How do you determine the binomial factors of x^3-5x^2+2x+8x3−5x2+2x+8?
1 Answer
Dec 28, 2016
Explanation:
Given:
x^3-5x^2+2x+8x3−5x2+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 = 0−1−5−2+8=0
Hence
x^3-5x^2+2x+8 = (x+1)(x^2-6x+8)x3−5x2+2x+8=(x+1)(x2−6x+8)
To factor the remaining quadratic note that
x^2-6x+8 = (x-4)(x-2)x2−6x+8=(x−4)(x−2)
Putting it all together:
x^3-5x^2+2x+8 = (x+1)(x-4)(x-2)x3−5x2+2x+8=(x+1)(x−4)(x−2)