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# Divide 3x^3+129+16x by (x+3)

0
127
2

I'm stumped as to how to arrange the exponents so that I have a number to the second power

EDIT: I'm using synthetic division here

Guest Feb 22, 2017
edited by Guest  Feb 22, 2017
#1
0

3x^3 +16x +129  factors to (x+3)(3x^2- 9x + 43)

I am unable to use the software on this site to show you,but it does divide out,as you can see.I used synthetic division. Sorry I can't be of more help.

Guest Feb 22, 2017
#2
0

Simplify the following:
(3 x^3 + 16 x + 129)/(x + 3)

The possible rational roots of 3 x^3 + 16 x + 129 are x = ± 1/3, x = ± 43/3, x = ± 1, x = ± 3, x = ± 43, x = ± 129. Of these, x = -3 is a root. This gives x + 3 as all linear factors:
(((x + 3) (3 x^3 + 16 x + 129))/(x + 3))/(x + 3)

| |
x | + | 3 | | 3 x^2 | - | 9 x | + | 43
3 x^3 | + | 0 x^2 | + | 16 x | + | 129
3 x^3 | + | 9 x^2 | | | |
| | -9 x^2 | + | 16 x | |
| | -9 x^2 | - | 27 x | |
| | | | 43 x | + | 129
| | | | 43 x | + | 129
| | | | | | 0:
(3 x^2 - 9 x + 43 (x + 3))/(x + 3)

((x + 3) (3 x^2 - 9 x + 43))/(x + 3) = (x + 3)/(x + 3)×(3 x^2 - 9 x + 43) = 3 x^2 - 9 x + 43:
Answer: |3 x^2 - 9 x + 43

Guest Feb 22, 2017

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