Question

If p(x) = 2x³ $-$ 3x² + 4x $-$ 5. Find the remainder when p(x) is divided by.

(i) x $-$ 2
(ii) x + 3
(iii) x $-$ 1

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}p\left(x\right)=2x^3-3x^2+4x-5 \\ \mathrm{Here},\mathrm{divisor}=x-2 \\ \mathrm{When}x-2=0,\mathrm{we}\mathrm{have}x=2 \\ \mathrm{Putting}x=2\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(2\right)=2{\left(2\right)}^3-3{\left(2\right)}^2+4\left(2\right)-5 \\ =2\times 8-3\times 4+8-5 \\ =16-12+8-5 \\ =7 \\ \therefore \mathrm{Remainder}=7 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}p\left(x\right)=2x^3-3x^2+4x-5 \\ \mathrm{Here},\mathrm{divisor}=x+3 \\ \mathrm{When}x+3=0,\mathrm{we}\mathrm{have}x=-3 \\ \mathrm{Putting}x=-3\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(-3\right)=2{\left(-3\right)}^3-3{\left(-3\right)}^2+4\left(-3\right)-5 \\ =2\times \left(-27\right)-3\times 9-12-5 \\ =-54-27-12-5 \\ =-98 \\ \therefore \mathrm{Remainder}=-98 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}p\left(x\right)=2x^3-3x^2+4x-5 \\ \mathrm{Here},\mathrm{divisor}=x-1 \\ \mathrm{When}x-1=0,\mathrm{we}\mathrm{have}x=1 \\ \mathrm{Putting}x=1\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(1\right)=2{\left(1\right)}^3-3{\left(1\right)}^2+4\left(1\right)-5 \\ =2\times 1-3\times 1+4-5 \\ =2-3+4-5 \\ =-2 \\ \therefore \mathrm{Remainder}=-2 \end{gathered}$$