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

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Answer 1 · 2016-12-28T21:31:36

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

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:

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

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