a little synthetic division shows
f(x) = (x-1)(x^2-2x+5)
And it's easy from there
Find the complex zeros of the polynomial function. Write f in the factored form. f(x)=x^3-3x^2+7x-5
f(x)=
2 answers
First we have to find the real root, (every cubic has at least one real root)
try x = 1 , y = 1 - 3 + 7 - 5 = 0 , so x-1 is a factor
by synthetic division I got
(x-1)(x^2 - 2x + 5) = 0
So the real zero is when x=1
for complex ...
x^2 - 2x + 5= 0
x^2 - 2x = -5
x^2 - 2x + 1-5 + 1
(x-1)^2 =-4
x - 1 = ± 2i
x = 1 ± 2i
try x = 1 , y = 1 - 3 + 7 - 5 = 0 , so x-1 is a factor
by synthetic division I got
(x-1)(x^2 - 2x + 5) = 0
So the real zero is when x=1
for complex ...
x^2 - 2x + 5= 0
x^2 - 2x = -5
x^2 - 2x + 1-5 + 1
(x-1)^2 =-4
x - 1 = ± 2i
x = 1 ± 2i