# How do you divide (6x^3+7x^2+12x-5)div(3x-1) using synthetic division?

Sep 7, 2016

Be very careful when the divisor has a leading coefficent not equal to one.

#### Explanation:

Rewrite the divisor $3 x - 1$ as $3 \left(x - \frac{1}{3}\right)$.

Then the problem becomes
(6x^3+7x^2+12x-5)/(3(x-1/3) or $\left(\frac{6 {x}^{3} + 7 {x}^{2} + 12 x - 5}{x - \frac{1}{3}}\right) \cdot \frac{1}{3}$

Do synthetic division using the divisor$\left(x - \frac{1}{3}\right)$. Pull down the 6, write it under the line and multiply it by 1/3. The product 2 is written under the next number, 7. Add the 7 and 2. The sum 9 is written below the line. Multiply 9 by 1/3. The product 3 is written under the next number, 12. Add 12 and 3. The sum 15 is written under the line. Multiply 15 by 1/3. The product 5 is written under the next number, -5. Add -5 and 5. The sum zero is written below the line. So you are multiplying then adding, multiplying then adding, etc.

1/3 | 6...7....12...-5
......|.......2....3.....5
......-------------------
........6....9.....15....0

This result can be written as $6 {x}^{2} + 9 x + 5$

But remember to multiply this result by $\frac{1}{3}$!!

$\frac{1}{3} \left(6 {x}^{2} + 9 x + 15\right) = 2 {x}^{2} + 3 x + 5$