Thread: Long Division Problem - 2

1. Long Division Problem - 2

$\displaystyle \dfrac{x^{3} - 4x - 7}{x^{2}-x-6}$

First step?

2. Re: Long Division Problem - 2

you can always just divide

$x(x^2-x-6)=x^3-x^2-6x$

$x^3-4x-7-(x^3-x^2-6x)=x^2+2x-7$

$x^2+2x-7-(x^2-x-6)=3x-1$

so your quotient is $(x+1)$ with remainder $3x-1$

3. Re: Long Division Problem - 2

Hello, Jason76!

$\displaystyle \dfrac{x^3 - 4x - 7}{x^2-x-6}$

. . $\displaystyle \begin{array}{cccccccccc} &&&&&& x & + & 1 \\ && --&--&--&--&--&--& -- \\ x^2-x-6 & | & x^3 &&& - & 4x & - & 7 \\ && x^3 &-& x^2 &-& 6x \\ && --&--&--&--&-- \\ &&&& x^2 &+& 2x &-& 7 \\ &&&& x^2 &-& x &-& 6 \\ &&&& --&--&--&--&-- \\ &&&&&& 3x &-& 1 \end{array}$

Therefore: .$\displaystyle \frac{x^3-4x-7}{x^2-x -6} \;=\;x+1 + \frac{3x-1}{x^2-x-6}$