Class 9 NCERT Solutions of Chapter 2-Polynomial

The NCERT solutions of class 9 maths chapter 2- Polynomial is a basic chapter of algebra. The NCERT solutions of chapter 2-Polynomial will help you in understanding the questions based on the different structures of polynomials. The NCERT solutions of chapter 2-Polynomial provide the use of all algebraic identities which are used in higher classes and also in other branches of mathematics. All the NCERT solutions of chapter 2-Polynomial are solved by an expert teacher of maths as per the CBSE norms.

Polynomial is an algebraic expression in which the degree of the variable is a whole number as an example x, x +1,x²+1, etc but constants like 1,2,3 3/2, etc are also supposed as a polynomial because we can write them x⁰,2x⁰,3x⁰,3x⁰/2etc.In terms of the number of terms, polynomials are classified as monomial(single), binomial(two terms), trinomial(three terms), etc. In terms of degree, the polynomials are classified as linear polynomials (degree 1), quadratic polynomials (degree 2), cubic polynomials (degree 3), etc.

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Chapter 2 Polynomial of Class 9 contains 5 exercises

Exercise 2.1

Exercise 2.2

Exercise 2.3

Exercise 2.4

Exercise 2.5

NCERT Solutions of all chapters of Maths for Class 10 from Chapters 1 to 15

NCERT Solutions of Class 9 Maths : from chapter 1 to 15

You can also see the video NCERT Solutions for Class 10 Maths Chapter 2 -Polinomial updated for 2023-24 CBSE Board Exams,these NCERT solutions are referred to Chapter- 2 Current maths NCERT text book.

Class 9 maths NCERT Solutions of the chapter 2-Polynomial

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Exercise 2.1

Q1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

(i) 4x² –3x + 7

(ii) y² + √2

(iii) 3√t + t√2

$\boldsymbol{\left ( iv \right )\: y + \frac{2}{y}}$

$\boldsymbol{\left ( v \right )\: x^{10}+y^{3}+t^{50}}$

Ans. (i) 4x² –3x + 7

Hint: Polynomial is the algebraic expression in which the power of the variable is a whole number

The given expression is a polynomial in one variable x since each power of x (2,1,0) is a whole number

(ii) y² + √2

The given expression is polynomial in one variable y since each power of y (2,0) is a whole number

(iii) 3√t + t√2

The given expression is not a polynomial since the power of variable t is (1/2, 1) in which 1/2 is not a whole number

$\boldsymbol{\left ( iv \right )\: y + \frac{2}{y}}$

The given expression is not a polynomial since the power of variable y is (1, –1) in which –1 is not a whole number.

$\boldsymbol{\left ( v \right )\: x^{10}+y^{3}+t^{50}}$

The given expression is a polynomial in three variables x,y, and t since each exponent of the variable is 10, 3, and 50 in which all are whole numbers.

Q2.Write the coefficient of x² in each of the following.

(i) 2 + x² + x

(ii) 2 – x² + x³

$\boldsymbol{\left ( iii \right )\: \frac{\pi }{2}x^{2}+x}$

$\boldsymbol{\left ( iv \right )\: \sqrt{2}x-1}$

Ans.

(i) 2 + x² + x

In the given expression the term containing x² has the coefficient 1

(ii) 2 – x² + x³

In the given expression the term containing x² has the coefficient –1

$\boldsymbol{\left ( iii \right )\: \frac{\pi }{2}x^{2}+x}$

In the given expression the term containing x² has the coefficient π/2.

$\boldsymbol{\left ( iv \right )\: \sqrt{2}x-1}$

In the given expression there is no any term x², so the coefficient of x² is 0.

Q3. Give one example each of a binomial of degree 35, and a monomial of degree 100.

Ans. Binomial of degree 35 is following

$\boldsymbol{x^{35}+2}$

Monomial of degree 100 is following

$\boldsymbol{x^{100}}$

Q4. Write the degree of the following polynomial

(i) 5x³ + 4x² + 7x

(ii) 4 – y²

(iii)5t –√7

(iv) 3

Ans.

(i) 5x³ + 4x² + 7x

The highest exponent of the variable x is 3, so the degree of the given polynomial is 3

(ii) 4 – y²

The highest exponent of the variable y in the expression is 2, so the degree of the polynomial is 2

(iii)5t –√7

The highest exponent of the variable t in the expression is 1, so the degree of the polynomial is 1

(iv) 3

There is no variable in the given term, so the degree of the polynomial is 0

Q5. Classify the following as linear, quadratic, and cubic polynomial.

(i) x² + x

(ii) x – x³

(iii) y + y² + 4

(iv) 1 + x

(v) 3t

(vi) r²

(vii) 7x³

Ans.

(i) x² + x

The given polynomial is of the degree 2, so it is a quadratic polynomial

(ii) x – x³

The given polynomial is of the degree 3, so it is a cubic polynomial

(iii) y + y² + 4

The given polynomial is of the degree 2, so the given polynomial is a quadratic polynomial

(iv) 1 + x

The given polynomial is of the degree 1, so the given polynomial is a linear polynomial

(v) 3t

The given polynomial is of the degree 1, so the given polynomial is a linear polynomial

(vi) r²

The given polynomial is of the degree 2, so the given polynomial is a quadratic polynomial

(vii) 7x³

The given polynomial is of the degree 3, so the given polynomial is a cubic polynomial

See the video for exercise 2.1 solutions

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

Exercise 2.2

Q1. Find the value of polynomial 5x –4x² +3 at

(i) x = 0 (ii)x = –1 (iii) x = 2

Ans. Let the given polynomial is expressed as p(x)

(i) x = 0

p(x) = 5x –4x² +3

p(0) = 5× 0 –4 × 0 + 3 = 3

(ii)x = –1

p(x) = 5x –4x² +3

p(–1) = 5 ×–1 – 4 (–1) ² + 3 = –5 –4 + 3 = –9 + 3 = –6

(iii) x = 2

p(x) = 5x –4x² +3

p(2) = 5 × 2 – 4 ×2² + 3 = 10 – 16 + 3 = –3

Q2.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)

Ans.

