Question

When x³ + ax² + 4x $-$ 5 is divided by x + 1, the remainder is 14. Find a.

Answer 1

$$\begin{gathered} \mathrm{Let}p\left(x\right)=x^3+a{x}^2+4x-5 \\ \mathrm{Here},\mathrm{divisor}=x+1 \\ \mathrm{When}x+1=0,\mathrm{we}\mathrm{have}x=-1 \\ \mathrm{Putting}x=-1\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ \mathrm{Remainder}=p\left(-1\right)={\left(-1\right)}^3+a{\left(-1\right)}^2+4\left(-1\right)-5 \\ =-1+a-4-5 \\ =a-10 \\ \mathrm{Given}:\mathrm{Remainder}=14 \\ \mathrm{i}.\mathrm{e}.,a-10=14 \\ \Rightarrow a=14+10 \\ \therefore a=24 \end{gathered}$$