Question

If a polynomial $x^4+7x^3+7x^2+px+1$ is exactly divisible by $x^2+7x+12$, then find the values of $p$.

Answer 1

Solution:

Let $f\left(x\right)=x^4+7x^3+7x^2+px+1$

and $g\left(x\right)=x^2+7x+12$

$g\left(x\right)=x^2+7x+12$

$=(x+4)(x+3)$

So, zeroes of $g\left(x\right)$ are $-3$ and $-4$

Since, $g\left(x\right)$ exactly divides $f\left(x\right).$

So, $f(-3)=0$ and $f(-4)=0$

or, $f(-3)=(-3{)}^4+7(-3{)}^3+7(-3{)}^2+p(-3)+1=0$

or, $81-189+63-3p+1=0$

or, $3p=-44$

or, $p=\frac{-44}{3}$