Question

Using remainder theorem, Find the remainder when:

(i) 3x² + x + 7 is divided by x + 2.
(ii) 4x³ + 5x $-$ 10 is divided by x $-$ 3.
(iii) x³ $-$ax² + 2x $-$ a is divided by x $-$ a

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let},p\left(x\right)=3x^2+x+7 \\ \mathrm{Here},\mathrm{divisor}=x+2 \\ \mathrm{When}x+2=0 \\ \Rightarrow x=-2 \\ \mathrm{By}\mathrm{remainder}\mathrm{theorem}\mathrm{putting}x=-2\mathrm{in}p\left(x\right)\mathrm{we}\mathrm{get}, \\ \mathrm{Remainder}=p\left(-2\right)=3{\left(-2\right)}^2+\left(-2\right)+7 \\ =3\times 4-2+7 \\ =12-2+7 \\ =17 \\ \therefore \mathrm{Remainder}=17 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let}p\left(x\right)=4x^3+5x-10 \\ \mathrm{Here},\mathrm{divisor}=x-3 \\ \mathrm{When}x-3=0,\mathrm{we}\mathrm{have}x=3 \\ \mathrm{Putting}x=3\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(3\right)=4{\left(3\right)}^3+5\left(3\right)-10 \\ =4\times 27+15-10 \\ =108+15-10 \\ =113 \\ \therefore \mathrm{Remainder}=113 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Let}p\left(x\right)=x^3-a{x}^2+2x-a \\ \mathrm{Here},\mathrm{divisor}=x-a \\ \mathrm{When}x-a=0,\mathrm{we}\mathrm{have}x=a \\ \mathrm{Putting}x=a\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(a\right)={\left(a\right)}^3-a{\left(a\right)}^2+2\left(a\right)-a \\ =a^3-a\times a^2+2a-a \\ =a^3-a^3+2a-a \\ =a \\ \therefore \mathrm{Remainder}=a \end{gathered}$$