NCERT Solutions Class 9 Maths Chapter 2 Polynomial

Chapter 2 of Class 9 Mathematics, "Polynomials," is an essential topic that helps students build a strong math foundation. This chapter covers the basics of polynomials, including their definitions, types, and operations. Understanding polynomials is important because they are the basis for many advanced math concepts studied in higher classes.

The NCERT Solutions for Class 9 Maths Chapter 2 provides clear explanations and step-by-step solutions to textbook exercises, making it easier for students to understand and solve problems related to polynomial expressions, factorisation, and identities. This blog is designed to help students master this chapter, boost their analytical skills, and gain confidence in math as they prepare for exams and future studies.

1.0Download Class 9 Maths Chapter 2 NCERT Solutions PDF Online

Downloading the Polynomials Class 9 PDF will help you understand the curriculum's key concepts. Below is the link to download the NCERT solutions for Class 9 Maths Chapter 2 PDF

2.0NCERT Solutions for Class 9 Maths Chapter 2 Polynomials: Overview

Chapter 2 Polynomials is crucial in the Class 9 Maths syllabus. Students are introduced to the basic ideas behind algebraic expressions and their operations. The NCERT Solutions for this chapter make it simpler for students to comprehend and grasp the concepts by offering thorough explanations, examples, and step-by-step solutions to the exercises. Following are the key topics covered in the Polynomial chapter.

3.0Chapter 2 Polynomials Subtopics of NCERT Class 9 Maths

The following is a list of the subjects addressed in CBSE Class 9 Maths Chapter 2:

• Polynomials in One Variable
• Zeroes of a Polynomial
• Factorisation of Polynomials 
• Remainder Theorem
• Factor Theorem
• Algebraic Identities

4.0Polynomials in NCERT Solutions for Class 9 Maths, Chapter 2: All Exercises

NCERT Solutions for Class 9 Maths, Chapter 2: Polynomials, covers essential topics such as polynomial definitions, types, and operations, providing a solid foundation for understanding more complex mathematical ideas. The solutions to all exercises are designed to guide students step-by-step. By working through these exercises, learners can enhance their problem-solving skills and gain confidence in their mathematical abilities. You must practice the following number of questions under the NCERT solution for class 9, chapter 2.

5.0NCERT Questions with Solutions for Class 9 Maths Chapter 2 - Detailed Solutions

Exercise : 2.1

1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) $4x^2-3x+7$ (ii) ${\mathrm{y}}^2+\sqrt{2}$ (iii) $3\sqrt{\mathrm{t}}+\mathrm{t}\sqrt{2}$ (iv) $y+\frac{2}{y}$ (v) ${\mathrm{x}}^{10}+{\mathrm{y}}^3+{\mathrm{t}}^{50}$ Sol. (i) $4x^2-3x+7$ This expression is a polynomial in one variable x because there is only one variable ( x ) in the expression. (ii) ${\mathrm{y}}^2+\sqrt{2}$ This expression is a polynomial in one variable y because there is only one variable (y) in the expression. (iii) $3\sqrt{\mathrm{t}}+\mathrm{t}\sqrt{2}$ The expression is not a polynomial because in the term $3\sqrt{t}$, the exponent of t is $\frac{1}{2}$, which is not a whole number. (iv) $y+\frac{2}{y}=y+2y^{-1}$ The expression is not a polynomial because exponent of $y$ is $(-1)$ in term $\frac{2}{y}$ which in not a whole number. (v) ${\mathrm{x}}^{10}+{\mathrm{y}}^3+{\mathrm{t}}^{50}$ The expression is not a polynomial in one variable, it is a polynomial in 3 variables $\mathrm{x},\mathrm{y}$ and t .
2. Write the coefficients of $x^2$ in each of the following : (i) $2+x^2+x$ (ii) $2-x^2+x^3$ (iii) $\frac{\pi }{2}{x}^2+x$ (iv) $\sqrt{2}x-1$ Sol. (i) $2+x^2+x$ Coefficient of $x^2=1$ (ii) $2-x^2+x^3$ Coefficient of $x^2=-1$ (iii) $\frac{\pi }{2}{x}^2+x$ Coefficient of $x^2=\pi /2$ (iv) $\sqrt{2}x-1$ Coefficient of $x^2=0$
3. Give one example each of a binomial of degree 35 , and of a monomial of degree 100. Sol. One example of a binomial of degree 35 is $3x^{35}-4$. One example of monomial of degree 100 is $5x^{100}$.
4. Write the degree of each of the following polynomials : (i) $5{\mathrm{x}}^3+4{\mathrm{x}}^2+7\mathrm{x}$ (ii) $4-y^2$ (iii) $5t-\sqrt{7}$ (iv) 3 Sol. (i) $5x^3+4x^2+7x$ Term with the highest power of $x=5x^3$ Exponent of x in this term $=3$ $\therefore$ Degree of this polynomial $=3$. (ii) $4-y^2$ Term with the highest power of $y=-y^2$ Exponent of $y$ in this term $=2$ $\therefore$ Degree of this polynomial $=2$ (iii) $5\mathrm{t}-\sqrt{7}$ Term with highest power of $t=5\mathrm{t}$. Exponent of t in this term $=1$ $\therefore$ Degree of this polynomial $=1$ (iv) 3 This is a constant which is non-zero $\therefore$ Degree of this polynomial $=0$
5. Classify the following as linear, quadratic and cubic polynomials : (i) $x^2+x$ (ii) $\mathrm{x}-{\mathrm{x}}^3$ (iii) $y+y^2+4$ (iv) $1+x$ (v) $3t$ (vi) ${\mathrm{r}}^2$ (vii) $7{\mathrm{x}}^3$ Sol. (i) Quadratic (ii) Cubic (iii) Quadratic (iv) Linear (v) Linear (vi) Quadratic (vii) Cubic

