Question

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

Answer 1

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

Explanation:

Rewrite the divisor `3x-1` as `3(x-1/3)`.

Then the problem becomes
`(6x^3+7x^2+12x-5)/(3(x-1/3)` or `((6x^3+7x^2+12x-5)/(x-1/3))*1/3`

Do synthetic division using the divisor`(x-1/3)`. 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 `6x^2+9x+5`

But remember to multiply this result by `1/3`!!

`1/3(6x^2+9x+15) = 2x^2 +3x+5`