Polynomials

Class 9 NCERT Maths

NCERT

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$$\\$ (ii)$y^2+\sqrt{2}$$\\$ (iii)$3\sqrt{t}+t\sqrt{2}$$\\$ (iv)$y+\dfrac{2}{y}$$\\$ (v)$y+2y^{-1}$

Solution :

(i)$4x^2-3x+7$$\\$ One variable is involved in given polynomial which is ‘x’ Therefore, it is a polynomial in one variable ‘x’.$\\$ (ii)$y^2+\sqrt{2}$$\\$ One variable is involved in given polynomial which is ‘y’ Therefore, it is a polynomial in one variable ‘y’. $\\$ (iii)$3\sqrt{t}+t\sqrt{2}$$\\$ No. It can be observed that the exponent of variable t in term $3\sqrt{t} \ is \ \dfrac{1}{2}$,which is nota whole number. Therefore, this expression is not a polynomial.$\\$ (iv)$y+\dfrac{2}{y}$$\\$ $=y+2y^{-1}$$\\$ The power of variable ‘y’ is -1 which is not a whole number. Therefore, it is not a polynomial in one variable$\\$ No. It can be observed that the exponent of variable y in term $\dfrac{2}{y}$ is $-1$ which is not a whole number. Therefore, this expression is not a polynomial.$\\$ (v)$x^{10}+y^3+t^{50}$$\\$ In the given expression there are 3 variables which are ‘x, y, t’ involved.$\\$ Therefore, it is not a polynomial in one variable.

2   Write the coefficients of $x^2$ in each of the following:$\\$ (i)$2+x^2+x$$\\$ (ii)$2-x^2+x^3$$\\$ (iii)$\dfrac{\pi}{2}x^2+x$$\\$ (iv)$\sqrt{2x}-1$

Solution :

(i)$2+x^2+x\\ =2+1(x^2)+x$$\\$ The coefficient of $x^2$ is $1$.$\\$ (ii)$2-x^2+x^3 =2-1(x^2)+x$$\\$ The coefficient of $x^2$ is $-1$.$\\$ (iii)$\dfrac{\pi}{2}x^2+x$$\\$ The coefficient $x^2$ of is $\dfrac{pi}{2}$$\\$ (iv)$\sqrt{2x}-1 =0x^2+\sqrt{2x}-1$$\\$ The coefficient of $x^2$ is $0$

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

Solution :

Binomial of degree $35$ means a polynomial is having$\\$ 1. Two terms$\\$ 2. Highest degree is $35$$\\$ Example: $x^{35}+x^{34}$$\\$ Monomial of degree $100$ means a polynomial is having $\\$ 1. One term$\\$ 2. Highest degree is $100 $$\\$ Example : $x^{100}$ .

4   Write the degree of each of the following polynomials:$\\$ (i)$5x^3+4x^2+7x$$\\$ (ii)$4-y^2$$\\$ (iii)$5t-\sqrt{7}$$\\$ (iv)$3$

Solution :

Degree of a polynomial is the highest power of the variable in the polynomial.$\\$ (i)$5x^3+4x^2+7x$$\\$ Highest power of variable ‘x’ is $3$. Therefore, the degree of this polynomial is $3$$\\$ (ii)$4-y^2$$\\$ Highest power of variable ‘y’ is $2$. Therefore, the degree of this polynomial is $2$.$\\$ (iii)$5t-\sqrt{7}$$\\$ Highest power of variable ‘t’ is $1$. Therefore, the degree of this polynomial is $1$.$\\$ (iv)$3$$\\$ This is a constant polynomial. Degree of a constant polynomial is always $0$.

5   Classify the following as linear, quadratic and cubic polynomial:$\\$ (i)$x^2+x$$\\$ (ii)$x-x^3$$\\$ (iii)$y+y^2+4$$\\$ (iv)$1+x$$\\$ (v)$3t$$\\$ (vi)$r^2$$\\$ (vii)$7x^2 7x^3$$\\$

Solution :

Linear polynomial - whose variable power is ‘1’$\\$ Quadratic polynomial - whose variable highest power is ‘2’$\\$ Cubic polynomial- whose variable highest power is ‘3’$\\$ (i) $x^ 2 + x$ is a quadratic polynomial as its highest degree is $2$.$\\$ (ii) $x - x^ 3$ is a cubic polynomial as its highest degree is $3.$$\\$ (iii) $y + y^2 + 4$ is a quadratic polynomial as its highest degree is $2.$$\\$ (iv) $1 + x$ is a linear polynomial as its degree is $1.$$\\$ (v)$ 3t$ is a linear polynomial as its degree is $1$.$\\$ (vi) $r^2$ is a quadratic polynomial as its degree is $2.$$\\$ (vii) $7x^2 7x^3$ is a cubic polynomial as highest its degree is $3.$

