f(±1) ≠ 0
f(±2) ≠ 0
but f(-4) = 0
so x+4 is a factor
I then used synthetic division to get
(x+4)(x^2 - 2x + 2) = 0
the real root is x = -4
Use the quadratic formula to get the two complex roots of
1+i and 1-i
Find the real and complex zeros of the following function. Please show all of your work.
f(x) = x^3 + 2x^2 - 6x + 8
2 answers
f(x)=x^3+4x^2-2x^2-8x+2x+8=(x^3+4x^2)-
(2x^2+8x)+(2x+8)=x^2(x+4)-2x(x+4)+2(x+4)=
(x+4)(x^2-2x+2)
x+4=0 => x=-4
or
x^2-2x+2=0 => x=1+i or x=1-i
(2x^2+8x)+(2x+8)=x^2(x+4)-2x(x+4)+2(x+4)=
(x+4)(x^2-2x+2)
x+4=0 => x=-4
or
x^2-2x+2=0 => x=1+i or x=1-i