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October 20, 2014

October 20, 2014

Posted by **sue** on Friday, December 9, 2011 at 4:56pm.

f(x) x^4 + 10x^3 + 12x^2 - 58x + 35

Someone anyone?????

- algebra -
**Reiny**, Friday, December 9, 2011 at 6:52pmif any factors exist then, then some of ±1, ±5, ±7 must be zeros of the function

it is easy to see that x=1 works

f(1) = 1+10+12-58+35 = 0

by synthetic division,

x^4 + 10x^3 + 12x^2 - 58x + 35 = (x-1)(x^3 + 11x^2 + 23x - 35)

mmmh, looks like x=1 is also a zero of the cubic facor

another sysnthetic division left me with

x^4 + 10x^3 + 12x^2 - 58x + 35 =(x-1)(x-1)(x^2 + 12x + 35)

and that quadratic obviously factors again, so we have

x^4 + 10x^3 + 12x^2 - 58x + 35 =(x-1)^2 (x+5)(x+7)

- algebra -
**anon**, Saturday, December 10, 2011 at 10:22pmThank you!

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