Question

Use the complex zeros(1/2, -3+7i, -3-7i) to factor f(x)=-2x^3 - 11x^2 -110x +58

Answer 1

I'd like to point out that Caylus' posting is correct using complex factors.

one may note that it could also be simplified to 

$\bf f(x)=(1-2x)(x+3+7i)(x+3-7i) \\\\\\ f(x)=(1-2x)[(x+3)+(7i)][(x+3)-(7i)] \\\\\\ f(x)=(1-2x)~~[\stackrel{\textit{difference of squares}}{(x+3)^2-(7i)^2}] \\\\\\ f(x)=(1-2x)[(x^2+6x+9)-(7^2i^2)] \\\\\\ f(x)=(1-2x)[(x^2+6x+9)-(49\cdot -1)] \\\\\\ f(x)=(1-2x)[(x^2+6x+9)-(-49)] \\\\\\ f(x)=(1-2x)[(x^2+6x+9)+49] \\\\\\ f(x)=(1-2x)(x^2+6x+58)$

Answer 2

P(x)=-(2x^3+11x^2+110x-58)
P(1/2)=-(2*1/8+11*1/4+110*1/2-58)=0
P(x)=(1/2-x)(2x²+12x+116) (using synthetic division)
P(x)=(1-2x)(x²+6x+58)=(1-2x)(x+3+7i)(x+3-7i)