(i) p(y) = y² –y + 1

p(0) = 0 – 0 +1 = 1

p(1) = 1² –1 + 1 = 1

p(2) = 2² –2 + 1 = 3

(ii) p(t) = 2 +t +2t² – t³

p(0) = 2 +0 +2×0² – 0³= 2

p(1) = 2 +1 +2×1² – 1³ = 3 +2 – 1 = 4

p(2)=2 +2 +2×2² – 2³ = 4 + 8 – 8 = 4

(iii) p(x) = x³

p(0) = 0³ = 0

p(1) = 1³ = 1

p(2) = 2³ = 8

(iv) p(x) = (x +1)(x – 1)

p(0) = (0+1)(0 – 1) = –1

p(1) = (1 +1)(1 – 1) = 0

p(2) = (2 +1)(2 – 1) = 3

Q3. Verify whether the following are zeroes of the polynomial , indicated against them.

$\boldsymbol{\left ( i \right )\: p\left ( x \right )=3x +1,x = -\frac{1}{3}}$

$\boldsymbol{\left ( ii \right )\: p(x)=5x - \pi ,x=\frac{4}{5}}$

$\boldsymbol{\left ( iii \right )\: p(x)=x^{2} - 1 ,x=1}$

$\boldsymbol{\left ( iv \right )\: p(x)=\left ( x+1 \right )\left ( x -2 \right ),x=-\frac{1}{2}}$

$\boldsymbol{\left ( v \right )\: p(x)=x^{2},x=0}$

$\boldsymbol{\left ( vi \right )\: p(x)=lx+m,x=-\frac{m}{l}}$

$\boldsymbol{\left ( vii \right )\: p\left ( x \right )=3x^{2}-1,x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}}$

$\boldsymbol{\left ( viii \right )\: p\left ( x \right )=2x +1, x=\frac{1}{2}}$

Ans.

$\boldsymbol{\left ( i \right )\: p\left ( x \right )=3x +1,x = -\frac{1}{3}}$

$\boldsymbol{p\left ( -\frac{1}{3} \right )=3\times -\frac{1}{3}+1=-1+1=0}$

Since the value of the polynomial at x = – 1/3 is zero, so x = – 1/3 is the zero of the given polynomial

$\boldsymbol{\left ( ii \right )\: p(x)=5x - \pi ,x=\frac{4}{5}}$

$\boldsymbol{p\left ( \frac{4}{5} \right )=5\times \frac{4}{5}+\pi =4+\pi }$

The value of the polynomial is not zero at the value ,x = 4/5, so x = 4/5 is not the zero of the given polynomial

$\boldsymbol{\left ( iii \right )\: p(x)=x^{2} - 1 ,x=1}$

p(1) = 1² – 1 = 0

The value of the polynomial at x = 1, is zero, so x = 1 is the zero of the given polynomial.

$\boldsymbol{\left ( iv \right )\: p(x)=\left ( x+1 \right )\left ( x -2 \right ),x=-\frac{1}{2}}$

$\boldsymbol{p\left ( -\frac{1}{2} \right )=\left ( -\frac{1}{2}+1 \right )\left ( -\frac{1}{2} -2\right ) =\frac{1}{2}\times -\frac{5}{2}=-\frac{5}{4}}$

The value of the polynomial at x = –1/2, is not zero, so x = –1/2 is not the zero of the given polynomial.

$\boldsymbol{\left ( v \right )\: p(x)=x^{2},x=0}$

p(0) = 0

The value of the polynomial at x = 0, is zero, so x = 0 is the zero of the given polynomial

$\boldsymbol{\left ( vi \right )\: p(x)=lx+m,x=-\frac{m}{l}}$

$\boldsymbol{P\left ( -\frac{m}{l} \right )\: =l\times -\frac{m}{l}+m=0}$

The value of the polynomial at x = –m/l is zero, so x = –m/l is the zero of the given polynomial

$\boldsymbol{\left ( vii \right )\: p\left ( x \right )=3x^{2}-1,x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}}$

$\boldsymbol{p\left ( -\frac{1}{\sqrt{3}} \right )\: =3\times \left ( -\frac{1}{\sqrt{3}} \right )^{2}-1=3\times \frac{1}{3}-1=1-1=0}$

The value of the polynomial at x = –1/l√3 is zero, so x = –1/√3 is the zero of the given polynomial

$\boldsymbol{p\left ( \frac{2}{\sqrt{3}} \right )\: =3\times \left ( \frac{2}{\sqrt{3}} \right )^{2}-1=4-1=3}$

The value of the polynomial at x = 2/√3 is not zero, so x = 2/√3 is not the zero of the given polynomial

$\boldsymbol{\left ( viii \right )\: p\left ( x \right )=2x +1, x=\frac{1}{2}}$

$\boldsymbol{p\left ( \frac{1}{2} \right )\: =2\times \frac{1}{2}+1=1+1=2}$

The value of the polynomial at x = 1/2 is not zero, so x = 1/2 is not the zero of the given polynomial

Q4. Find the zero of the polynomial in each of the given cases.

(i) p(x) = x +5

(ii) p(x) = x – 5

(iii) p(x) = 2 x + 5

(iv) p(x) = 3x – 2

(v) p(x) = 3x

(vi) p(x) = ax, a ≠ 0

(vii) p(x) = cx + d, c ≠ 0 where c and d are real numbers

Ans.

(i) p(x) = x +5

x +5 = 0

x = – 5

Hence the zero of the given polynomial is x =– 5

(ii) p(x) = x – 5

x – 5 = 0

x = 5

Hence the zero of the given polynomial is x = 5

(iii) p(x) = 2 x + 5

2 x + 5 = 0

$\boldsymbol{x=-\frac{5}{2}}$

Hence the zero of the given polynomial is – 5/2

(iv) p(x) = 3x – 2

3x – 2 = 0

$\boldsymbol{x=\frac{2}{3}}$

Hence the zero of the given polynomial is 2/3

(v) p(x) = 3x

3x = 0

x = 0

Hence the zero of the given polynomial is 0

(vi) p(x) = ax, a ≠ 0

ax = 0

x = 0

Hence the zero of the given polynomial is 0

vii) p(x) = cx + d

cx + d = 0

Hence zero of the given polynomial is – d/c

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

EXERCISE-2.3

Q1.Find the remainder when $x^{3}+3x^{2}+3x+1$ is divided by

(i) x $+1$

$\boldsymbol{\left ( ii \right )\: x-\frac{1}{2}}$

(iii) x

(iv) x+π

(v) 5+2x

Ans.

