How do you use synthetic division to divide #12x^3 + 19x^2 - 13x - 6# by #2x-1#?

1 Answer
Sep 6, 2015

Answer:

#(12x^3+19x^2-13x-6)/(2x-1)#

#= (6x^2+13/2x-13/4) -37/(4(2x-1))#

Explanation:

Synthetic division of polynomials is similar to division of integers.

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Write the coefficients of your dividend under the bar and the coefficients of the divisor to the left.

Choose the first number for the quotient so that when multiplied by the divisor it will match the first number of the dividend. In our case we write #6# above the bar since #6 xx (2, -1) = (12, -6)# matches #(12, 19, -13, -6)# in its first component.

Write the product #(12, -6)# below the dividend and subtract it to give #13#.

Bring down the next number #-13# from the dividend alongside it.

Choose the next number for the quotient so that when multiplied by the divisor it matches the leading number in the remainder. In our case we write #13/2# since the first component of #13/2 xx (2, -1) = (13, -13/2)# matches the first component of #(13, -13)#.

Write the product #(13, -13/2)# below the remainder and subtract it to get #-13/2#, etc.

We end up with a quotient #(6, 13/2, -13/4)# meaning #6x^2+13/2x-13/4# and a remainder #-37/4#

So:

#12x^3+19x^2-13x-6#

#= (2x-1)(6x^2+13/2x-13/4) -37/4#

Or:

#(12x^3+19x^2-13x-6)/(2x-1)#

#= (6x^2+13/2x-13/4) -37/(4(2x-1))#