Exercise : 2.2

1. Find the value of the polynomial $5x-4x^2$ +3 at (i) $\mathrm{x}=0$ (ii) $\mathrm{x}=-1$ (iii) $x=2$ Sol. Let $\mathrm{f}(\mathrm{x})=5\mathrm{x}-4{\mathrm{x}}^2+3$ (i) Value of $\mathrm{f}(\mathrm{x})$ at $\mathrm{x}=0=\mathrm{f}(0)$ $=5(0)-4(0{)}^2+3=3$ (ii) Value of $\mathrm{f}(\mathrm{x})$ at $\mathrm{x}=-1=\mathrm{f}(-1)$ $=5(-1)-4(-1{)}^2+3=-5-4+3=-6$ (iii) Value of $\mathrm{f}(\mathrm{x})$ at $\mathrm{x}=2=\mathrm{f}(2)$ $=5(2)-4(2{)}^2+3$ $=10-16+3=-3$
2. Find $p(0),p(1)$ and $p(2)$ for each of the following polynomials : (i) $p(y)=y^2-y+1$ (ii) $\mathrm{p}(\mathrm{t})=2+\mathrm{t}+2{\mathrm{t}}^2-{\mathrm{t}}^3$ (iii) $p(x)=x^3$ (iv) $\mathrm{p}(\mathrm{x})=(\mathrm{x}-1)(\mathrm{x}+1)$ Sol. (i) $p(y)=y^2-y+1$ $\therefore \mathrm{p}(0)=(0{)}^2-(0)+1=1$, $p(1)=(1{)}^2-(1)+1=1$, $p(2)=(2{)}^2-(2)+1=4-2+1=3$. (ii) $\mathrm{p}(\mathrm{t})=2+\mathrm{t}+2{\mathrm{t}}^2-{\mathrm{t}}^3$ $p(0)=2+0+2(0{)}^2-(0{)}^3=2$ $p(1)=2+1+2(1{)}^2-(1{)}^3=2+1+2-1=4$ $p(2)=2+2+2(2{)}^2-(2{)}^3=2+2+8-8=4$ (iii) $\mathrm{p}(\mathrm{x})={\mathrm{x}}^3$ $p(0)=(0{)}^3=0$ $p(1)=(1{)}^3=1$ $p(2)=(2{)}^3=8$ (iv) $\mathrm{p}(\mathrm{x})=(\mathrm{x}-1)(\mathrm{x}+1)$ $p(0)=(0-1)(0+1)=(-1)(1)=-1$ $p(1)=(1-1)(1+1)=0(2)=0$ $p(2)=(2-1)(2+1)=(1)(3)=3$
3. Verify whether the following are zeroes of the polynomial, indicated against them. (i) $\mathrm{p}(\mathrm{x})=3\mathrm{x}+1,\mathrm{x}=-\frac{1}{3}$ (ii) $\mathrm{p}(\mathrm{x})=5\mathrm{x}-\pi ,\mathrm{x}=\frac{4}{5}$ (iii) $\mathrm{p}(\mathrm{x})={\mathrm{x}}^2-1,\mathrm{x}=1,-1$ (iv) $p(x)=(x+1)(x-2),x=-1,2$ (v) $p(x)=x^2,x=0$ (vi) $p(x)=\ell x+m,x=-\frac{m}{\ell }$ (vii) $p(x)=3x^2-1,x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$ (viii) $p(x)=2x+1,x=\frac{1}{2}$ Sol. (i) $\mathrm{p}(\mathrm{x})=3\mathrm{x}+1,\mathrm{x}=-\frac{1}{3}$ $$\begin{gathered} \mathrm{p}\left(-\frac{1}{3}\right)=3\left(-\frac{1}{3}\right)+1=-1+1=0 \\ -\frac{1}{3}\text{ is a zero of }\mathrm{p}(\mathrm{x}) \end{gathered}$$ (ii) $\mathrm{p}(\mathrm{x})=5\mathrm{x}-\pi ,\mathrm{x}=\frac{4}{5}$ $p\left(\frac{4}{5}\right)=5\left(\frac{4}{5}\right)-\pi =4-\pi \ne 0$ $\therefore \frac{4}{5}$ is not a zero of $\mathrm{p}(\mathrm{x})$ (iii) $\mathrm{p}(\mathrm{x})={\mathrm{x}}^2-1,\mathrm{x}=1,-1$ $\mathrm{p}(1)=(1{)}^2-1=1-1=0$ $\mathrm{p}(-1)=(-1{)}^2-1=1-1=0$ $\therefore 1,-1$ are zeroes of $p(x)$ (iv) $p(x)=(x+1)(x-2),x=-1,2$ $p(-1)=(-1+1)(-1-2)=(0)(-3)=0$ $p(2)=(2+1)(2-2)=(3)(0)=0$ $\therefore -1,2$ are zeroes of $p(x)$ (v) $p(x)=x^2,x=0$ $\mathrm{p}(0)=0$ $\therefore 0$ is a zero of $\mathrm{p}(\mathrm{x})$ (vi) $\mathrm{p}(\mathrm{x})=\ell \mathrm{x}=\mathrm{m},\mathrm{x}=\frac{-\mathrm{m}}{\ell }$ $\mathrm{p}\left(\frac{-\mathrm{m}}{\ell }\right)=\ell \left(\frac{-\mathrm{m}}{\ell }\right)+\mathrm{m}=-\mathrm{m}+\mathrm{m}=0$ $\therefore \frac{-\mathrm{m}}{\ell }$ is a zero of $\mathrm{p}(\mathrm{x})$. (vii) $p(x)=3x^2-1,x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$ $p\left(-\frac{1}{\sqrt{3}}\right)=3{\left(-\frac{1}{\sqrt{3}}\right)}^2-1=3\left(\frac{1}{3}\right)-1$ $=1-1=0$ $\mathrm{p}\left(\frac{2}{\sqrt{3}}\right)=3{\left(\frac{2}{\sqrt{3}}\right)}^2-1=3\left(\frac{4}{3}\right)-1$ $=4-1=3\ne 0$ So, $-\frac{1}{\sqrt{3}}$ is a zero of $\mathrm{p}(\mathrm{x})$ and $\frac{2}{\sqrt{3}}$ is not a zero of $p(x)$. (viii) $p(x)=2x+1,x=\frac{1}{2}$ $p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)+1=1+1=2\ne 0$ $\therefore \frac{1}{2}$ is not a zero of $\mathrm{p}(\mathrm{x})$.
4. Find the zero of the polynomial in each of the following cases : (i) $\mathrm{p}(\mathrm{x})=\mathrm{x}+5$ (ii) $p(x)=x-5$ (iii) $p(x)=2x+5$ (iv) $p(x)=3x-2$ (v) $\mathrm{p}(\mathrm{x})=3\mathrm{x}$ (vi) $p(x)=ax,a\ne 0$ (vii) $p(x)=cx+d,c\ne 0,c,d$ are real numbers. Sol. (i) $p(x)=x+5$ $\mathrm{p}(\mathrm{x})=0$ $\Rightarrow \mathrm{x}+5=0\Rightarrow \mathrm{x}=-5$ $\therefore -5$ is zero of the polynomial $\mathrm{p}(\mathrm{x})$. (ii) $\mathrm{p}(\mathrm{x})=\mathrm{x}-5$ $\mathrm{p}(\mathrm{x})=0$ $\mathrm{x}-5=0$ or $x=5$ $\therefore 5$ is zero of polynomial $\mathrm{p}(\mathrm{x})$. (iii) $\mathrm{p}(\mathrm{x})=2\mathrm{x}+5$ $\mathrm{p}(\mathrm{x})=0$ $2x+5=0$ $2\mathrm{x}=-5\Rightarrow \mathrm{x}=-\frac{5}{2}$ $\therefore -\frac{5}{2}$ is zero of polynomial $\mathrm{p}(\mathrm{x})$. (iv) $\mathrm{p}(\mathrm{x})=3\mathrm{x}-2$ $\mathrm{p}(\mathrm{x})=0\Rightarrow 3\mathrm{x}-2=0$ or $\mathrm{x}=\frac{2}{3}$ $\therefore \frac{2}{3}$ is zero of polynomial $\mathrm{p}(\mathrm{x})$. (v) $p(x)=3x$ $\mathrm{p}(\mathrm{x})=0\Rightarrow 3\mathrm{x}=0$ or $x=0$ $\therefore 0$ is zero of polynomial $\mathrm{p}(\mathrm{x})$. (vi) $\mathrm{p}(\mathrm{x})=\mathrm{ax},\mathrm{a}\ne 0$ $\Rightarrow \mathrm{ax}=0$ or $\mathrm{x}=0$ $\therefore 0$ is zero of $\mathrm{p}(\mathrm{x})$ (vii) $p(x)=cx+d,c\ne 0,c,d$ are real numbers $cx+d=0\Rightarrow cx=-d$ $x=\frac{-d}{c}$ $\therefore \frac{-\mathrm{d}}{\mathrm{c}}$ is zero of polynomial $\mathrm{p}(\mathrm{x})$.

