Question

How do you find the factors of `f(x) = x^3-9x^2+8x+60`?

Answer 1

`f(x)=(x-5)(x-6)(x+2)`

Explanation:

A cubic equation always has at least one real root. So we can find at least one linear factor.

ie

`f(x)=(x+a)(x^2+bx+c)`

when multiplying out we have the constant term as being `60` so `a" "` must be a factor of `60`

using the factor theorem we try factors of 60 to find one root

`f(5)=5^3-9xx5^2+8xx5+60=0`

`:. x=5` is a root`=>(x-5)`a factor

`f(x)=(x-5)(x^2+bx+c)==x^3-9x^2+8x+60`

comparing

coefficients of `x^2`

`LHS=-5+b`

`RHS=-9`

`=>b=-4`

comparing coefficients of `x`

`LHS=c-5b`

`RHS=8`

`:.c-5b=8`

`c-5xx-4=8`

`c=8-20=-12`

so
`f(x)=(x-5)(x^2-4x-12)`

the quadratic will factorise

`x^2-4x-12=(x" ")(x" ")`

we need factors of `-12` that add up to `-4`

by trial and error we find `-6 " & "2`

so `x^2-4x-12=(x-6)(x+2)`

the final result is:

`f(x)=(x-5)(x-6)(x+2)`