Question

rewrite x^3+5x^2+3x-10/x+4 in the form q(x)+r(x)/b(x)

Answer 1

One way to do it: Find coefficients $a,b,c$ such that

$x^3+5x^2+3x-10=(x+4)^3+a(x+4)^2+b(x+4)+c$
$=x^3+(a+12)x^2+(8a+b+48)x+(16a+4b+c+64)$
$\implies\begin{cases}5=a+12\\3=8a+b+48\\-10=16a+4b+c+64\end{cases}\implies a=-7,b=11,c=-6$

So, we have

$\dfrac{x^3+5x^2+3x-10}{x+4}=\dfrac{(x+4)^3-7(x+4)^2+11(x+4)-6}{x+4}$
$=(x+4)^2-7(x+4)+11-\dfrac6{x+4}$
$=x^2+x-1-\dfrac6{x+4}$

This can also be done with long or synthetic division. Just to verify the result above, synthetic division yields

-4  |  1   5   3   -10
.    |      -4  -4      4
- - - - - - - - - - - - - -
.    |  1   1  -1     -6

which translates to

$\dfrac{x^3+5x^2+3x-10}{x+4}=x^2+x-1-\dfrac6{x+4}$

as required.

Here, $q(x)=x^2+x-1$, $r(x)=-6$, and $b(x)=x+4$.