How do you factor $(x^{3} - 7 x^{2} + 14 x - 8)$ $(x^3 - 7x^2 + 14x - 8)$ ?

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Answer 1 · 2015-04-08T20:25:40

There doesn't seem to be a grouping that will work.

The rational zero theorem tells us that tlhe possible rational zeros are $\pm 1, \pm 2, \pm 4, \pm 8$ $+-1, +-2, +-4, +-8$ .

We can see that ${(1)}^{3} - 7 {(1)}^{2} + 14 (1) - 8 = 1 - 7 + 14 - 8 = 0$ $(1)^3-7(1)^2+14(1)-8=1-7+14-8=0$ , so 1 is a zero.

The factor theorem then tells us that $x - 1$ $x-1$ is a factor.
Do the division:

$\frac{x^{3} - 7 x^{2} + 14 x - 8}{x - 1} = x^{2} - 6 x + 8$ $(x^3-7x^2+14x-8)/(x-1) = x^2-6x+8$

So
$(x^{3} - 7 x^{2} + 14 x - 8) = (x - 1) (x^{2} - 6 x + 8) = (x - 1) (x - 2) (x - 4)$ $(x^3-7x^2+14x-8) =(x-1) (x^2-6x+8)=(x-1)(x-2)(x-4)$

How do you factor (x3−7x2+14x−8)(x^3 - 7x^2 + 14x - 8)?