(i) We have to divide the polynomial $x^{3}+3x^{2}+3x+1$ by the polynomial x + 1

Let’s find the zero of the x $+1$ = 0

x = – 1

Putting the value of x = – 1, in $x^{3}+3x^{2}+3x+1$

$\boldsymbol{\left ( -1 \right )^{3}+3\times \left ( -1 \right )^{2}+3\times -1+1}$

= – 1 + 3 – 3 + 1 = 0

(ii) Dividing the polynomial $x^{3}+3x^{2}+3x+1$ by $\boldsymbol{x-\frac{1}{2}}$ , we can get the reminder in the following way

$\boldsymbol{x-\frac{1}{2}=0}$

$\boldsymbol{x=\frac{1}{2}}$

Putting the value $\boldsymbol{x=\frac{1}{2}}$ in $x^{3}+3x^{2}+3x+1$

$\boldsymbol{\left ( \frac{1}{2} \right )^{3}+3\left ( \frac{1}{2} \right )^{2}+3\times \frac{1}{2}+1}$

$\boldsymbol{=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1}$

$\boldsymbol{=\frac{27}{8}}$

(iii) Dividing the polynomial $x^{3}+3x^{2}+3x+1$ by x, we can get the reminder by the following way

Putting x = 0 the zero of the polynomial x, in the given polynomial $x^{3}+3x^{2}+3x+1$

0 + 0 + 0 +1 = 1

(iv) Dividing the polynomial $x^{3}+3x^{2}+3x+1$ by x + π, we can get the reminder by the following way

x + π = 0

x = – π , is the zero of the polynomial x + π

Putting the value x = – π, in the given polynomial $x^{3}+3x^{2}+3x+1$

The reminder is

$\boldsymbol{=\left ( -\pi \right )^{3}+3\times \left ( -\pi \right )^{2}+3\times -\pi +1}$

$\boldsymbol{=-\pi ^{3}+3\pi ^{2}-3\pi +1}$

(v) Dividing the polynomial $x^{3}+3x^{2}+3x+1$ by 5 + 2x, we can get the reminder by the following way

5+2x = 0

$\boldsymbol{x=-\frac{5}{2}}$

$\boldsymbol{x=-\frac{5}{2}}$ , is the zero of the polynomial 5+2x

Putting the value of $\boldsymbol{x=-\frac{5}{2}}$ in the polynomial $x^{3}+3x^{2}+3x+1$ , we get the reminder

$\boldsymbol{=\left ( -\frac{5}{2} \right )^{3}+3\times \left ( -\frac{5}{2} \right )^{2}+3\times -\frac{5}{2}+1}$

$\boldsymbol{=-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1}$

$\boldsymbol{=-\frac{27}{8}}$

Q2.Find the remainder when $x^{3}-ax^{2}+6x-a$ is divided by $x-a$ .

The zero the polynomial x – a, is

x – a = 0

x = a

Putting this value in $x^{3}-ax^{2}+6x-a$

a³ – a × a² + 6 × a – a

= a³ – a³ +6a – a = 5a

Therefore the required reminder is 5a

Q3.Check whether $7+3x$ is a factor of $3x^{3}+7x$ .

The zero of the polynomial 7 + 3x is as follows

7 + 3x = 0

$\boldsymbol{x=-\frac{7}{3}}$

Putting this value of x in $3x^{3}+7x$ , we shall get the required reminder

$\boldsymbol{3\times \left ( -\frac{7}{3} \right )^{3}+7\times -\frac{7}{3}}$

$\boldsymbol{=-\frac{49}{9}-\frac{49}{3}}$

$\boldsymbol{=-\frac{343}{9}-\frac{49}{3}=-\frac{490}{9}}$

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

EXERCISE-2.4

Q1.Determine which of the following polynomials has $(x+1)$ a factor.

(i) $x^{3}+x^{2}+x+1$

(ii) $x^{4}+x^{3}+x^{2}+x+1$

(iii) $x^{4}+3x^{3}+3x^{2}+x+1$

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

Ans.

(i) $x^{3}+x^{2}+x+1$

The zero of the polynomial x + 1 is as follows

x + 1 = 0

x = – 1

Putting this value of x = – 1, in the given polynomial $x^{3}+x^{2}+x+1$

(–1)³ + 1² + 1 + 1

– 1 + 1 + 1 + 1 = 2

Since value of the polynomial at x = – 1, is not zero so (x +1) is not the factor of the given polynomial

(ii) $x^{4}+x^{3}+x^{2}+x+1$

The zero of the polynomial (x +1) is x + 1 = 0 ⇒ x = – 1

Putting x = – 1, in the given polynomial

$\boldsymbol{\left ( -1 \right )^{4}+\left ( -1 \right )^{3}+\left ( -1 \right )^{2}-1+1}$

= 1 – 1 + 1 – 1 + 1 = 1

Since value of the given polynomial at x = 1, is not zero,therefore (x +1) is not the factor of the given polynomial

(iii) $x^{4}+3x^{3}+3x^{2}+x+1$

The zero of the polynomial (x +1) is x + 1 = 0 ⇒ x = – 1

Putting x = – 1, in the given polynomial

$\boldsymbol{\left ( -1 \right )^{4}+3\left ( -1 \right )^{3}+3\left ( -1 \right )^{2}-1+1}$

= 1 – 3 + 3 –1 +1 = 1

Since value of the given polynomial at x = 1, is not zero,therefore (x +1) is not the factor of the given polynomial

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

The zero of the polynomial (x +1) is x + 1 = 0 ⇒ x = – 1

Putting x = – 1, in the given polynomial

$\boldsymbol{\left ( -1 \right )^{3}-\left ( -1 \right )^{2}-\left ( 2+\sqrt{2} \right )\times -1+\sqrt{2}}$

