How do you divide (6x^4+6x^3-12x^2-7x-7)/(x-2) ?

Aug 6, 2016

I got:

$= 6 {x}^{3} + 18 {x}^{2} + 24 x + 41 + \frac{75}{x - 2}$

A fairly straightforward way to do it is synthetic division.

For this, you use the coefficients of each term. It's like regular long division, but faster/more compact.

$6 {x}^{4} + 6 {x}^{3} - 12 {x}^{2} - 7 x - 7$

$\implies$ $6 \text{ "6" "-12" "-7" } - 7$

The factor divided by is as if you were setting it equal to $0$ and solved for it. If we have $x - 2$, the factor in the upper-left square is $2$, because $x - 2 = 0 \implies x = 2$.

So, we begin with:

$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" }}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{}\right)\right]}$

Then, the basic steps are:

1. Bring down the first polynomial term.
2. Multiply by the divisor, place under the next polynomial term, moving left to right.
3. Add the current polynomial term and the current product from step 2, place underneath.
4. Repeat steps 2 and 3 until you've reached the last column and added the terms together already.

So, we would get:

$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" }}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{}\right)\right]}$

$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul"12"),color(black)(ul" "),color(black)(ul" "),color(black)(ul" }}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{18} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{}\right)\right]}$

$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul"12"),color(black)(ul"36"),color(black)(ul" "),color(black)(ul" }}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{18} , \textcolor{b l a c k}{24} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{}\right)\right]}$

$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul"12"),color(black)(ul"36"),color(black)(ul"48"),color(black)(ul" }}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{18} , \textcolor{b l a c k}{24} , \textcolor{b l a c k}{41} , \textcolor{b l a c k}{}\right)\right]}$

$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul"12"),color(black)(ul"36"),color(black)(ul"48"),color(black)(ul"82}}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{18} , \textcolor{b l a c k}{24} , \textcolor{b l a c k}{41} , \textcolor{b l a c k}{75}\right)\right]}$

Now just re-assign the coefficients to the polynomial solution. Our answer had to have gone down one degree from the quartic that we started with.

Therefore, our answer is a cubic:

$= q \left(x\right) + r \left(x\right)$

$= \textcolor{b l u e}{\stackrel{q \left(x\right) , \text{quotient")overbrace(6x^3 + 18x^2 + 24x + 41) + stackrel(r(x), "remainder}}{\overbrace{\frac{75}{x - 2}}}}$