ + 12a²b + 6ab²
(ii) 8a³ – b³ – 12a²b + 6ab²
(iii) 27 – 125a³ -135a + 225a²
(iv) 64a³ – 27b³ -144a²b + 108ab²
(v) \(27p^3- \frac{1}{216} – \frac{9}{2} p^2 + \frac{1}{4}p\)
Answer
(i) 8a³ + b³ + 12a²b + 6ab²
[A³ + B³ + 3A²B + 3AB² = (A + B)³]
= (2a)³ + b³ + 3(2a)²(b) + 3(2a)(b)²
= (2a + b)³ = (2a + b)(2a + b)(2a + b)

(ii) 8a³ – b³ – 12a²b + 6ab²
[A³ – B³ – 3A²B + 3AB² = (A – B)³]
= (2a)³ – b³ – 3(2a)²(b) + 3(2a)(b)²
= (2a – b)³ = (2a – b)(2a – b)(2a – b)

(iii) 27 – 125a³ -135a + 225a²
[A³ – B³ – 3A²B + 3AB² = (A – B)³]
= (3a)³ – (5b)³ – 3(3a)²(b) + 3(2a)(b)²
= (3a – 5b)³ = (3a – 5b)(3a – 5b)(3a – 5b)

(iv) 64a³ – 27b³ -144a²b + 108ab²
[A³ – B³ – 3A²B + 3AB² = (A – B)³]
= (4a)³ – (3b)³ – 3(4a)²(3b) + 3(4a)(3b)²
= (4a – 3b)³ = (4a – 3b)(4a – 3b)(4a – 3b)

(v) \(27p^3- \frac{1}{216} – \frac{9}{2} p^2 + \frac{1}{4}p\)
[A³ – B³ – 3A²B + 3AB² = (A – B)³]
= (3p)³ – (\(\frac{1}{6}\))³ – 3(3p)²(\(\frac{1}{6}\)) + 3(3p)(\(\frac{1}{6}\))²
= (3p – \(\frac{1}{6}\))³ = (3p – \(\frac{1}{6}\))(3p – \(\frac{1}{6}\))(3p – \(\frac{1}{6}\))

Question 9:
Verify : (i) x³ + y³ = (x + y)(x² – xy + y²) (ii) x³ – y³ = (x – y)(x² + xy + y²)
Answer
(i) x³ + y³ = (x + y)(x² – xy + y²)
RHS
= (x + y)(x² – xy + y²)
= x(x² – xy + y²) + y(x² – xy + y²)
= x³ – x²y + xy² + x²y – xy² + y³
= x³ + y³
= LHS

(ii) x³ – y³ = (x – y)(x² + xy + y²)
RHS
= (x – y)(x² + xy + y²)
= x(x² + xy + y²) – y(x² + xy + y²)
= x³ + x²y + xy² – x²y – xy² – y³
= x³ – y³
= LHS

Question 10:
Factorise each of the following:
(i) 27y³ + 125z³ (ii) 64m³ – 343n³
Answer
(i) 27y³ + 125z³
= (3y)³ + (5z)³
[A³ + B³ = (A + B)(A² – AB + B²)]
= (3y + 5z)((3y)² – (3y)(5z) + (5z)²)
= (3y + 5z)(9y² – 15yz + 25z²)

(ii) 64m³ – 343n³
= (4m)³ – (7n)³
[A³ – B³ = (A – B)(A² + AB + B²)]
= (4m – 7n)(4m)² + (4m)(7n) + (7n)²)
= (4m – 7n)(16m² + 28mn + 49n²)

Question 11:
Factorise : 27x³ + y³ + z³ – 9xyz
Answer
A³ + B³ + C³ – 3ABC = (A +B +C)(A² + B² + C² – AB – BC – CA)
27x³ + y³ + z³ – 9xyz = (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)
= (3x + y + z) (9x² + y² + z²– 3xy – yz – 3zx)

Question 12:
Verify that x³ + y³ + z³ – 3xyz = \(\frac{1}{2}\)(x + y + z)[(x-y)² + (y-z)² + (z-x)²] [Imp.]
सत्यापित कीजिए x³ + y³ + z³ – 3xyz = \(\frac{1}{2}\)(x + y + z)[(x-y)² + (y-z)² + (z-x)²]
Answer
RHS
= \(\frac{1}{2}\)(x + y + z)[(x-y)² + (y-z)² + (z-x)²]
= \(\frac{1}{2}\)(x + y + z)[x² + y² – 2xy + y² + z² – 2yz + z² + x² – 2zx]
= \(\frac{1}{2}\)(x + y + z)[2x² + 2y² + 2z²– 2xy – 2yz – 2zx]
= \(\frac{1}{2} \times {2}\) (x + y + z) [x² + y² + z²– xy – yz – zx]
= (x + y + z) [x² + y² + z²– xy – yz – zx]
= x³ + xy²+ xz²– x²y -xyz -zx² + x²y +y³ + yz² – xy² – y²z – xyz + x²z + y²z + z³ – xyz – yz² – z²x
= x³ + y³ + z³ – 3xyz
= LHS

Question 13:
If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.
Answer
We know that x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
If x + y + z = 0,
then x³ + y³ + z³ – 3xyz = (0)(x² + y² + z² – xy – yz – zx) = 0
\(\Rightarrow\) x³ + y³ + z³ – 3xyz = 0
\(\Rightarrow\) x³ + y³ + z³ = 3xyz

Question 14:
Without actually calculating the cubes, find the value of each of the following:
(i) (–12)³ + (7)³ + (5)³
(ii) (28)³ + (–15)³ + (–13)³
Answer
We know that if a + b + c = 0, then a³ + b³ + c³ = 3abc
(i) (–12)³ + (7)³ + (5)³
-12 + 7 + 5 = 0
(–12)³ + (7)³ + (5)³
= \(3\times (-12)\times(7)\times(5)\) = -1260
(ii) (28)³ + (–15)³ + (–13)³
28 + (-15) + (-13) = 0
(28)³ + (–15)³ + (–13)³
= \(3\times (28)\times(-15)\times(-13)\) = 16380

Question 15:
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : 25a² – 35a + 12
(ii) Area : 35y² + 13y – 12
Answer
(i) Area of rectangle = 25a² – 35a + 12
= (5a – 4) (5a -3)
Hence, the possible dimensions of the cuboids are (5a – 4) and (5a -3)

(ii) Area : 35y² + 13y – 12
= (5y + 4) (7y -3)
Hence, the possible dimensions of the cuboids are (5y + 4) and (7y -3)

Question 16:
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x² – 12x
(ii) Volume : 12ky² + 8ky – 20k
Answer
(i) Volume of cuboid = 3x² – 12x
= 3x (x – 4)
= (3) (x) (x – 4)
Hence, the possible dimensions of the cuboids are 3, x and (x-4).
(ii) Volume of cuboid = 12ky² + 8ky – 20k
= 4k (3y² + 2y – 5)
= 4k (3y + 5) (y – 1)
Hence, the possible dimensions of the cuboids are 4k, (3y + 5) and (y – 1).

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NCERT Solutions of Class 9 Maths Ch-2 Polynomials बहुपद

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