$\boldsymbol{=-1-1+2+\sqrt{2}+\sqrt{2}}$

$\boldsymbol{=-2+2+2\sqrt{2}}$

$\boldsymbol{=2\sqrt{2}}$

The value of the polynomial at x = – 1 is not zero, so (x + 1) is not the factor of the given polynomial

Q2. 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

Ans. (i) p (x)= 2x³ + x² – 2x – 1, g (x) = x + 1

x + 1 = 0

x = – 1

Putting the value of x = – 1 in the given polynomial

p(x) = 2x³ + x² – 2x – 1

$\boldsymbol{p(-1)}\boldsymbol{=2\left ( -1 \right )^{3}+\left ( -1 \right )^{2}-2\times -1 -1}$

= 2× – 1 + 1 + 2 – 1

p(–1) = – 2 + 3 – 1 = 0

Since value of p(–1) is zero ,therefore g(x) is the factor of the given polynomial

(ii) p(x)= x³ + 3x² + 3x + 1, g (x) = x + 2

x + 2 = 0

x = – 2

p(x)= x³ + 3x² + 3x + 1

p(– 2) = (– 2)³ + 3(–2)² + 3 (– 2) +1 = -8 +12 -6 +1 = -1

Since the value of p(– 2) is not zero, therefore the given polynomial is not the factor of (x + 2)

(iii) p (x) = x³ – 4x² + x + 6, g (x) = x – 3

x – 3 = 0

x = 3

Putting the value of x in the given polynomial

p(3) = 3³ –4× 3² + 3 + 6

= 27– 36 + 9

= 0

Since the value of p(3) is zero, so g(x) is the factor of p(x)

Q3.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 + √2
(iii) p (x) = kx² – √2 x + 1
(iv) p (x) = kx² – 3x + k
Ans.
For (x – 1) to be a factor of p(x), p(1) should be equal to 0.

(i) Here, p(x) = x² + x + k
Since, p(1) = (1)² +1 + k
⇒ p(1) = k + 2 = 0
⇒ k = -2.

(ii) Here, p (x) = 2x² + kx + √2
Since, p(1) = 2(1)² + k(1) + √2
= 2 + k + √2 =0
k = -2 – √2 = -(2 + √2)

(iii) Here, p (x) = kx² – √2 x + 1
Since, p(1) = k(1)² – (1) + 1
= k – √2 + 1 = 0
⇒ k = √2 -1

(iv) Here, p(x) = kx² – 3x + k
p(1) = k(1)² – 3(1) + k
= k – 3 + k
= 2k – 3 = 0
⇒ k =

Q4. Factorise
(i) 12x² – 7x +1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x – 6
(iv) 3x² – x – 4
Ans.
(i) We have,
12x² – 7x + 1 = 12x² – 4x- 3x + 1
= 4x (3x – 1 ) -1 (3x – 1)
= (3x -1) (4x -1)
Thus, 12x² -7x + 3 = (2x – 1) (x + 3)

(ii) We have, 2x² + 7x + 3 = 2x² + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)
Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)

(iii) We have, 6x² + 5x – 6 = 6x² + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)
Thus, 6x² + 5x – 6 = (2x + 3)(3x – 2)

(iv) We have, 3x² – x – 4 = 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)
Thus, 3x² – x – 4 = (3x – 4)(x + 1)

Q5. Factorise
(i) x³ – 2x² – x + 2
(ii) x³ – 3x² – 9x – 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² – 2y – 1
Ans.
(i) We have, x³ – 2x² – x + 2
Rearranging the terms, we have x³ – x – 2x² + 2
= x(x² – 1) – 2(x² -1) = (x² – 1)(x – 2)
= [(x)² – (1)²](x – 2)
= (x – 1)(x + 1)(x – 2)
[∵ (a² – b²) = (a + b)(a-b)]
Therefore , x³ – 2x² – x + 2 = (x – 1)(x + 1)(x – 2)

(ii) We have, x³ – 3x² – 9x – 5
= x³ + x² – 4x² – 4x – 5x – 5 ,
= x² (x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1)(x² – 4x – 5)
= (x + 1)(x² – 5x + x – 5)
= (x + 1)[x(x – 5) + 1(x – 5)]
= (x + 1)(x – 5)(x + 1)
Therefore, x³ – 3x² – 9x – 5 = (x + 1)(x – 5)(x +1)

(iii) We have, x³ + 13x² + 32x + 20
= x³ + x² + 12x² + 12x + 20x + 20
= x²(x + 1) + 12x(x +1) + 20(x + 1)
= (x + 1)(x² + 12x + 20)
= (x + 1)(x² + 2x + 10x + 20)
= (x + 1)[x(x + 2) + 10(x + 2)]
= (x + 1)(x + 2)(x + 10)
Therefore, x³ + 13x² + 32x + 20
= (x + 1)(x + 2)(x + 10)

(iv) We have, 2y³ + y² – 2y – 1
= 2y³ – 2y² + 3y² – 3y + y – 1
= 2y²(y – 1) + 3y(y – 1) + 1(y – 1)
= (y – 1)(2y² + 3y + 1)
= (y – 1)(2y² + 2y + y + 1)
= (y – 1)[2y(y + 1) + 1(y + 1)]
= (y – 1)(y + 1)(2y + 1)
Therefore, 2y³ + y² – 2y – 1
= (y – 1)(y + 1)(2y +1)

See the video for Question (4) and Question (5)

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

Exercise 2.5

Q1.Use suitable identities to find the following products

(i) $(x+4)(x+10)$

(ii) $(x+8)(x-10)$

(iii) $(3x+4)(3x-5)$

(iv) $(y^{2}+\frac{3}{2})(y^{2}-\frac{3}{2})$

Ans.

