Answer:
f(x) = (x − 1 + √3) (x − 1 − √3) (x + 2) (x − 1) (x − 5)
Step-by-step explanation:
Use rational root theorem to check for rational roots. Possible roots are:
±1, ±2, ±4, ±5, ±10, ±20
Plug each one into f(x). We find that three of the possibilities are roots: x = -2, x = 1, and x = 5.
f(x) is a 5th order polynomial, so there are 5 total roots. To find the other 2, we divide f(x) by the product of (x + 2) (x − 1) (x − 5). That product is:
(x + 2) (x − 1) (x − 5)
= (x² + x − 2) (x − 5)
= x³ + x² − 2x − 5x² − 5x + 10
= x³ − 4x² − 7x + 10
Using long division, we find that the other factor of f(x) is x² − 2x − 2. We can find the roots of this with quadratic formula:
x = [ -b ± √(b² − 4ac) ] / 2a
x = [ 2 ± √(4 − 4(1)(-2)) ] / 2
x = (2 ± √12) / 2
x = (2 ± 2√3) / 2
x = 1 ± √3
So f(x) can be written as:
f(x) = (x − 1 + √3) (x − 1 − √3) (x + 2) (x − 1) (x − 5)