6   Find the value of the polynomial at $5 x - 4 x ^2 + 3$ at $\\$ (i) $x = 0$$\\$ (ii) $x = -1$$\\$ (iii) $x = 2$

Solution :

(i) $p(x) = 5x - 4x^ 2 + 3$$\\$ $p(0) = 5(0) - 4(0) ^2 +3 = 3$$\\$ (ii)$p(x) = 5x - 4x ^2 + 3$$\\$ $p(-1) = 5(-1) - 4(-1)^ 2 + 3$$\\$ $= -5 - 4(1) + 3 =-6$$\\$ (iii)$p(x) = 5x - 4x^ 2 + 3$$\\$ $p(2) = 5(2) - 4(2)^ 2 + 3 = 10 - 16 + 3 =-3$$\\$

7   Find $p(0), p(1)$ and $p(2)$ for each of the following polynomials:$\\$ (i)$p(y) = y ^2 - y + 1$$\\$ (ii)$p(t) = 2 + t + 2t ^2 - t3$$\\$ (iii)$p(x) = x^ 3$$\\$ (iv)$p(x) = (x - 1) (x + 1)$$\\$

Solution :

$\\$$(i) p(y) = y^ 2 - y + 1\\ \bullet p(0) = (0) ^2 - (0) + 1 = 1\\ \bullet p(1) = (1) ^2 - (1) + 1 = 1 - 1 + 1 = 1\\ \bullet p(2) = (2) ^2 - (2) + 1 = 4 -2 +1 = 3\\ \\ \\ (ii) p(t) = 2 + t + 2t ^2 - t ^3\\ \bullet p(0) = 2 + 0 + 2 (0) ^2 - (0) ^3 = 2\\ \bullet p(1) = 2 + (1) + 2(1) ^2 - (1) ^3 = 2 + 1 + 2 - 1 = 4\\ \bullet p(2) = 2 + 2 + 2(2) ^2 - (2)^ 3\\ = 2 + 2 + 8 -8 = 4\\ \\ \\ (iii) p(x) = x^ 3\\ \bullet p(0) = (0)^ 3 = 0\\ \bullet p(1) = (1) ^3 = 1\\ \bullet p(2) = (2) ^3 = 8\\ \\ \\ (v)p(x) = (x -1) (x + 1)\\ \bullet p(0) = (0 -1) (0 + 1) = (- 1) (1) = - 1\\ \bullet p(1) = (1 - 1) (1 + 1) = 0 (2) = 0\\ \bullet p(2) = (2 -1 ) (2 + 1) = 1(3) = 3$$\\$

8   Verify whether the following are zeroes of the polynomial, indicated against them.$\\$ (i) $p ( x )= 3x+1,x=-\dfrac{1}{3}$$\\$ (ii)$p(x)=5x-\pi ,x=\dfrac{4}{5}$$\\$ (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$$\\$ (vi)$p(x)=lm+m,x=-\dfrac{m}{l}$$\\$ (vii)$p(x)=3x^2-1,x=-\dfrac{1}{\sqrt{3}},\dfrac{2}{\sqrt{3}}$$\\$ (viii)$p(x)=2x+1,x=\dfrac{1}{2}$$\\$

Solution :

(i) If $x =-\dfrac{1}{3}$ is a zero of given polynomial $p(x)=3x+1,$ then $ p(-\dfrac{1}{3}) $ should be $0$.$\\$ Here,$p(-\dfrac{1}{3})=3(-\dfrac{1}{3})+1=-1+1=0$$\\$ Therefore, $x =-\dfrac{1}{3}$ is a zero of given polynomial $\\$ (ii)If $x=\dfrac{4}{5}$ is a zero of polynomial $p(x)=5x-\pi $, then $ p(\dfrac{4}{5})$ should be $ 0$$\\$ Here,$p(\dfrac{4}{5})=5(\dfrac{4}{5})-\pi =4-\pi$$\\$ As$ p(\dfrac{4}{5})\neq 0$$\\$ Therefore, $ x=\dfrac{4}{5}$ is not a zero of the given polynomial.