(i) (x + 4)( x+ 10)

$\boldsymbol{\because \left ( x+a \right )\left ( x+b) \right )=x^{2}+\left ( a+b \right )x+ab}$

$\boldsymbol{\therefore \left ( x+4 \right )\left ( x+10) \right )=x^{2}+\left ( 4+10 \right )x+4\times 10}$

$\boldsymbol{=x^{2}+14x+20}$

$\boldsymbol{\left ( ii \right )\: \left ( x+8 \right )\left ( x-10 \right )}$

$\boldsymbol{\because \left ( x+a \right )\left ( x+b) \right )=x^{2}+\left ( a+b \right )x+ab}$

$\boldsymbol{\therefore \left ( x+8 \right )\left ( x-10 \right )=x^{2}+\left ( 8-10 \right )x+8\times -10}$

$\boldsymbol{=x^{2}-x-80}$

$\boldsymbol{\left ( iii \right )\: \left ( 3x+4 \right )\left ( 3x-5 \right )}$

$\boldsymbol{\because \left ( x+a \right )\left ( x+b) \right )=x^{2}+\left ( a+b \right )x+ab}$

$\boldsymbol{\: \left ( 3x+4 \right )\left ( 3x-5 \right )=\left ( 3x \right )^{2}+\left ( 4-5 \right )3x+4\times -5}$

$\boldsymbol{\left ( iv \right )\: \left ( y^{2}+\frac{3}{2} \right )\left ( y^{2}-\frac{3}{2} \right )}$

$\boldsymbol{\because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2}}$

$\boldsymbol{\: \left ( y^{2}+\frac{3}{2} \right )\left ( y^{2}-\frac{3}{2} \right )=\left ( y^{2} \right )^{2}-\left ( \frac{3}{2} \right )^{2}}$

$\boldsymbol{=y^{4}-\frac{9}{4}}$

Q2.Evaluate the following products without multiplying directly:

(i)103×107

(ii)95×96

(iii)104× 96

Ans.

(i)103×107

Modifying the given expression in form of (x + a)(x +b) as following

(103)×(107) = (100 +3)(100 +7)

$\boldsymbol{\because \left ( x+a \right )\left ( x+b) \right )=x^{2}+\left ( a+b \right )x+ab}$

(100 +3)(100 +7) = 100² +(3 +7)100 + 3×7 = 10000 +1000 + 21 = 110021

(ii) 95×96

95×96 = (100 – 5)(100 – 4)

$\boldsymbol{\because \left ( x+a \right )\left ( x+b) \right )=x^{2}+\left ( a+b \right )x+ab}$

(100 – 5)(100 – 4) = 100² +( – 5 – 4)100 +( – 4)( – 5)

10000 – 900 + 20 = 9120

(iii) 104× 96

(100 +4)(100 – 4)

(100 +4)(100 – 4) = 100² – 4² = 10000 – 16 = 9984

Q3.Factorize the following using appropriate identities

(i) $9x^{2}+6xy+y^{2}$

(ii) $4y^{2}-4y+1$

(iii) $x^{2}-y^{2}/100$

Ans.

(i) $9x^{2}+6xy+y^{2}$

Using the following identity

$\boldsymbol{\because \left ( a+b \right )^{2}=\left ( a+b \right )(a+b)}$

$9x^{2}+6xy+y^{2}=(3x)^{2}+(2)(3x)(y)+(y)^{2}$

$(3x+y)^{2}=(3x+y)(3x+y)$

(ii) $4y^{2}-4y+1$

$\boldsymbol{\because \left ( a-b \right )^{2}=\left ( a-b \right )(a-b)}$

$4y^{2}-4y+1=(2y)^{2}-(2)(2y)(1)+(1)^{2}$

$(2y-y)^{2}=(2y-1)(2y-1)$

(iii) $x^{2}-y^{2}/100$

$x^{2}-\frac{y^{2}}{100}=(x)^{2}-(\frac{y}{10})^{2}$

$\boldsymbol{\because a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )}$

$=(x+\frac{y}{10})(x-\frac{y}{10})$

Q4.Expand each of the following, using a suitable identity

(i) $(x+2y+4z)^{2}$

(ii) $(2x-y+z)^{2}$

(iii) $(-2x+3y+2z)^{2}$

(iv) $(3a-3b-c)^{2}$

(v) $(-2x+5y-3z)^{2}$

(iv) $[\frac{1}{4a}-\frac{1}{4b}+1^{}]^{2}$

Ans.

(i) $(x+2y+4z)^{2}$

$\because \boldsymbol{\left ( a+b+c \right )^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac}$

$\boldsymbol{\left ( x+2y+4z \right )^{2}=x^{2}+\left ( 2y \right )^{2}+\left ( 4z \right )^{2}+2\times x\times 2y+2\times 2y\times 4z+2\times x\times 4z}$

$\boldsymbol{=x^{2}+4y^{2}+16z^{2}+4xy+16yz+8xz}$

(ii) $(2x-y+z)^{2}$

$\because \boldsymbol{\left ( a+b+c \right )^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac}$

$\boldsymbol{\left ( 2x-y+z \right )^{2}=\left ( 2x \right )^{2}+\left ( -y \right )^{2}+z^{2}+2\times 2x\times -y+2\times -y\times z+2\times 2x\times z}$

$\boldsymbol{=4x^{2}+y^{2}+z^{2}-4xy-2yz+4xz}$

(iii) $(-2x+3y+2z)^{2}$

$\because \boldsymbol{\left ( a+b+c \right )^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac}$

$\boldsymbol{\left ( -2x+3y+2z \right )^{2}=\left ( -2x \right )^{2}+\left ( 3y \right )^{2}+\left ( 2z \right )^{2}+2\times -2x\times 3y+2\times 3y\times 2z+2\times 2z\times -2x}$

$\boldsymbol{=4x^{2}+9y^{2}+4z^{2}-12xy+12yz-8xz}$

(iv) $(3a-3b-c)^{2}$

$\because \boldsymbol{\left ( a+b+c \right )^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac}$

$\boldsymbol{\left ( 3a-3b-c \right )^{2}=\left ( 3a \right )^{2}+\left ( -3b \right )^{2}+\left ( -c \right )^{2}+2\times 3a\times -3b+2\times -3b\times -c+2\times -c\times 3a}$

