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

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Answer 1 · 2016-12-28T21:31:36

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

Given:

$x^{3} - 5 x^{2} + 2 x + 8$ $x^3-5x^2+2x+8$

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:

$- 1 - 5 - 2 + 8 = 0$ $-1-5-2+8 = 0$

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

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

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

$x^{2} - 6 x + 8 = (x - 4) (x - 2)$ $x^2-6x+8 = (x-4)(x-2)$

Putting it all together:

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

How do you determine the binomial factors of x^3-5x^2+2x+8x3−5x2+2x+8x^3-5x^2+2x+8?