Exercise : 2.3

1. Find the remainder when $x^3+3x^2+3x+1$ is divided by : (i) $\mathrm{x}+1$ (ii) $\mathrm{x}-\frac{1}{2}$ (iii) x (iv) $x+\pi$ (v) $5+2x$ Sol. (i) $x+1$ $x+1=0$ $\Rightarrow \mathrm{x}=-1$ $\therefore$ Remainder $=\mathrm{p}(-1)=(-1{)}^3+3(-1{)}^2+3(-$ 1) $+1=-1+3-3+1=0$ (ii) $\mathrm{x}-\frac{1}{2}$ $x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}$ $\therefore$ Remainder $=\mathrm{p}\left(\frac{1}{2}\right)$ $={\left(\frac{1}{2}\right)}^3+3{\left(\frac{1}{2}\right)}^2+3\left(\frac{1}{2}\right)+1=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$ $=\frac{27}{8}$ (iii) x $\mathrm{x}=0$ Remainder $=p(0)$ $=(0{)}^3+3(0{)}^2+3(0)+1=1$ (iv) $\mathrm{x}+\pi$ $\mathrm{x}+\pi =0\Rightarrow \mathrm{x}=-\pi$ $\therefore$ Remainder $=\mathrm{p}(-\pi )$ $=(-\pi {)}^3+3(-\pi {)}^2+3(-\pi )+1$ $=-{\pi }^3+3{\pi }^2-3\pi +1$ (v) $5+2x$ $5+2\mathrm{x}=0\Rightarrow \mathrm{x}=-5/2$ $\therefore$ Remainder $=\mathrm{p}(-5/2)$ $={\left(\frac{-5}{2}\right)}^3+3{\left(\frac{-5}{2}\right)}^2+3\left(\frac{-5}{2}\right)+1$ $=\frac{-125}{8}+\frac{75}{4}-\frac{15}{2}+1=-\frac{27}{8}$
2. Find the remainder when $x^3-a{x}^2+6x-a$ is divided by $\mathrm{x}-\mathrm{a}$. Sol. Let $\mathrm{p}(\mathrm{x})={\mathrm{x}}^3-{\mathrm{ax}}^2+6\mathrm{x}-\mathrm{a}$ $\mathrm{x}-\mathrm{a}=0\Rightarrow \mathrm{x}=\mathrm{a}$ $\therefore$ Remainder $=(a{)}^3-a(a{)}^2+6(a)-a$ $=a^3-a^3+6a-a=5a$
3. Check whether $7+3x$ is a factor of $3x^3+$ 7 x . Sol. $7+3x$ will be a factor of $3x^3+7x$ only if 7 +3 x divides $3{\mathrm{x}}^3+7\mathrm{x}$ leaving 0 as remainder. Let $\mathrm{p}(\mathrm{x})=3{\mathrm{x}}^3+7\mathrm{x}$ $7+3\mathrm{x}=0\Rightarrow 3\mathrm{x}=-7\Rightarrow \mathrm{x}=-7/3$ $\therefore$ Remainder $3{\left(-\frac{7}{3}\right)}^3+7\left(-\frac{7}{3}\right)=\frac{-343}{9}-\frac{49}{3}=\frac{-490}{9}\ne 0$ so, $7+3x$ is not a factor of $3x^3+7x$.