$\boldsymbol{=9a^{2}+9b^{2}+c^{2}-18ab+6bc-6ac}$

(v) $(-2x+5y-3z)^{2}$

$\because \boldsymbol{\left ( a+b+c \right )^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac}$

$\boldsymbol{\left ( -2x+5y-3z \right )^{2}=\left ( -2x \right )^{2}+\left ( 5y \right )^{2}+\left ( -3z \right )^{2}+2\times -2x\times 5y+2\times 5y\times -3z+2\times -2x\times -3z}$

$\boldsymbol{=4x^{2}+5y^{2}+9z^{2}-20xy-30yz+12xz}$

(iv) $[\frac{1}{4a}-\frac{1}{4b}+1^{}]^{2}$

$\because \boldsymbol{\left ( a+b+c \right )^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac}$

$\boldsymbol{\left ( \frac{1}{4a}-\frac{1}{4b} +1\right )^{2}=\left ( \frac{1}{4a} \right )^{2}+\left (- \frac{1}{4b} \right )^{2}+1^{2}+2\times \frac{1}{4a}\times -\frac{1}{4b}+2\times -\frac{1}{4b}\times 1+2\times 1\times \frac{1}{4a}}$

$\boldsymbol{=\frac{1}{16a^{2}}+\frac{1}{16b^{2}}+1-\frac{1}{8ab}-\frac{1}{2b}+\frac{1}{2a}}$

Q5.Factorise:

(i) $4x^{2}+9y^{2}+16z^{2}+12xy-24yz-16xz$

(ii) $2x^{2}+y^{2}+8z^{2}-2\sqrt{2}xy+4\sqrt{2}yz-8xz$

Ans.

(i) $4x^{2}+9y^{2}+16z^{2}+12xy-24yz-16xz$

$\boldsymbol{=\left ( 2x \right )^{2}+\left ( 3y \right )^{2}+\left ( 4z \right )^{2}+12xy-24yz-16xz}$

$\because \boldsymbol{\left ( a+b+c \right )^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac}$

Since the terms –24yz and –16xz, both are having negative sign, so the term z which is common between them must be negative

$\boldsymbol{=\left ( 2x \right )^{2}+\left ( 3y \right )^{2}+\left ( -4z \right )^{2}+12xy-24yz-16xz}$

$\boldsymbol{=\left ( 2x+3y-4z \right )^{2}=\left ( 2x+3y-4z \right )\left ( 2z+3y-4z \right )}$

(ii) $2x^{2}+y^{2}+8z^{2}-2\sqrt{2}xy+4\sqrt{2}yz-8xz$

$2x^{2}+y^{2}+8z^{2}-2\sqrt{2}xy+4\sqrt{2}yz-8xz$

$\boldsymbol{=\left ( \sqrt{2}x \right )^{2}+y^{2}+\left ( 2\sqrt{2}z \right )^{2}+2\sqrt{2}xy+4\sqrt{2}yz+8xz}$

$\because \boldsymbol{\left ( a+b+c \right )^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac}$

$\boldsymbol{=\left ( \sqrt{2}x+y+2\sqrt{2}z \right )^{2}=\left ( \sqrt{2}x+y+2\sqrt{2z} \right )\left ( \sqrt{2}x+y+2\sqrt{2}z \right )}$

Q6.Write the following cubes in expanded form

$\boldsymbol{\left ( i \right )\: (2x+1)^{3}}$

$\mathbf{(ii)}\mathbf{(2a-3b)^{3}}$

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

Ans.

$\boldsymbol{\left ( i \right )\: (2x+1)^{3}}$

Applying the following identity

$(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$

$(2x+1)^{3}=(2x)^{3}+(1)^{3}+(3)(2x)(1)(2x+1)$

= $8x^{3}+1+6x(2x+1)$

= $8x^{3}+12x^{2}+6x+1$

$\mathbf{(ii)}\mathbf{(2a-3b)^{3}}$

Applying the following identity

$\mathbf{(a-b)^{3}=a^{3}-b^{3}-3ab(a-b\mathbf{})}$

$\mathbf{(2a-3b)^{3}=(2a)^{3}-(3b)^{3}-(3)(2a)(3b)(2a-3b)}$

$\mathbf{\mathbf{\mathbf{}=}}\mathbf{8a^{3}-27b^{3}-18ab(2a-3b)}$

$\mathbf{=8a^{3}-27b^{3}-36a^{2}b+54ab^{2}}$

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

Using the following identity

$\mathbf{(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)}$

$\boldsymbol{\left ( \frac{3}{2}x+1 \right )^{3}=\left ( \frac{3}{2}x \right )^{3}+1^{3}+3\times \frac{3}{2}x\times 1\left ( \frac{3}{2}x+1 \right )}$

$\boldsymbol{=\frac{27}{8}x^{3}+1+\frac{9}{2}x\left ( \frac{3}{2}x+1 \right )}$

$\boldsymbol{=\frac{27}{8}x^{3}+1+\frac{27}{4}x^{2}+\frac{9}{2}x}$

$\boldsymbol{\left ( iv \right )\: \left ( x-\frac{2}{3}y \right )^{3}}$

$\mathbf{(a-b)^{3}=a^{3}-b^{3}-3ab(a-b\mathbf{})}$

$\boldsymbol{\: \left ( x-\frac{2}{3}y \right )^{3}=x^{3}-\left ( \frac{2}{3}y \right )^{3}-3\times x\times \frac{2}{3}y\left ( x-\frac{2}{3}y \right )}$

$\boldsymbol{\: =x^{3}- \frac{8}{27}y ^{3}-2x^{2}y+2y^{2}x}$

Q7.Evaluate the following using suitable identities

(i)99³ (ii)102³ (iii) 998³

Ans.

