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

1 Answer
Sep 7, 2016

Answer:

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#