Exercise : 2.4

1. Determine which of the following polynomials has $(x+1)$ a factor : (i) ${\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}+1$ (ii) $x^4+x^3+x^2+x+1$ (iii) ${\mathrm{x}}^4+3{\mathrm{x}}^3+3{\mathrm{x}}^2+\mathrm{x}+1$ (iv) $x^3-x^2-(2+\sqrt{2})x+\sqrt{2}$ Sol. (i) $x^3+x^2+x+1$ Let $p(x)=x^3+x^2+x+1$ The zero of $x+1$ is -1 $\mathrm{p}(-1)=(-1{)}^3+(-1{)}^2+(-1)+1$ $=-1+1-1+1=0$ By Factor theorem $x+1$ is a factor of $p(x)$. (ii) ${\mathrm{x}}^4+{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}+1$ Let $p(x)=x^4+x^3+x^2+x+1$ The zero of $x+1$ is -1 $p(-1)=(-1{)}^4+(-1{)}^3+(-1{)}^2+(-1)+1=1\ne 0$ By Factor theorem $x+1$ is not a factor of $\mathrm{p}(\mathrm{x})$ (iii) $x^4+3x^3+3x^2+x+1$ Let $p(x)=x^4+3x^3+3x^2+x+1$ Zero of $x+1$ is -1 $\mathrm{p}(-1)=(-1{)}^4+3(-1{)}^3+3(-1{)}^2+(-1)+1$ $=1-3+3-1+1=1\ne 0$ By Factor theorem $x+1$ is not a factor of $\mathrm{p}(\mathrm{x})$ (iv) Let $p(x)=x^3-x^2-(2+\sqrt{2})x+\sqrt{2}$ zero of $x+1$ is -1 $p(-1)=(-1{)}^3-(-1{)}^2-(2+\sqrt{2})(-1)+\sqrt{2}$ $=-1-1+2+\sqrt{2}+\sqrt{2}=2\sqrt{2}\ne 0$ By Factor theorem, $x+1$ is not a factor of $\mathrm{p}(\mathrm{x})$.
2. Use the factor theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases : (i) $\mathrm{p}(\mathrm{x})=2{\mathrm{x}}^3+{\mathrm{x}}^2-2\mathrm{x}-1,\mathrm{g}(\mathrm{x})=\mathrm{x}+1$. (ii) $p(x)=x^3+3x^2+3x+1,g(x)=x+2$. (iii) $\mathrm{p}(\mathrm{x})={\mathrm{x}}^3-4{\mathrm{x}}^2+\mathrm{x}+6$; $\mathrm{g}(\mathrm{x})=\mathrm{x}-3$ Sol. (i) $p(x)=2x^3+x^2-2x-1,g(x)=x+1$. $\mathrm{g}(\mathrm{x})=0\Rightarrow \mathrm{x}+1=0\Rightarrow \mathrm{x}=-1$ $\therefore$ Zero of $\mathrm{g}(\mathrm{x})$ is -1 Now, $p(-1)=2(-1{)}^3+(-1{)}^2-2(-1)-1$ $=-2+1+2-1=0$ $\therefore$ By Factor theorem, $\mathrm{g}(\mathrm{x})$ is a factor of $\mathrm{p}(\mathrm{x})$. (ii) Let $\mathrm{p}(\mathrm{x})={\mathrm{x}}^3+3{\mathrm{x}}^2+3\mathrm{x}+1$, $\mathrm{g}(\mathrm{x})=\mathrm{x}+2$ $\mathrm{g}(\mathrm{x})=0\Rightarrow \mathrm{x}+2=0\Rightarrow \mathrm{x}=-2$ $\therefore$ Zero of $\mathrm{g}(\mathrm{x})$ is -2 Now, $p(-2)=(-2{)}^3+3(-2{)}^2+3(-2)+1$ $=-8+12-6+1=-1$ $\therefore$ By Factor theorem, $\mathrm{g}(\mathrm{x})$ is not a factor of $\mathrm{p}(\mathrm{x})$ (iii) $p(x)=x^3-4x^2+x+6,g(x)=x-3$ $\mathrm{g}(\mathrm{x})=0$ $\Rightarrow \mathrm{x}-3=0\Rightarrow \mathrm{x}=3$ $\therefore$ Zero of $\mathrm{g}(\mathrm{x})=3$ Now $p(3)=3^3-4(3{)}^2+3+6$ $=27-36+3+6=0$ $\therefore$ By Factor theorem, $\mathrm{g}(\mathrm{x})$ is a factor of $\mathrm{p}(\mathrm{x})$.