(i)99³

Modifying 99 into (100– 1) and using the following identity

(x +y)³ = x³ + 3x²y + 3y²x + y³

$\boldsymbol{\left ( 100-1 \right )^{3}=100^{3}-3\times 100^{2}\times 1+3\times 1^{2}\times 100-1^{3}}$

= 1000000 – 30000 + 300 – 1

= 970299

(ii)102³

Modifying 102 into ( 100 +2) and using the following identity

(x +y)³ = x³ + 3x²y + 3y²x + y³

(100 +2)³ = 100³ + 3× 100² ×2 + 3×2²× 100 + 2³

= 1000000 + 60000 +1200 +8

=1061208

(iii) 998³

Modifying the number 998 into (1000 – 2) and using the following identity

(x +y)³ = x³ + 3x²y + 3y²x + y³

(1000 –2)³ = 1000³ – 3× 1000² ×2 + 3×2²× 1000 – 2³

1000000000 –6000000 +12000 – 8

=994011992

Q8.Factorise each of the following

(1) 8a³ + b³ +12a²b + 6ab²

(ii) 8a³ – b³ – 12a²b + 6ab²

(iii) 27 – 125a³ – 135a + 225a²

(iv) 64a³ – 27b³ – 144a²b + 108b²a

$\boldsymbol{\left ( iv \right )\: 27p^{3}-\frac{1}{216}-\frac{9}{2}p^{2}+\frac{1}{4}p}$

Ans.

(1) 8a³ + b³ +12a²b + 6ab²

Using the following identity

(x + y)³ = x³ + y³ +3x²y + 3y²x

Modifying the given expression as following

(2a)³ + b³ +3×(2a)²×b + 3×b²×2a

= (2a + b)³

(ii) 8a³ – b³ – 12a²b + 6ab²

Using the following identity

(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x

Modifying the given expression as following

(2a)³ – b³ –3×(2a)²×b + 3×b²×2a

= (2a – b)³

(iii) 27 – 125a³ – 135a + 225a²

Using the following identity

(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x

27 – 125a³ – 135a + 225a²

= 3³ – (5a)³ –3×(3)²×5a + 3×(5a)²×5a

= (3 – 5a)³

(iv) 64a³ – 27b³ – 144a²b + 108b²a

(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x

Modifying the given expression as following

(4a)³ – (3b)³ – 3×(4a)²×3b + 3×(3b)²× 4a

(4a – 3b)³

$\boldsymbol{\left ( iv \right )\: 27p^{3}-\frac{1}{216}-\frac{9}{2}p^{2}+\frac{1}{4}p}$

(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x

Modifying the given expression as following

$\boldsymbol{=\left ( 3p \right )^{3}-\left ( \frac{1}{6} \right )^{3}-3\times \left ( 3p \right )^{2}\times \frac{1}{6}+3\times \left ( \frac{1}{6} \right )^{2}\times 3p}$

$\boldsymbol{=\left ( 3p-\frac{1}{6} \right )^{3}}$

Q9.Verify

$\boldsymbol{\left ( i \right )\: x^{3}+y^{3}=\left ( x+y \right )\left ( x^{2}+xy+y^{2} \right )}$

$\boldsymbol{\left ( ii \right )\: x^{3}-y^{3}}\boldsymbol{=\left ( x-y \right )\left ( x^{2}+xy+y^{2} \right )}$

Ans.

$\boldsymbol{\left ( i \right )\: x^{3}+y^{3}=\left ( x+y \right )\left ( x^{2}+xy+y^{2} \right )}$

Using the following identity

$\boldsymbol{\because \left ( x+y \right )^{3}=x^{3}+y^{3}+3xy\left ( x+y \right )}$

$\boldsymbol{\Rightarrow \left ( x+y \right )^{3}-3xy\left ( x+y \right )=x^{3}+y^{3}}$

$\boldsymbol{x^{3}+y^{3}}\boldsymbol{=(x+y)\left [ \left ( x+y \right )^{2} -3xy\right ]}$

$\boldsymbol{x^{3}+y^{3}}\boldsymbol{=\left ( x+y \right )\left ( x^{2}+y^{2}+2xy-3xy \right )}$

$\boldsymbol{x^{3}+y^{3}}\boldsymbol{=\left ( x+y \right )\left ( x^{2}-xy+y^{2} \right )}$

$\boldsymbol{\left ( ii \right )\: x^{3}-y^{3}}\boldsymbol{=\left ( x-y \right )\left ( x^{2}+xy+y^{2} \right )}$

Using the following identity

$\boldsymbol{\left ( x-y \right )^{3}=x^{3}-y^{3}-3xy\left ( x-y \right )}$

$\boldsymbol{\left ( x-y \right )^{3}+3xy\left ( x-y \right )=x^{3}-y^{3}}$

$\boldsymbol{x^{3}-y^{3}}=\boldsymbol{\left ( x-y \right )\left [ \left ( x-y \right )^{2}+3xy \right ]}$

$\boldsymbol{x^{3}-y^{3}}=\boldsymbol{\left ( x-y \right )\left [ x^{2}+y^{2}-2xy+3xy \right ]}$

$\boldsymbol{x^{3}-y^{3}}=\boldsymbol{\left ( x-y \right )( x^{2}+xy+y^{2} )}$

Hence Proved

Q10.Factorise each of the following

$\boldsymbol{\left ( i \right )\: 27y^{3}+125z^{3}}$

$\boldsymbol{\left ( ii \right )\: 64m^{3}-343n^{3}}$

Ans.

$\boldsymbol{\left ( i \right )\: 27y^{3}+125z^{3}}$

$\boldsymbol{\because x^{3}+y^{3}}=\boldsymbol{\left ( x+y \right )( x^{2}-xy+y^{2} )}$

Modifying the given expression according to the identity

$\boldsymbol{=\left ( 3y \right )^{3}+\left ( 5z \right )^{3}}$

$\boldsymbol{\left ( 3y \right )^{3}+\left ( 5z \right )^{3}}=\boldsymbol{\left ( 3y+5z \right )\left [ \left ( 3y \right )^{2}-3y\times 5z+\left ( 5z \right )^{2} \right ]}$

$\boldsymbol{27y^{3}+125z^{3}=\left ( 3y+5z \right )\left ( 9y^{2}-15yz+25z^{2} \right )}$

$\boldsymbol{\left ( ii \right )\: 64m^{3}-343n^{3}}$

$\boldsymbol{\because x^{3}-y^{3}}=\boldsymbol{\left ( x-y \right )( x^{2}+xy+y^{2} )}$

