Question

The polynomials $k{x}^3+4x^2+3x-4$ and $x^3-4x+k$ leave the same remainder when divided by $(x-3)$, find the value of $k$.

Answer 1

Step 1: Determine the remainder.

If $p\left(x\right)$ is divided by $x+c$, then the remainder is $p\left(-c\right)$.$[\because x+c=x-(-c\left)\right]$

The divisor here is $x-3$.

Let $f\left(x\right)=k{x}^3+4x^2+3x-4$ and $g\left(x\right)=x^3-4x+k$

By remainder theorem, when $f\left(x\right)$ and $g\left(x\right)$ is divided by $x-3$, then the remainder is $f\left(3\right)$ and $g\left(3\right)$.

Step 2: Determine the value of $k$.

It is given that the remainder is the same for both the given polynomial.

$\begin{array}{rcl}f\left(3\right)& =& g\left(3\right)\\ k{\left(3\right)}^3+4{\left(3\right)}^2+3\left(3\right)-4& =& {\left(3\right)}^3-4\left(3\right)+k\\ 27k+36+9-4& =& 27-12+k\\ 26k& =& -26\\ k& =& -1\end{array}$

Hence, the remainder is $-1$