(iii) If $x = 1$ and $ x = -1 $ are zeroes of polynomial $ p(x) = x^ 2 - 1,$ then $p(1)$ and $p(-1)$should be $0.$$\\$ Here, $p(1) = (1)^ 2 - 1 = 0$, and $p(-1) = (-1) ^2 - 1 = 0$$\\$ Hence, $x = 1$ and $-1$ are zeroes of the given polynomial.$\\$ (iv) If $x = -1$ and $x = 2$ are zeroes of polynomial $p(x) = (x +1) (x - 2),$ then $p(-1)$ and $p(2)$should be $0.$$\\$ Here, $p(-1) = (-1 + 1) (-1 - 2) = 0(-3) = 0$, and $p(2) = (2 + 1) (2 - 2) = 3(0) = 0$$\\$ Therefore, $x = -1$ and $x = 2$ are zeroes of the given polynomial.$\\$ (v) If $x = 0$ is a zero of polynomial $p(x) = x ^2 $, then $p(0)$ should be zero.$\\$ Here, $p(0) = (0)^ 2 = 0$$\\$ Hence,$ x = 0$ is a zero of the given polynomial.

(vi) If $ p(\dfrac{-m}{l})$ is a zero of polynomial $p(x) = lx + m,$ then $p(\dfrac{-m}{l})$ should be $0$.$\\$ Here,$p(\dfrac{-m}{l})=l(\dfrac{-m}{l})+m=-m+m=0$$\\$ Therefore , $x=\dfrac{-m}{l}$ is a zero of the given polynomial.$\\$ (vii)If$ x=\dfrac{-1}{\sqrt{3}}$ and $ x=\dfrac{2}{\sqrt{3}}$ are zeroes of polynomial $p(x) = 3x 2 - 1$, then $\\$ $p(\dfrac{-1}{\sqrt{3}})$ and $ p(\dfrac{2}{\sqrt{3}})$ should be $ 0$.$\\$ Here,$ p(\dfrac{-1}{\sqrt{3}})=3(\dfrac{-1}{\sqrt{3}})^2-1=3(\dfrac{1}{3})-1=1-1=0,$ and$\\$ $p(\dfrac{2}{\sqrt{3}})=3(\dfrac{2}{\sqrt{3}})^2-1=3(\dfrac{4}{3})-1=4-1=3$$\\$ Hence,$ x=\dfrac{2}{\sqrt{3}}$ is a zero of the given polynomial.$\\$ However,$x=\dfrac{2}{\sqrt{3}} $ is not a zero of the given polynomial.$\\$ (viii) If $ x=\dfrac{1}{2}$ is a zero of polynomial $p(x) = 2x + 1,$ then $p(\dfrac{1}{2})$ should be $0$.$\\$ Here,$ p(\dfrac{1}{2})=2(\dfrac{1}{2})+1=1+1=2$$\\$ As $ p(\dfrac{1}{2})\neq 0,$$\\$ Therefore ,$ x=\dfrac{1}{2} $ is not a zero of the given polynomial.

9   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$$\\$ (vi) $p(x) = ax, a \neq 0$$\\$ (vii) $p(x) = cx + d, c \neq 0, c, d$ are real numbers.

Solution :

Zero of a polynomial is that value of the variable at which the value of thepolynomial is obtained as $ 0.$$\\$ (i)$ p(x) = x + 5$$\\$ Let $p(x) = 0$$\\$ $x + 5 = 0$$\\$ $x = -5$$\\$ Therefore, for $x = -5,$ the value of the polynomial is $0$ and hence, $x = -5$ is a zero of the given polynomial.$\\$ (ii) $p(x) = x - 5$$\\$ Let $p(x) = 0$$\\$ $x - 5 = 0$$\\$ $x = 5$$\\$ Therefore, for $x = 5$, the value of the polynomial is $0$ and hence,$x = 5$ is a zero of the given polynomial.$\\$ (iii) $p(x) = 2x + 5$$\\$ Let $p(x) = 0$$\\$ $2x + 5 = 0$$\\$ $2x = - 5$$\\$ $x=-\dfrac{5}{2}$$\\$ Therefore, for $x =-\dfrac{5}{2}$ , the value of the polynomial is $0$ and hence, $x =-\dfrac{5}{2}$ is a zero of the given polynomial.$\\$ (iv) $p(x) = 3x - 2$$\\$ $p(x) = 0$$\\$ $3x -2 = 0$$\\$ Therefore, for $x =\dfrac{2}{3}$ , the value of the polynomial is 0 and hence, $x=\dfrac{2}{3}$ is a zero of the given polynomial. $\\$

