Find the complex zeros of the polynomial function. Write f in factored form.
f(x)=x^3-8x^2+29x-52
First we try to find any real rational zeros.
The only possible rational zeros are ± the factors of 52. They are
±1, ±2, ±4, ±13, ±26, ±52
Try 1 using synthetic division
1 | 1 -8 29 -52
| 1 -7 22
1 -7 22 -30 Nope. f(1)=-30, not 0.
Try -1 using synthetic division
-1 | 1 -8 29 -52
| -1 9 -38
1 -9 38 -90 Nope. f(-1)=-90, not 0.
Try 4 using synthetic division
4 | 1 -8 29 -52
| 4 -16 52
1 -4 13 0 Yes! f(4)=0.
That means (x-4) is a factor of f(x), and the quotient is
1x²-4x+13 (the numbers left of the 0 in the synthetic division.
So we have this factorization of f(x)
f(x) = (x-4)(x²-4x+13) <-- that's the simplest factored form without
using complex numbers!
We set that = 0
(x-4)(x²-4x+13) = 0
x-4=0; x²-4x+13 = 0
x=4
So f(x) has real zero 4, and complex zeros 2+3i and 2-3i.
So allowing complex numbers we can find another factored form
for f(x) of only linear factors:
f(x) = (x-4)[x-(2+3i)][x-(2-3i)]
f(x) = (x-4)(x-2-3i)(x-2+3i)
Edwin