NCERT Exemplar · Class 9 Mathematics Polynomials

Which of the following expressions are polynomials? Justify your answer: (i) $\displaystyle 8$ (ii) $\displaystyle \sqrt{3} x^{2}-2 x$ (iii) $\displaystyle 1-\sqrt{5 x}$ (iv) $\displaystyle \frac{1}{5 x^{-2}}+5 x+7$ (v) $\displaystyle \frac{(x-2)(x-4)}{x}$ (vi) $\displaystyle \frac{1}{x+1}$ (vii) $\displaystyle \frac{1}{7} a^{3}-\frac{2}{\sqrt{3}} a^{2}+4 a-7$ (viii) $\displaystyle \frac{1}{2 x}$

Write whether the following statements are True or False. Justify your answer. (i) A binomial can have atmost two terms (ii) Every polynomial is a binomial (iii) A binomial may have degree $\displaystyle 5$ (iv) Zero of a polynomial is always $\displaystyle 0$ (v) A polynomial cannot have more than one zero (vi) The degree of the sum of two polynomials each of degree $\displaystyle 5$ is always $\displaystyle 5$ . (D) Short Answer Questions Sample Question $\displaystyle 1$ : (i) Check whether $\displaystyle p(x)$ is a multiple of $\displaystyle g(x)$ or not, where $$p(x)=x^{3}-x+1, \quad g(x)=2-3 x$$ (ii) Check whether $\displaystyle g(x)$ is a factor of $\displaystyle p(x)$ or not, where $$p(x)=8 x^{3}-6 x^{2}-4 x+3, \quad g(x)=\frac{x}{3}-\frac{1}{4}$$ Solution : (i) $\displaystyle p(x)$ will be a multiple of $\displaystyle g(x)$ if $\displaystyle g(x)$ divides $\displaystyle p(x)$. Now, $\displaystyle g(x)=2-3 x=0$ gives $\displaystyle x=\frac{2}{3}$ Remainder $$\begin{aligned} & =p\left(\frac{2}{3}\right)=\left(\frac{2}{3}\right)^{3}-\left(\frac{2}{3}\right)+1 \\ & =\frac{8}{27}-\frac{2}{3}+1=\frac{17}{27} \end{aligned}$$ Since remainder $\displaystyle \neq 0$, so, $\displaystyle p(x)$ is not a multiple of $\displaystyle g(x)$. (ii) $\displaystyle g(x)=\frac{x}{3}-\frac{1}{4}=0$ gives $\displaystyle x=\frac{3}{4}$ $\displaystyle g(x)$ will be a factor of $\displaystyle p(x)$ if $\displaystyle p\left(\frac{3}{4}\right)=0$ (Factor theorem) Now, $\displaystyle p\left(\frac{3}{4}\right)=8\left(\frac{3}{4}\right)^{3}-6\left(\frac{3}{4}\right)^{2}-4\left(\frac{3}{4}\right)+3$ $$=8 \times \frac{27}{64}-6 \times \frac{9}{16}-3+3=0$$ Since, $\displaystyle p\left(\frac{3}{4}\right)=0$, so, $\displaystyle g(x)$ is a factor of $\displaystyle p(x)$. Sample Question $\displaystyle 2$ : Find the value of $\displaystyle a$, if $\displaystyle x-a$ is a factor of $\displaystyle x^{3}-a x^{2}+2 x+a-1$. Solution : Let $\displaystyle p(x)=x^{3}-a x^{2}+2 x+a-1$ Since $\displaystyle x-a$ is a factor of $\displaystyle p(x)$, so $\displaystyle p(a)=0$. i.e., $\displaystyle a^{3}-a(a)^{2}+2 a+a-1=0$ $$\begin{aligned} & a^{3}-a^{3}+2 a+a-1=0 \\ & 3 a=1 \end{aligned}$$ Therefore, $\displaystyle a=\frac{1}{3}$ Sample Question $\displaystyle 3$ : (i)Without actually calculating the cubes, find the value of $\displaystyle 48^{3}-30^{3}-18^{3}$. (ii) Without finding the cubes, factorise $\displaystyle (x-y)^{3}+(y-z)^{3}+(z-x)^{3}$. Solution : We know that $\displaystyle x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$. If $\displaystyle x+y+z=0$, then $\displaystyle x^{3}+y^{3}+z^{3}-3 x y z=0$ or $\displaystyle x^{3}+y^{3}+z^{3}=3 x y z$. (i) We have to find the value of $\displaystyle 48^{3}-30^{3}-18^{3}=48^{3}+(-30)^{3}+(-18)^{3}$. Here, $\displaystyle 48+(-30)+(-18)=0$ So, $\displaystyle 48^{3}+(-30)^{3}+(-18)^{3}=3 \times 48 \times(-30) \times(-18)=77760$ (ii) Here, $\displaystyle (x-y)+(y-z)+(z-x)=0$ Therefore, $\displaystyle (x-y)^{3}+(y-z)^{3}+(z-x)^{3}=3(x-y)(y-z)(z-x)$.