(v)$ p(x) = 3x$$\\$ Let $p(x) = 0$$\\$ $3x = 0$$\\$ $x = 0$$\\$ Therefore, for $x = 0$, the value of the polynomial is $0$ and hence, $x = 0$ is a zero of the given polynomial.$\\$ (vi) $p(x) = ax$$\\$ Let $p(x) = 0$$\\$ $ax = 0$$\\$ $x = 0$$\\$ Therefore, for $x = 0$, the value of the polynomial is $0$ and hence,$ x = 0$ is a zero of the given polynomial.$\\$ (vii) $p(x) = cx + d$$\\$ Let $p(x) = 0$$\\$ $cx + d = 0$$\\$ $x=\dfrac{-d}{c}$$\\$ Therefore, for $x=\dfrac{-d}{c}$ , the value of the polynomial is $0$ and hence, $x =\dfrac{-d}{c}$ is a zero of the given polynomial.

10   Find the remainder when $x ^3 + 3x ^2 + 3x + 1$ is divided by$\\$ (i)$ x + 1$$\\$ (ii) $x-\dfrac{1}{2}$$\\$ (iii) $x$$\\$ (iv) $x + \pi $$\\$ (v) $5 + 2x$

Solution :

(i)$x ^3 + 3x^ 2 + 3x + 1 \div x + 1$$\\$ By long division, we get $\\$ $$\require{enclose} \begin{array}{r} x^2+2x+1 \\[-3pt] x+1\enclose{longdiv}{x^3+3x^2+3x+1} \\[-3pt] x^3+x^2 \phantom{000000000} \\[-3pt] \underline{- \quad -\phantom{00000000000}}\\[-3pt] 2x^2+3x+1 \\[-3pt] 2x^2+2x \phantom{000}\\[-3pt] \underline{- \quad - \phantom{00000}}\\[-3pt] x+1\\[-3pt] x+1\\[-3pt] \underline{- \quad -\phantom{0}}\\[-3pt] \underline{0\phantom{0}} \end{array}$$ Therefore, the remainder is $0.$

(ii) $x ^3 + 3x ^2 + 3x + 1\div x-\dfrac{1}{2}$$\\$ By long division,$\\$ $$\require{enclose} \begin{array}{r} x^2+\dfrac{7}{2}x+\dfrac{19}{4} \\[-3pt] x-\dfrac{1}{2} \enclose{longdiv}{x^3+3x^2+3x+1}\\[-3pt] x^3-\dfrac{x^2}{2}\phantom{00000000}\\[-3pt] \underline{-\quad+ \phantom{00000000000}}\\[-3pt] \dfrac{7}{2}x^2+3x+1 \quad\\[-3pt] \dfrac{7}{2}x^2-\dfrac{7}{4}x\phantom{00000}\\[-3pt] \underline{- \qquad +\phantom{0000000}}\\[-3pt] \dfrac{19}{4}x+1 \phantom{00}\\[-3pt] \dfrac{19}{4}x-\dfrac{19}{8}\\[-3pt] \underline{-\qquad+ \phantom{00}}\\[-3pt] \underline{\phantom{000}\dfrac{27}{8}} \end{array}$$ Therefore, the remainder is $\dfrac{27}{8}$

(iii) $x ^3 + 3x ^2 + 3x + 1 \div x$$\\$ By long division,$\\$ $$ \require{enclose} \begin{array}{r} x^2+3x+3\\[-3pt] x\enclose{longdiv}{x^3+3x+3x+1}\\[-3pt] x^3\qquad \quad \phantom{0000000}\\[-3pt] \underline{- \qquad \quad \phantom{000000000}}\\[-3pt] 3x^2+3x+1\\[-3pt] 3x^2\qquad \quad \phantom{00}\\[-3pt] \underline{-\qquad\phantom{0000000}}\\[-3pt] 3x+1\\[-3pt] 3x \quad \phantom{00}\\[-3pt] \underline{-\qquad \phantom{0}}\\[-3pt] \underline{\phantom{000000}1} \end{array}$$ Therefore, the remainder is $1.$

