College Algebra & Trigonometry, 2018a

2.7. SYNTHETIC DIVISION 129
Then, we proceed as usual:
− 1 2
6 1 9 1 −2
↓ −3 1 −5 2
6 −2 10 −4 0
Notice that, again, we have a zero remainder. Also, notice that each coefficient
in our answer has a common factor of 2, which was the coefficient of the x in
2x +1, which we orginally were going to divide by. What we’ve done here is not
division by 2x +1, but division by x + 1 2 .
So, in the end, our work tells us that:
6x 4 + x 3 +9x 2 + x − 2
x + 1 2
=6x 3 − 2x 2 +10x − 4
6x 4 + x 3 +9x 2 + x − 2=
and
(
x + 1 )
(6x 3 − 2x 2 +10x − 4)
2
Notice that if we factor out the common factor of 2 from our answer, we can
multiply it back into the x + 1 and get an answer for our original problem:
2
6x 4 + x 3 +9x 2 + x − 2=
(
x + 1 )
2(3x 3 − x 2 +5x − 2)
2
=(2x + 1)(3x 3 − x 2 +5x − 2)
This means that:
6x 4 + x 3 +9x 2 + x − 2
2x +1
=3x 3 − x 2 +5x − 2