How do you determine the binomial factors of x^3-5x^2+2x+8x35x2+2x+8?

1 Answer
Dec 28, 2016

x^3-5x^2+2x+8 = (x+1)(x-4)(x-2)x35x2+2x+8=(x+1)(x4)(x2)

Explanation:

Given:

x^3-5x^2+2x+8x35x2+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 = 0152+8=0

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

x^3-5x^2+2x+8 = (x+1)(x^2-6x+8)x35x2+2x+8=(x+1)(x26x+8)

To factor the remaining quadratic note that 6=4+26=4+2 and 8=4*28=42, so we find:

x^2-6x+8 = (x-4)(x-2)x26x+8=(x4)(x2)

Putting it all together:

x^3-5x^2+2x+8 = (x+1)(x-4)(x-2)x35x2+2x+8=(x+1)(x4)(x2)