College Algebra & Trigonometry, 2018a

120 CHAPTER 2. POLYNOMIAL AND RATIONAL FUNCTIONS
This indicates that we’ll have the 2x 4 we’ll need in our answer, as well six of the
seven x 3 ’s and two of the four x 2 ’s. We’ll now need 1x 3 next:
2x 2
x 2 +3x +1 ) 2x 4 +7x 3 +4x 2 − 2x − 1
− 2x 4 − 6x 3 − 2x 2
This means we’ll need to multiply by 1x:
x 3 +2x 2 − 2x
2x 2
+ x
x 2 +3x +1 ) 2x 4 +7x 3 +4x 2 − 2x − 1
− 2x 4 − 6x 3 − 2x 2
x 3 +2x 2 − 2x
− x 3 − 3x 2 − x
− x 2 − 3x − 1
Here, we still need to pick up a −1x 2 , which means that our next multiplication
will be with −1:
2x 2 + x − 1
x 2 +3x +1 ) 2x 4 +7x 3 +4x 2 − 2x − 1
− 2x 4 − 6x 3 − 2x 2
x 3 +2x 2 − 2x
− x 3 − 3x 2 − x
− x 2 − 3x − 1
x 2 +3x +1
Because x 2 +3x +1divides evenly into 2x 4 +7x 3 +4x 2 − 2x − 1 we have a zero
remainder. In the next example there will be a non-zero remainder:
Example
Divide:
3x 4 − 8x 3 +19x 2 − 15x +10
x 2 − x +4
0