Question

If the polynomials $2x^3+a{x}^2+3x-5$ and $x^3+x^2-4x+a$ leave the same remainder when divided by $x-2$, Find the value of $a$.

Answer 1

Step 1: Determine the value of $x$

Take $2x^3+a{x}^2+3x-5=g\left(x\right)$

$x^3+x^2-4x+a=f\left(x\right)$.

On dividing $g\left(x\right)$ and $f\left(x\right)$ by $x-2$ gives the same remainder.

$\therefore x-2$ is a factor of $g\left(x\right)$ and $f\left(x\right)$.

$$\begin{gathered} x-2=0\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{}\mathbf{is}\mathbf{}\mathbf{a}\mathbf{}\mathbf{factor}\mathbf{]} \\ \Rightarrow x=2 \end{gathered}$$

Step 2: Equate the functions

By the remainder theorem, the remainder of $f\left(x\right)$ when divided by $x-a$ is $f\left(a\right)$

So, the remainder of $f\left(x\right)$when divided by $x-2$ is $f\left(2\right)$

and the remainder of $g\left(x\right)$when divided by $x-2$ is $g\left(2\right)$

Now, since the remainder of $f\left(x\right)$ and $g\left(x\right)$ when divided by $x-2$ is equal

,$$\begin{gathered} g\left(2\right)=f\left(2\right) \\ \Rightarrow 2{\left(2\right)}^3+a{\left(2\right)}^2+3\left(2\right)-5={\left(2\right)}^3+{\left(2\right)}^2-4\left(2\right)+a \\ \Rightarrow 16+4a+6-5=8+4-8+a \\ \Rightarrow 4a+17=a+4 \\ \Rightarrow 3a=-13 \\ \Rightarrow a=\frac{-13}{3} \end{gathered}$$

Hence, the value of $a$ is $\frac{-13}{3}$.