Posted by sue on Friday, December 9, 2011 at 4:56pm.
if 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)
Thank you!
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