Question

If r and s be the remainders when the polynomials (x³ + 2x² − 5ax − 7) and (x³ + ax² − 12ax + 6) are divided by (x + 1) and (x − 2), respectively and 2r + s = 6, find the value of a.

Answer 1

$$\begin{gathered} \mathrm{Let}f\left(x\right)=x^3+2x^2-5ax-7. \\ \text{When}f\left(x\right)\mathrm{is}\mathrm{divided}\mathrm{by}\left(x+1\right),\text{the}\mathrm{remainder}\mathrm{is}r. \\ \mathrm{Therefore},f\left(-1\right)=r \\ \Rightarrow {\left(-1\right)}^3+2{\left(-1\right)}^2-5a\left(-1\right)-7=r \\ \Rightarrow -1+2+5a-7=r \\ \Rightarrow 5a-6=r \\ \Rightarrow r=5a-6-------\left(1\right) \\ \mathrm{Again},\mathrm{let}g\left(x\right)=x^3+a{x}^2-12ax+6. \\ \text{When}g\left(x\right)\mathrm{is}\mathrm{divided}\mathrm{by}\left(x-2\right),\text{the}\mathrm{remainder}\mathrm{is}s. \\ \mathrm{Therefore},g\left(2\right)=s \\ \Rightarrow {\left(2\right)}^3+a{\left(2\right)}^2-12a\left(2\right)+6=s \\ \Rightarrow 8+4a-24a+6=s \\ \Rightarrow s=-20a+14-----\left(2\right) \\ \mathrm{From}\left(1\right)\times 2+\left(2\right),\mathrm{we}\mathrm{have}: \\ 2r+s=2\left(5a-6\right)+\left(-20a+14\right) \\ \Rightarrow 6=10a-12-20a+14 \\ \Rightarrow 6=-10a+2 \\ \Rightarrow -10a=4 \\ \Rightarrow a=\frac{-2}{5} \end{gathered}$$