How do you determine the binomial factors of x^3-5x^2+2x+8?

1 Answer
Dec 28, 2016

x^3-5x^2+2x+8 = (x+1)(x-4)(x-2)

Explanation:

Given:

x^3-5x^2+2x+8

Notice that if we reverse the signs on the coefficients of the terms with odd degree then the sum of the coefficients is zero.

That is:

-1-5-2+8 = 0

Hence x=-1 is a zero and (x+1) a factor:

x^3-5x^2+2x+8 = (x+1)(x^2-6x+8)

To factor the remaining quadratic note that 6=4+2 and 8=4*2, so we find:

x^2-6x+8 = (x-4)(x-2)

Putting it all together:

x^3-5x^2+2x+8 = (x+1)(x-4)(x-2)