Question

If $x^3+p{x}^2+qx+6$ has $x-2$ as a factor and leaves remainder $3$ when divided by $x-3$ find the value of $p$ and $q$

Answer 1

If the given polynomial is $f\left(x\right)=x^3+p{x}^2+qx+6$ then solution will follows as:

When $f\left(x\right)$ is divided by $x-3$ and $x-2$, the remainders are 3 and 0 respectively.

$\therefore f\left(3\right)=3$ and $f\left(2\right)=0$

$\Rightarrow (3{)}^3+p(3{)}^2+q\left(3\right)+6=3and(2{)}^3+p(2{)}^2+q\left(2\right)+6=0$

$\Rightarrow 27+9p+3q+6=3$ and $8+4p+2q+6=0$

$\Rightarrow 9p+3q+33=3$ and $4p+2q+14=0$

$\Rightarrow 9p+3q=-30$ and $4p+2q=-14$

$\Rightarrow 3p+q=-10$ ... (1) and $2p+q=-7$ .... (2)

On subtracting (1) and (2), we get

$3p+q-(2p+q)=-10-(-7)$

$\Rightarrow p=-10+7=-3$

On putting $p$ in (2), we get

$2(-3)+q=-7$

$\Rightarrow q=-7+6=-1$

$\Rightarrow p=-3$ and $q=-1$