(iv) $x ^3 + 3x ^2 + 3x + 1 \div x + \pi $$\\$ By long division, we get$\\$ $$ \require{enclose} \begin{array}{r} x^2+(3-\pi)x+(3-3\pi+\pi^2)\\[-3pt] x+\pi \enclose{longdiv}{x^3+3x^2+3x+1 \qquad \qquad \quad\phantom{000000} }\\[-3pt] x^3+\pi x^2 \qquad \qquad \qquad \phantom{000000000000}\\[-3pt] \underline{- \quad -\qquad \qquad \qquad \quad \phantom{000000000000}}\\[-3pt] (3-\pi)x^2+3x+1 \phantom{0000000000000}\\[-3pt] (3-\pi)x^2+(3-\pi)\pi x \phantom{0000000000}\\[-3pt] \underline{- \qquad \qquad -\qquad \qquad \phantom{00000000000}}\\[-3pt] [3-3\pi+\pi^2]x+1 \phantom{0000000000000}\\[-3pt] [3-3\pi+\pi^2]x+(3-3\pi+\pi^2)\pi \phantom{0}\\[-3pt] \underline{-\qquad \qquad \qquad - \phantom{00000000000000}}\\[-3pt] \underline{\phantom{000000000000}[1-3\pi+3\pi^2-\pi^3] \quad}\\[-3pt] \end{array}$$ Therefore, the remainder is $-\pi^3+3\pi^2-3\pi+1$

(v) $5 + 2x$$\\$ By long division, we get$\\$ $$ \require{enclose} \begin{array}{r} \dfrac{x^2}{2}+\dfrac{x}{4}+\dfrac{7}{8}\\[-3pt] 2x+5 \enclose{longdiv}{x^3+3x^2+3x+1}\\[-3pt] x^3+\dfrac{5}{2}x^2 \phantom{00000000}\\[-3pt] \underline{- \quad - \phantom{00000000000}}\\[-3pt] \dfrac{x^2}{2}+3x+1 \phantom{0}\\[-3pt] \dfrac{x^2}{2}+\dfrac{5x}{4}\phantom{0000}\\[-3pt] \underline{- \qquad -\phantom{000000}}\\[-3pt] \dfrac{7x}{4}+1\phantom{0}\\[-3pt] \dfrac{7}{4}x+\dfrac{35}{8}\\[-3pt] \underline{-\qquad -\phantom{00}}\\[-3pt] \underline{\phantom{0000}-\dfrac{27}{8}\phantom{0}} \end{array}$$ Therefore, the remainder is $-\dfrac{27}{8}.$

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

Solution :

$x^ 3 - ax ^2 + 6x - a \div x - a$ $\\$ By long division,$\\$ $$\require{enclose} \begin{array}{r} x^2+6\\[-3pt] x-a \enclose{longdiv}{x^3-ax^2+6x-a}\\[-3pt] x^3-ax^2 \phantom{00000000}\\[-3pt] \underline{- \quad + \phantom{0000000000}}\\[-3pt] 6x-a\\[-3pt] 6x-6a\\[-3pt] \underline{- \quad + \phantom{00}}\\[-3pt] \underline{\phantom{000000}5a} \end{array}$$ Therefore, when $x^ 3 - ax ^2 + 6x - a$ s divided by $ x - a.$ the remainder obtained is $5a.$

12   Check whether $7 + 3x$ is a factor of $3x ^3 + 7x.$$\\$

Solution :

Let us divide $(3x^ 3 + 7x)$ by $(7 + 3x).$$\\$ By long division, we get$\\$ $$ \require{enclose} \begin{array}{r} x^2-\dfrac{7}{3}x+\dfrac{70}{9}\\[-3pt] 3x+7 \enclose{longdiv}{3x^3+0x^2+7x\qquad \phantom{00}} \\[-3pt] 3x^3+7x^2\phantom{00000000000}\\[-3pt] \underline{- \qquad - \phantom{0000000}}\phantom{000000}\\[-3pt] -7x^2+7x \phantom{0000000}\\[-3pt] -7x^2-\dfrac{49x}{3}\phantom{00000}\\[-3pt] \underline{+ \qquad + \phantom{000000000}}\\[-3pt] \dfrac{10x}{3}\phantom{00000}\\[0.3pt] \dfrac{70x}{3}+\dfrac{490}{9}\\[-3pt] \underline{- \qquad -\phantom{0000}}\\[-3pt] \underline{\phantom{00000}-\dfrac{490}{9}} \end{array}$$ The remainder is not zero,$\\$ Therefore, $7 + 3x$ is not a factor of $3x ^3 + 7x.$