Question

If p(x) = 2x³ + ax² + 3x − 5 and q(x) = 2x³ + ax² + 3x − 5 leave the same remainder when divided by (x − 2), show that $a=\frac{-13}{3}$.

Answer 1

Let:
$p\left(x\right)=2x^3+a{x}^2+3x-5$
Now,
$x-2=0\Rightarrow x=2$
When p(x) is divided by (x $-$ 2), then the remainder is p(2).
$$\begin{gathered} \mathrm{Here}, \\ p\left(2\right)=2\times 2^3+a\times 2^2+3\times 2-5 \\ =16+4a+6-5 \\ =4a+17 \end{gathered}$$

Let:
$q\left(x\right)=x^3+x^2-4x+a$
When q(x) is divided by (x $-$ 2), then the remainder is q(2).
$$\begin{gathered} q\left(2\right)=2^3+2^2-4\times 2+a \\ =8+4-8+a \\ =a+4 \end{gathered}$$
According to the question, p(2) and q(2) are the same.
Thus, we have:
$$\begin{gathered} 4a+17=a+4 \\ \Rightarrow 3a=-13 \\ \Rightarrow a=\frac{-13}{3} \end{gathered}$$