Question

R₁ and R₂ are the remainders when the polynomial ax³ + 3x^₂ $-$ 3 and 2x^₃ $-$ 5x + 2a are divided by (x $-$ 4) respectively. If 2R₁ $-$ R₂ = 0, then find the value of a.

Answer 1

$$\begin{gathered} \mathrm{Let}p\left(x\right)=a{x}^3+3x^2-3 \\ \mathrm{Here},\mathrm{divisor}=x-4 \\ \mathrm{When}x-4=0,\mathrm{we}\mathrm{have}x=4 \\ \mathrm{Given}:\mathrm{When}p\left(x\right)\mathrm{is}\mathrm{divided}\mathrm{by}\left(x-4\right),\mathrm{the}\mathrm{remainder}\mathrm{is}{R}_{1.} \\ \therefore \mathrm{Remainder}=R_1=p\left(4\right) \\ \Rightarrow R_1=a{\left(4\right)}^3+3{\left(4\right)}^2-3 \\ \Rightarrow R_1=64a+48-3 \\ \Rightarrow R_1=64a+45...\left(1\right) \\ \mathrm{Again},\mathrm{let}q\left(x\right)=2x^3-5x+2a \\ \mathrm{Here},\mathrm{divisor}=x-4 \\ \mathrm{When}x-4=0,\mathrm{we}\mathrm{have}x=4 \\ \mathrm{Given}:\mathrm{When}q\left(x\right)\mathrm{is}\mathrm{divided}\mathrm{by}\left(x-4\right),\mathrm{the}\mathrm{remainder}\mathrm{is}{R}_{2.} \\ \therefore \mathrm{Remainder}=R_2=q\left(4\right) \\ \Rightarrow R_2=2{\left(4\right)}^3-5\left(4\right)+2a \\ \Rightarrow R_2=128-20+2a \\ \Rightarrow R_2=2a+108...\left(2\right) \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}2R_1-R_2=0 \\ \mathrm{Putting}\mathrm{the}\mathrm{values}\mathrm{of}{R}_1\mathrm{and}{R}_2\mathrm{from}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}: \\ 2\left(64a+45\right)-\left(2a+108\right)=0 \\ \Rightarrow 128a+90-2a-108=0 \\ \Rightarrow 126a-18=0 \\ \Rightarrow 126a=18 \\ \Rightarrow a=\frac{18}{126}=\frac{1}{7} \\ \therefore a=\frac{1}{7} \end{gathered}$$