How do you factor $x^4-2x^3-13x^2+38x-24$ ?

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Answer 1 · 2015-07-18T10:58:19

Use sum of coefficients short cut, then rational roots theorem to find:

$x^4-2x^3-13x^2+38x-24$

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

Let $f(x) = x^4-2x^3-13x^2+38x-24$

First notice that the sum of the coefficients is $0$ , so $f(1) = 0$ and $(x-1)$ is a factor of $f(x)$ .

$x^4-2x^3-13x^2+38x-24$

$=(x^4-x^3)-(x^3-x^2)-(14x^2-14x)+(24x-24)$

$=x^3(x-1)-x^2(x-1)-14x(x-1)+24(x-1)$

$=(x^3-x^2-14x+24)(x-1)$

Let $g(x) = x^3-x^2-14x+24$

By the rational roots theorem, any rational root of $g(x) = 0$ must be a factor of $24$ and hence one of:

$+-1$ , $+-2$ , $+-3$ , $+-4$ , $+-6$ , $+-12$ , $+-24$

We soon find $g(2) = 8-4-28+24 = 0$ ,

so $(x-2)$ is a factor of $g(x)$

$x^3-x^2-14x+24$

$= (x^3-2x^2) + (x^2-2x) - (12x-24)$

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

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

Then $x^2+x-12 = (x+4)(x-3)$ by finding two numbers whose product is $12$ and whose difference is $1$

Putting it all together:

$x^4-2x^3-13x^2+38x-24$

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

How do you factor x^4-2x^3-13x^2+38x-24x^4-2x^3-13x^2+38x-24?