# 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) $x = 2$

(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.

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 $5a.$