3. Find the value of $k$, if $x-1$ is a factor of $p(x)$ in each of the following cases : (i) $\mathrm{p}(\mathrm{x})={\mathrm{x}}^2+\mathrm{x}+\mathrm{k}$ (ii) $\mathrm{p}(\mathrm{x})=2{\mathrm{x}}^2+\mathrm{kx}+\sqrt{2}$ (iii) $\mathrm{p}(\mathrm{x})={\mathrm{kx}}^2-\sqrt{2}\mathrm{x}+1$ (iv) $p(x)=k{x}^2-3x+k$ Sol. (i) $p(x)=x^2+x+k$ If $x-1$ is a factor of $p(x)$, then $p(1)=0$ $\Rightarrow (1{)}^2+(1)+\mathrm{k}=0$ $\Rightarrow 1+1+\mathrm{k}=0$ $\Rightarrow 2+\mathrm{k}=0$ $\Rightarrow \mathrm{k}=-2$ (ii) $\mathrm{p}(\mathrm{x})=2{\mathrm{x}}^2+\mathrm{kx}+\sqrt{2}$ If $(x-1)$ is a factor of $p(x)$ then $p(1)=0$ $\Rightarrow 2(1{)}^2+\mathrm{k}(1)+\sqrt{2}=0$ $\Rightarrow 2+\mathrm{k}+\sqrt{2}=0$ $\Rightarrow \mathrm{k}=-(2+\sqrt{2})$ (iii) $\mathrm{p}(\mathrm{x})={\mathrm{kx}}^2-\sqrt{2}\mathrm{x}+1$ If $(x-1)$ is a factor of $p(x)$ then $p(1)=0$ $\mathrm{k}(1{)}^2-\sqrt{2}(1)+1=0$ $\Rightarrow \mathrm{k}-\sqrt{2}+1=0$ $\mathrm{k}=\sqrt{2}-1$ (iv) $\mathrm{p}(\mathrm{x})={\mathrm{kx}}^2-3\mathrm{x}+\mathrm{k}$ If $(x-1)$ is a factor of $p(x)$ then $p(1)=0$ $\Rightarrow \mathrm{k}(1{)}^2-3(1)+\mathrm{k}=0$ $2\mathrm{k}=3$ $\mathrm{k}=3/2$
4. Factorise : (i) $12{\mathrm{x}}^2-7\mathrm{x}+1$ (ii) $2{\mathrm{x}}^2+7\mathrm{x}+3$ (iii) $6x^2+5x-6$ (iv) $3x^2-x-4$ Sol. (i) $12x^2-7x+1$ $$\begin{gathered} =12x^2-4x-3x+1 \\ =4x(3x-1)-1(3x-1) \\ =(3x-1)(4x-1) \end{gathered}$$ (ii) $2x^2+7x+3$ $=2x^2+6x+x+3$ $=2x(x+3)+1(x+3)$ $=(\mathrm{x}+3)(2\mathrm{x}+1)$ (iii) $6x^2+5x-6=6x^2+9x-4x-6$ $=3x(2x+3)-2(2x+3)$ $=(3x-2)(2x+3)$ (iv) $3x^2-x-4=3x^2-4x+3x-4$ $=x(3x-4)+1(3x-4)$ $=(x+1)(3x-4)$
5. Factorise : (i) $x^3-2x^2-x+2$ (ii) $x^3-3x^2-9x-5$ (iii) ${\mathrm{x}}^3+13{\mathrm{x}}^2+32\mathrm{x}+20$ (iv) $2y^3+y^2-2y-1$ Sol. (i) $x^3-2x^2-x+2$ Let $p(x)=x^3-2x^2-x+2$ By trial, we find that $p(1)=(1{)}^3-2(1{)}^2-(1)+2$ $=1-2-1+2=0$ $\therefore$ By Factor Theorem, $(x-1)$ is a factor of $\mathrm{p}(\mathrm{x})$. Now, $x^3-2x^2-x+2$ $=x^2(x-1)-x(x-1)-2(x-1)$ $=(x-1)\left(x^2-x-2\right)$ $=(x-1)\left(x^2-2x+x-2\right)$ $=(x-1)\{x(x-2)+1(x-2)\}$ $=(x-1)(x-2)(x+1)$ (ii) $x^3-3x^2-9x-5$ Let $p(x)=x^3-3x^2-9x-5$ By trial, we find that $\mathrm{p}(-1)=(-1{)}^3-3(-1{)}^2-9(-1)-5$ $=-1-3+9-5=0$ $\therefore$ By Factor Theorem, $\mathrm{x}=-1$ or $\mathrm{x}+1$ is factor of $p(x)$. Now, $x^3-3x^2-9x-5$ $=x^2(x+1)-4x(x+1)-5(x+1)$ $=(x+1)\left(x^2-4x-5\right)$ $=(x+1)\left(x^2-5x+x-5\right)$ $=(x+1)\{x(x-5)+1(x-5)\}$ $=(x+1{)}^2(x-5)$ (iii) ${\mathrm{x}}^3+13{\mathrm{x}}^2+32\mathrm{x}+20$ Let $p(x)=x^3+13x^2+32x+20$ By trial, we find that $p(-1)=(-1{)}^3+13(-1{)}^2+32(-1)+20$ $=-1+13-32+20=0$ $\therefore$ By Factor Theorem, $\mathrm{x}=-1$ or $\mathrm{x}+1$ is a factor of p ( x ) $x^3+13x^2+32x+20$ $=x^2(x+1)+12(x)(x+1)+20(x+1)$ $=(x+1)\left(x^2+12x+20\right)$ $=(x+1)\left(x^2+2x+10x+20\right)$ $=(x+1)\{x(x+2)+10(x+2)\}$ $=(x+1)(x+2)(x+10)$ (iv) $2y^3+y^2-2y-1$ $p(y)=2y^3+y^2-2y-1$ By trial, we find that $p(1)=2(1{)}^3+(1{)}^2-2(1)-1=0$ $\therefore$ By Factor Theorem, $(y-1)$ is a factor of $p(y)=2y^3+y^2-2y-1$ $=2y^2(y-1)+3y(y-1)+1(y-1)$ $=(y-1)\left(2y^2+3y+1\right)$ $=(y-1)\left(2y^2+2y+y+1\right)$ $=(y-1)\{2y(y+1)+1(y+1)\}$ $=(y-1)(2y+1)(y+1)$

