Question

Use long division to divide x^3+x^2-2x+14 by x+3

Answer 1

First, $x^3=x^2\cdot x$, and if we multiply $x+3$ by $x^2$ we get

$x^3+3x^2$

Subtracting this from the numerator gives a remainder of

$-2x^2-2x+14$

Next, $-2x^2=-2x\cdot x$, and if we multiply $x+3$ by $-2x$ we have

$-2x^2-6x$

and subtracting this from the previous remainder, we end up with a new remainder of

$4x+14$

Next, $4x=4\cdot x$, and if we multiply $x+3$ by $4$ we get

$4x+12$

Subtracting from the previous remainder, we get a new remainder of

$2$

which contains no more factors of $x$, so we're done.

So,

$\dfrac{x^3+x^2-2x+14}{x+3}=x^2-2x+4+\dfrac2{x+3}$