Modifying the given expression according to the identity

$\boldsymbol{64m^{3}-343n^{3}=\left ( 4m \right )^{3}-\left ( 7n \right )^{3}}$

$\boldsymbol{\left ( 4m \right )^{3}-\left ( 7n \right )^{3}=\left ( 4m-7n \right )\left [ \left ( 4m \right )^{2}+\left ( 7n \right )^{2}+28mn \right ]}$

$\boldsymbol{=\left ( 4m-7n \right )\left ( 16m^{2}+28mn+49n^{2} \right )}$

Q11.Factorise $27x^{3}+y^{3}+z^{3}-9xyz$

Using the following identity

$\boldsymbol{\because a^{3}+b^{3}+c^{3}-3abc=\left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2}-ab-bc-ac \right )}$

Modifying the given expression according to the identity

$\boldsymbol{27x^{3}+y^{3}+z^{3}-9xyz=\left ( 3x \right )^{3}+y^{3}+z^{3}-3\times 3x\times y\times z}$

$\boldsymbol{\left ( 3x \right )^{3}+y^{3}+z^{3}-3\times 3x\times y\times z=\left ( 3x+y+z \right )\left [ \left ( 3x \right )^{2}+y^{2}+z^{2}-3x\times y-yz-z\times 3x \right ]}$

$\boldsymbol{27x^{3}+y^{3}+z^{3}-9xy z=\left ( 3x+y+z \right )\left ( 9x^{2}+y^{2}+z^{2}-3x y-yz-3z x \right ]}$

Q12.Verify that

$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}(x+y+z)[(x-y^{2})+(y-z)^{2}+(z-x)^{2}]$

Applying the following identity

$\boldsymbol{x^{3}+y^{3}+z^{3}-3xyz=\left ( x+y+z \right )\left ( x^{2}+y^{2}+z^{2}-xy-yz-zx \right )}$

Dividing and multiplying the RHS by 2

$\boldsymbol{x^{3}+y^{3}+z^{3}-3xyz=\frac{2}{2}\left ( x+y+z \right )\left ( x^{2}+y^{2}+z^{2}-xy-yz-zx \right )}$

$\boldsymbol{=\frac{1}{2}\left ( x+y+z \right )\left (2 x^{2}+2y^{2}+2z^{2}-2xy-2yz-2zx \right )}$

$\boldsymbol{=\frac{1}{2}\left ( x+y+z \right )\left (x^{2}+ x^{2}+y^{2}+y^{2}+z^{2}+z^{2}-2xy-2yz-2zx \right )}$

$\boldsymbol{=\frac{1}{2}\left ( x+y+z \right )\left [ \left ( x-y \right )^{2} +\left ( y-z \right )^{2}+\left ( z-x \right )^{2}\right ]}$

Q13. If x + y + z = 0, show that x³ + y³ + z³ = 3 xyz.
Ans.
Using the following identity

$\boldsymbol{x^{3}+y^{3}+z^{3}-3xyz=\left ( x+y+z \right )\left ( x^{2}+y^{2}+z^{2}-xy-yz-zx \right )}$

Since, x + y + z = 0
x³ + y³ + z³ – 3 xyz = 0

x³ + y³ + z³ = 3 xyz
Hence if x + y + z = 0, then x³ + y³ + z³ = 3 xyz

See the video for Solutions of Question 1 to Question 13

Q14.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

Ans.
(i) We have, (– 12)³ + (7)³ + (5)³
Let x = – 12, y = 7 and z = 5
Then, x + y + z = – 12 + 7 + 5 = 0
We know that if x + y + z = 0, then, x³ + y³ + z³ = 3xyz
∴ (-12)³ + (7)³ + (5)³ = 3×– 12×7×5
= – 1260

(ii) We have, (28)³ + (–15)³ + (–13)³
Let x = 28, y = –15 and z = – 13
Then, x + y + z = 28 – 15 – 13 = 0
We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz
∴ (28)³ + (–15)³ + (–13)³ = 3×28×–15×–13
= 16380

Q15.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

Ans.
Area of a rectangle = Length x Breadth
(i) 25a² – 35a + 12 = 25a² – 20a – 15a + 12

= 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)

Thus, the possible length and breadth are (5a – 3) and (5a – 4).

(ii) 35y²+ 13y -12 = 35y² + 28y – 15y -12
= 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3)
Thus, the possible length and breadth are (7y – 3) and (5y + 4).

Q16.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
Ans.
Volume of a cuboid = Length x Breadth x Height
(i) We have, 3x² – 12x = 3(x² – 4x)
= 3 x x x (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).

(ii) We have, 12ky² + 8ky – 20k
= 4[3ky² + 2ky – 5k] = 4[k(3y² + 2y – 5)]
= 4 x k x (3y² + 2y – 5)
= 4k[3y² – 3y + 5y – 5]
= 4k[3y(y – 1) + 5(y – 1)]
= 4k[(3y + 5) x (y – 1)]
= 4k x (3y + 5) x (y – 1)
Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution

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Class 9 NCERT Solutions of Chapter 2-Polynomial

Class 9 maths NCERT Solutions of the chapter 2-Polynomial

Exercise 2.1

See the video for exercise 2.1 solutions

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

Exercise 2.2

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

EXERCISE-2.4

See the video for Question (4) and Question (5)

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

Exercise 2.5

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

NCERT Solutions for class 9 science

NCERT Solutions for class 10 maths

NCERT Solutions for Class 10 Science

NCERT Solutions for class 11 maths

NCERT solutions for class 12 maths

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Class 9 NCERT Solutions of Chapter 2-Polynomial

Class 9 maths NCERT Solutions of the chapter 2-Polynomial

Exercise 2.1

See the video for exercise 2.1 solutions

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

Exercise 2.2

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

EXERCISE-2.4

See the video for Question (4) and Question (5)

Class 9 Maths NCERT Solutions of Chapter 2-Polynomial

Exercise 2.5

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

NCERT Solutions for class 9 science

NCERT Solutions for class 10 maths

NCERT Solutions for Class 10 Science

NCERT Solutions for class 11 maths

NCERT solutions for class 12 maths

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