How do you find the factors of f(x) = x^3-9x^2+8x+60f(x)=x3−9x2+8x+60?

1 Answer
sjc
May 11, 2017

f(x)=(x-5)(x-6)(x+2)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)f(x)=(x+a)(x2+bx+c)

when multiplying out we have the constant term as being 6060 so a" "a must be a factor of 6060

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

f(5)=5^3-9xx5^2+8xx5+60=0f(5)=53−9×52+8×5+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)

Related questions
Impact of this question
2843 views around the world
You can reuse this answer
Creative Commons License