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October 1, 2014

October 1, 2014

Posted by **Math factoring** on Thursday, April 8, 2010 at 10:44am.

y=x^4-4x^3+16

i could only get up to:

y=(x-2)(x^3-2x^2-4x-8)

- Mica -
**Reiny**, Thursday, April 8, 2010 at 11:08amThat's as far as you can go using rational numbers

(I ran x^3-2x^2-4x-8 = 0 through my "cubic equation solver and got one messy real root and 2 complex roots)

http://www.1728.com/cubic.htm

- Mica -
**Math factoring**, Thursday, April 8, 2010 at 11:11amwould there be another way to find the x intercepts? Cause thats what i was trying to achieve

- Mica -
**Reiny**, Thursday, April 8, 2010 at 11:16amyou could sketch it and see appr. where it crosses the x-axis.

let f(x) = x^3 - 2x^2 - 4x - 8

If there are any rational factors, then test for

f(x) = 0 using x = ±1, ±2,±4, and ±8

none work, so there are no rational factors.

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