Question

3x^3+5x^2-48x-80 divided by x+4

Answer 1

One way to do is to write the first polynomial in terms of the second; this means find $a,b,c,d$ so that

$3x^3+5x^2-48x-80=a(x+4)^3+b(x+4)^2+c(x+4)+d$

Expanding the right side and matching up coefficients of terms with the same power of $x$ gives

$\begin{cases}a=3\\12a+b=5\\48a+8b+c=-48\\64a+16b+4c+d=-80\end{cases}\implies a=3,b=-31,c=56,d=0$

So we have

$3x^3+5x^2-48x-80=3(x+4)^3-31(x+4)^2+56(x+4)$

and in particular we can see $x+4$ divides this exactly, giving us

$\dfrac{3x^3+5x^2-48x-80}{x+4}=3(x+4)^2-31(x+4)+56$

and expanding gives

$\dfrac{3x^3+5x^2-48x-80}{x+4}=\boxed{3x^2-7x-20}$