Question

When 2x² $-$ ax + 7 and ax² + 7x + 12 are divided by (x $-$ 3) and (x + 1) respectively, the remainder is same. Find a.

Answer 1

$$\begin{gathered} \mathrm{Let}p\left(x\right)=2x^2-ax+7 \\ \mathrm{Here},\mathrm{divisor}=x-3 \\ \mathrm{When}x-3=0,\mathrm{we}\mathrm{have}x=3 \\ \mathrm{Putting}x=3\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ \mathrm{Remainder}=p\left(3\right)=2{\left(3\right)}^2-a\left(3\right)+7 \\ =18-3a+7 \\ =-3a+25 \\ \mathrm{Similarly},\mathrm{let}q\left(x\right)=a{x}^2+7x+12 \\ \mathrm{Here},\mathrm{divisor}=x+1 \\ \mathrm{When}x+1=0,\mathrm{we}\mathrm{have}x=-1 \\ \mathrm{Putting}x=-1\mathrm{in}q\left(x\right),\mathrm{we}\mathrm{get}: \\ \mathrm{Remainder}=q\left(-1\right)=a{\left(-1\right)}^2+7\left(-1\right)+12 \\ =a-7+12 \\ =a+5 \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{the}\mathrm{remainders}\mathrm{in}\mathrm{both}\mathrm{the}\mathrm{cases}\mathrm{are}\mathrm{the}\mathrm{same}. \\ \mathrm{i}.\mathrm{e}.,-3a+25=a+5 \\ \Rightarrow -3a-a=5-25 \\ \Rightarrow -4a=-20 \\ \therefore a=5 \end{gathered}$$