how do i factor f(x)=x^3-x^2-4x+4

Question

how do i factor f(x)=x^3-x^2-4x+4

Answer 1

x^3-x^2-4x+4=f(x)

(x^3-x^2)-(4x-4)=f(x)

x^2(x-1)-4(x-1)=f(x)

(x-1)(x^2-4)=f(x)

(x-1)(x-2)(x+2)=0

x-1=0 ---> x=1

x-2=0 ---> x=2

x+2=0 ---> x=-2

Answer 2

Let's try grouping them, the first two terms and the last two terms:

f(x) = (x³ - x²) + (-4x + 4)

We factor out x2 in the first and 4 out of the second:

f(x) = x²(x - 1) + 4(-x + 1)

Looks at the second half. if we factor out a -1 as well, then our factors will have the same parentheses:

f(x) = x²(x - 1) - 4(x - 1)

 

Now we put the factored out parts in their own parentheses:

f(x) = (x² - 4)(x - 1)

Take a look at the first parentheses set--we have a subtraction of two squares, which means that too can be factored:

f(x) = (x + 2)(x - 2)(x - 1)

 

And that's as factored as it's going to get. Distribute everything back together, and you'll see we will return to our original cube equation.