How do you factor $(x^5) - (x^4) + (x^3) - (x^2) - 12x + 12$ ?

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Answer 1 · 2015-06-16T23:43:13

$f(x) = x^5-x^4+x^3-x^2-12x+12$

$=(x-1)(x^4+x^2-12)$

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

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

The sum of the coefficients is zero, that is:

$1-1+1-1-12+12 = 0$

So $f(1) = 0$ and $(x-1)$ is a factor of $f(x)$

The remaining quartic factor is actually quadratic in $x^2$ , so factors just as easily into two quadratic factors:

$x^4+x^2-12 = (x^2)^2+(x^2)-12$

$=(x^2-3)(x^2+4)$

The first of these can be viewed as a difference of squares:

$x^2-3 = x^2-(sqrt(3))^2 = (x-sqrt(3))(x+sqrt(3))$

...using the difference of squares identity:

$a^2-b^2 = (a-b)(a+b)$

The remaining quadratic factor has no linear factors with real coefficients, since $x^2+4 >= 4 > 0$ for all $x in RR$ .

How do you factor (x^5) - (x^4) + (x^3) - (x^2) - 12x + 12(x^5) - (x^4) + (x^3) - (x^2) - 12x + 12?