Exercise : 2.5

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) $\left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right)$ (v) $(3-2x)(3+2x)$ Sol. (i) $(x+4)(x+10)$ $=x^2+(4+10)x+(4)(10)$ $=x^2+14x+40$ (ii) $(x+8)(x-10)$ $=(x+8)\{x+(-10)\}$ $=x^2+\{8+(-10)\}x+8(-10)$ $=x^2-2x-80$ (iii) $(3x+4)(3x-5)$ $=(3x+4)(3x-5)=(3x+4)(3x+(-5))$ $=(3\mathrm{x}{)}^2+\{4+(-5)\}(3\mathrm{x})+4(-5)$ $=9x^2-3x-20$ (iv) $\left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right)$ Let, ${\mathrm{y}}^2=\mathrm{x}$ $$\begin{gathered} \Rightarrow \left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right)=\left(x+\frac{3}{2}\right)\left(x-\frac{3}{2}\right) \\ =x^2-\frac{9}{4} \end{gathered}$$ [using identity $(a+b)(a-b)=a^2-b^2$ ] Substituting $x=y^2$, we get $={\left(y^2\right)}^2-\frac{9}{4}$ $={\mathrm{y}}^4-\frac{9}{4}$ (v) $(3-2x)(3+2x)$ $(3{)}^2-(2x{)}^2=9-4x^2$ [using identity $(a+b)(a-b)=a^2-b^2$ ]
2. Evaluate the following products without multiplying directly: (i) $103\times 107$ (ii) $95\times 96$ (iii) $104\times 96$ Sol. (i) $103\times 107=(100+3)\times (100+7)$ $=(100{)}^2+(3+7)(100)+(3)(7)$ $=10000+1000+21=11021$ Alternate solution: $103\times 107=(105-2)\times (105+2)$ $=(105{)}^2-(2{)}^2=(100+5{)}^2-4$ $=(100{)}^2+2(100)(5)+(5{)}^2-4$ $=10000+1000+25-4$ $=11021$. (ii) $95\times 96$ $=(90+5)\times (90+6)$ $=(90{)}^2+(5+6)90+(5)(6)$ $=8100+990+30=9120$ (iii) $104\times 96$ $=(100+4)\times (100-4)$ [using identity $(a+b)(a-b)=a^2-b^2$ ] $=(100{)}^2-(4{)}^2=10000-16$ $=9984$
3. Factorise the following using appropriate identities: (i) $9x^2+6xy+y^2$ (ii) $4y^2-4y+1$ (iii) ${\mathrm{x}}^2-\frac{{\mathrm{y}}^2}{100}$ Sol. (i) $9x^2+6xy+y^2$ $=(3x{)}^2+2(3x)(y)+(y{)}^2$ $=(3x+y{)}^2$ $=(3x+y)(3x+y)$ (ii) $4y^2-4y+1$ $=(2y{)}^2-2(2y)(1)+(1{)}^2$ $=(2y-1{)}^2=(2y-1)(2y-1)$ (iii) $x^2-\frac{y^2}{100}$ [using identity $(a+b)(a-b)=a^2-b^2$ ] $x^2-{\left(\frac{y}{10}\right)}^2=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)$
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) ${\left(\frac{1}{4}a-\frac{1}{2}b+1\right)}^2$ Sol. (i) $(x+2y+4z{)}^2=(x{)}^2+(2y{)}^2+(4z{)}^2+$ $2(x)(2y)+2(2y)(4z)+2(4z)(x)$ $=x^2+4y^2+16z^2+4xy+16yz+8zx$ (ii) $(2x-y+z{)}^2$ $=(2x-y+z)(2x-y+z)$ $=(2x{)}^2+(-y{)}^2+(z{)}^2+2(2x)(-y)+2(-$ y) $(\mathrm{z})+2(\mathrm{z})(2\mathrm{x})$ $=4x^2+y^2+z^2-4xy-2yz+4zx$ (iii) $(-2x+3y+2z{)}^2$ $=(-2x{)}^2+(3y{)}^2+(2z{)}^2+2(-2x)(3y)+$ $2(-2x)(2z)+2(3y)(2z)$ $=4x^2+9y^2+4z^2-12xy-8xz+12yz$ (iv) $(3a-7b-c{)}^2=(3a-7b-c)(3a-7b-c)$ $=(3a{)}^2+(-7b{)}^2+(-c{)}^2+2(3a)(-7b)+$ $2(3a)(-c)+2(-7b)(-c)$ $=9a^2+49b^2+c^2-42ab-6ac+14bc$ (v) $(-2x+5y-3z{)}^2$ $=(-2x+5y-3z)(-2x+5y-3z)$ $=(-2x{)}^2+(5y{)}^2+(-3z{)}^2+2(-2x)(5y)+$ $2(-2x)(-3z)+2(-3z)(5y)$ $=4x^2+25y^2+9z^2-20xy+12xz-30yz$ (vi) ${\left(\frac{1}{4}a-\frac{1}{2}b+1\right)}^2$ $=\left(\frac{1}{4}\mathrm{a}-\frac{1}{2}\mathrm{b}+1\right)\left(\frac{1}{4}\mathrm{a}-\frac{1}{2}\mathrm{b}+1\right)$ $={\left(\frac{1}{4}\mathrm{a}\right)}^2+{\left(-\frac{1}{2}\mathrm{b}\right)}^2+(1{)}^2+2\left(\frac{1}{4}\mathrm{a}\right)\left(-\frac{1}{2}\mathrm{b}\right)$ $+2\left(\frac{1}{4}\mathrm{a}\right)(1{)}^2+2\left(-\frac{1}{2}\mathrm{b}\right)(1)$ $=\frac{1}{16}{\mathrm{a}}^2+\frac{1}{4}{\mathrm{b}}^2+1-\frac{1}{4}\mathrm{ab}-\mathrm{b}+\frac{1}{2}\mathrm{a}$
