Question

if the polynomial ax³+4x²+3x-4 and x³-4x+a leaves same remainder when divided by (x-3) . find the value of a.

Answer 1

Dear Student,

Here is the solution of your asked query:

$$\begin{gathered} \mathrm{Let}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{ax}}^3+4{\mathrm{x}}^2+3\mathrm{x}-4 \\ \mathrm{Let}\mathrm{q}\left(\mathrm{x}\right)={\mathrm{x}}^3-4\mathrm{x}+\mathrm{a} \\ \mathrm{Now},\mathrm{by}\mathrm{remainder}\mathrm{theorem},\mathrm{when}\mathrm{p}\left(\mathrm{x}\right)\mathrm{and}\mathrm{q}\left(\mathrm{x}\right)\mathrm{is}\mathrm{divided}\mathrm{by}\left(\mathrm{x}-3\right),\mathrm{then}\mathrm{remainder}\mathrm{is}\mathrm{p}\left(3\right)\mathrm{and}\mathrm{q}\left(3\right). \\ \mathrm{Now},\mathrm{p}\left(3\right)=\mathrm{a}{\left(3\right)}^3+4\times 3^2+3\times 3-4=27\mathrm{a}+36+9-4=27\mathrm{a}+41 \\ \mathrm{Now},\mathrm{q}\left(3\right)={\left(3\right)}^3-4\times 3+\mathrm{a}=27-12+\mathrm{a}=15+\mathrm{a} \\ \mathrm{Now},\mathrm{p}\left(3\right)=\mathrm{q}\left(3\right) \\ \Rightarrow 27\mathrm{a}+41=15+\mathrm{a} \\ \Rightarrow 27\mathrm{a}-\mathrm{a}=15-41 \\ \Rightarrow 26\mathrm{a}=-26 \\ \Rightarrow \boxed{\mathbf{a}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{1}} \end{gathered}$$

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