5. Factorise : (i) $4x^2+9y^2+16z^2+12xy-24yz-16xz$ (ii) $2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz$ Sol. (i) $4x^2+9y^2+16z^2+12xy-24yz-16xz$ $=(2x{)}^2+(3y{)}^2+(-4z{)}^2+2(2x)(3y)+$ $2(3y)(-4z)+2(-4z)(2x)$ $=\{2\mathrm{x}+3\mathrm{y}+(-4\mathrm{z}){\}}^2=(2\mathrm{x}+3\mathrm{y}-4\mathrm{z}{)}^2$ $=(2x+3y-4z)(2x+3y-4z)$ (ii) $2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz$ $=(-\sqrt{2}x{)}^2+y^2+(2\sqrt{2}z{)}^2+2(-\sqrt{2}x)y+$ $2y(2\sqrt{2}z)+2(2\sqrt{2}z)(-\sqrt{2}x)$ $=(-\sqrt{2}x+y+2\sqrt{2}z{)}^2$
6. Write the following cubes in expanded form : (i) $(2x+1{)}^3$ (ii) $(2a-3b{)}^3$ (iii) ${\left(\frac{3}{2}x+1\right)}^3$ (iv) ${\left(x-\frac{2}{3}y\right)}^3$ Sol. (i) $(2x+1{)}^3=(2x{)}^3+(1{)}^3+3(2x)(1)(2x+1)$ $=8x^3+1+6x(2x+1)$ $=8{\mathrm{x}}^3+1+12{\mathrm{x}}^2+6\mathrm{x}$ $=8{\mathrm{x}}^3+12{\mathrm{x}}^2+6\mathrm{x}+1$ (ii) $(2a-3b{)}^3=(2a{)}^3-(3b{)}^3-3(2a)(3b)(2a-3b)$ $=8a^3-27b^3-18ab(2a-3b)$ $=8a^3-27b^3-36a^{2}b+54a{b}^2$ (iii) ${\left(\frac{3}{2}\mathrm{x}+1\right)}^3={\left(\frac{3}{2}\mathrm{x}\right)}^3+(1{)}^3+3\left(\frac{3}{2}\mathrm{x}\right)(1)\left(\frac{3}{2}\mathrm{x}+1\right)$ $=\frac{27}{8}{x}^3+1+\frac{27}{4}{x}^2+\frac{9}{2}x$ $=\frac{27}{8}{x}^3+\frac{27}{4}{x}^2+\frac{9}{2}x+1$ (iv) ${\left(x-\frac{2}{3}y\right)}^3=x^3-{\left(\frac{2}{3}y\right)}^3-3x\left(\frac{2}{3}y\right)\left(x-\frac{2}{3}y\right)$ $={\mathrm{x}}^3-\frac{8}{27}{\mathrm{y}}^3-2\mathrm{xy}\left(\mathrm{x}-\frac{2}{3}\mathrm{y}\right)$ $=x^3-\frac{8}{27}{y}^3-2x^{2}y+\frac{4}{3}{x}^2$
7. Evaluate the following using suitable identities : (i) $(99{)}^3$ (ii) $(102{)}^3$ (iii) $(998{)}^3$ Sol. (i) $(99{)}^3=(100-1{)}^3$ $$\begin{gathered} =(100{)}^3-(1{)}^3-3(100)(1)(100-1) \\ =1000000-1-300(100-1) \\ =1000000-1-30000+300 \\ =970299 \end{gathered}$$ (ii) $(102{)}^3=(100+2{)}^3$ $$\begin{gathered} =(100{)}^3+(2{)}^3+3(100)(2)(100+2) \\ =1000000+8+600(100+2) \\ =1000000+8+60000+1200 \\ =1061208 \end{gathered}$$ (iii) $(998{)}^3=(1000-2{)}^3$ $$\begin{gathered} =(1000{)}^3-(2{)}^3-3(1000)(2)(1000-2) \\ =1000000000-8-6000(1000-2) \\ =994011992 \end{gathered}$$
8. Factorise each of the following : (i) $8a^3+b^3+12a^{2}b+6a{b}^2$ (ii) $8a^3-b^3-12a^{2}b+6a{b}^2$ (iii) $27-125a^3-135a+225a^2$ (iv) $64a^3-27b^3-144a^{2}b+108a^2$ (v) $27p^3-\frac{1}{216}-\frac{9}{2}{p}^2+\frac{1}{4}p$ Sol. (i) $8a^3+b^3+12a^{2}b+6a{b}^2$ $$\begin{gathered} =(2a{)}^3+(b{)}^3+3(2a)(b)(2a+b) \\ =(2a+b{)}^3=(2a+b)(2a+b)(2a+b) \end{gathered}$$ (ii) $8a^3-b^3-12a^{2}b+6a{b}^2$ $=(2a{)}^3+(-b{)}^3+3(2a{)}^2(-b)+3(2a)(-b{)}^2$ $=(2a-b{)}^3$ (iii) $27-125a^3-135a+225a^2$ $=3^3-(5a{)}^3-3(3)(5a)(3-5a)$ $=(3-5a{)}^3$ (iv) $64a^3-27b^3-144a^{2}b+180a{b}^2$ $=(4a{)}^3-(3b{)}^3-3(4a)(3b)(4a-3b)$ $=(4a-3b{)}^3$ (v) $27p^3-\frac{1}{216}-\frac{9}{2}{p}^2+\frac{1}{4p}$ $=(3p{)}^3-{\left(\frac{1}{6}\right)}^3-3(3p)\left(\frac{1}{6}\right)\left(3p-\frac{1}{6}\right)$ $={\left(3p-\frac{1}{6}\right)}^3=\left(3p-\frac{1}{6}\right)\left(3p-\frac{1}{6}\right)\left(3p-\frac{1}{6}\right)$
9. Verify: (i) $x^3+y^3=(x+y)\left(x^2-xy+y^2\right)$ (ii) $x^3-y^3=(x-y)\left(x^2+xy+y^2\right)$ Sol. (i) $(x+y{)}^3=x^3+y^3+3xy(x+y)$ $\Rightarrow x^3+y^3=(x+y{)}^3-3xy(x+y)$ $\Rightarrow x^3+y^3=(x+y)\left\{(x+y{)}^2-3xy\right\}$ $\Rightarrow x^3+y^3=(x+y)\left(x^2+2xy+y^2-3xy\right)$ $\Rightarrow x^3+y^3=(x+y)\left(x^2-xy+y^2\right)$ (ii) $(x-y{)}^3=x^3-y^3-3xy(x-y)$ $\Rightarrow x^3-y^3=(x-y{)}^3+3xy(x-y)$ $\Rightarrow x^3-y^3=(x-y)\left[(x-y{)}^2+3xy\right]$ $\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2-2xy+3xy\right)$ $\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2+xy\right)$
10. Factorise each of the following : (i) $27y^3+125z^3$ (ii) $64m^3-343n^3$ Sol. (i) $27y^3+125z^3=(3y{)}^3+(5z{)}^3$ $$\begin{gathered} =(3y+5z)\left\{(3y{)}^2-(3y)(5z)+(5z{)}^2\right\} \\ =(3y+5z)\left(9y^2-15yz+25z^2\right) \end{gathered}$$ (ii) $64m^3-343n^3$ $=(4m{)}^3-(7n{)}^3$ $=(4m-7n)\left\{16m^2+4m.7n+(7n{)}^2\right\}$ $=(4m-7n)\left(16m^2+28mn+49n^2\right)$
11. Factorise: $27x^3+y^3+z^3-9xyz$ Sol. $27x^3+y^3+z^3-9xyz$ $=(3x{)}^3+(y{)}^3+(z{)}^3-3(3x)(y)(z)$ $=(3x+y+z)((3x{)}^2+(y{)}^2+(z{)}^2-(3x)-$ (y) $(z)-(z)(3x))$ $=(3x+y+z)\left(9x^2+y^2+z^2-3xy-yz-3zx\right)$
12. Verify that $x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)$ $\left[(x-y{)}^2+(y-z{)}^2+(z-x{)}^2\right]$ Sol. $\frac{1}{2}(x+y+z)\left[(x-y{)}^2+(y-z{)}^2+(z-x{)}^2\right]$ $=\frac{1}{2}(x+y+z)[\left(x^2-2xy+y^2\right)+(y^2-2yz$ $+z^2)+\left(z^2-2zx+x^2\right)]$ $=\frac{1}{2}(x+y+z)\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)$ $=\frac{1}{2}(x+y+z)2\left(x^2+y^2+z^2-xy-yz-zx\right)$ $=\frac{1}{2}(x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)$ $=x^3+y^3+z^3-3xyz$
13. If $x+y+z=0$, show that $x^3+y^3+z^3=3xyz$ Sol. We know $x^3+y^3+z^3-3xyz$ $=(x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)$ Given: $x+y+z=0$ $=(0)\left(x^2+y^2+z^2-xy-yz-zx\right)=0$ or $x^3+y^3+z^3=3xyz$
14. Without actually calculating the cubes, find the value of each of the following : (i) $(-12{)}^3+(7{)}^3+(5{)}^3$ (ii) $(28{)}^3+(-15{)}^3+(-13{)}^3$ Sol. (i) $(-12{)}^3+(7{)}^3+(5{)}^3$ $-12+7+5=0$ $(-12{)}^3+(7{)}^3+(5{)}^3$ $=3(-12)(7)(5)=-1260$ [using identity] if $\mathrm{a}+\mathrm{b}+\mathrm{c}=0\Rightarrow {\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3=3\mathrm{abc}$ (ii) $(28{)}^3+(-15{)}^3+(-13{)}^3$ $28-15-13=0$ $(28{)}^3+(-15{)}^3+(-13{)}^3$ $=3(28)(-15)(-13)=16380$ [using identity] if $\mathrm{a}+\mathrm{b}+\mathrm{c}=0\Rightarrow {\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3=3\mathrm{abc}$
15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given : (i) Area $=25a^2-35a+12$ (ii) Area $=35y^2+13y-12$ Sol. (i) Area $=25a^2-35a+12$ $$\begin{gathered} =25a^2-20a-15a+12 \\ =5a(5a-4)-3(5a-4) \\ =(5a-3)(5a-4) \end{gathered}$$ $\text{ Here, Length }=5a-3\text{, Breadth }=5a-4$ (ii) $35y^2+13y-12$ $=35y^2+28y-15y-12$ $=7y(5y+4)-3(5y+4)$ $=(5y+4)(7y-3)$ Here, Length $=5y+4$, Breadth $=7y-3$.
16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume: $3{\mathrm{x}}^2-12\mathrm{x}$ (ii) Volume: $12{\mathrm{ky}}^2+8\mathrm{xy}-20\mathrm{k}$ Sol. (i) Volume $=3x^2-12x$ $=3x(x-4)=3\times x\times (x-4)$ $\therefore$ Dimensions are 3 units, x -units and ( $x-4$ ) units (ii) $12{\mathrm{ky}}^2+8\mathrm{ky}-20\mathrm{k}$ $=4\mathrm{k}\left(3{\mathrm{y}}^2+2\mathrm{y}-5\right)=4\mathrm{k}\left(3{\mathrm{y}}^2+5\mathrm{y}-3\mathrm{y}-5\right)$ $=4\mathrm{k}\{\mathrm{y}(3\mathrm{y}+5)-1(3\mathrm{y}+5)\}$ $=4\mathrm{k}(3\mathrm{y}+5)(\mathrm{y}-1)$ $\therefore$ Dimensions of cuboid are $4\mathrm{k},3\mathrm{y}+5,\mathrm{y}-1$