How do you factor $P(x)= x^4 - 6x^3 - 8x^2$ ?

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Answer 1 · 2018-04-30T08:42:26

See explanation.

First you can notice that $x^2$ is a common factor, so we can write that:

$P(x)=x^2*(x^2-6x-8)$

To check if the second expression can be factorized we have to calculate its discriminant:

The expression is: $x^2-6x-8$ , so:

$a=1$ , $b=-6$ , $c=-8$ ,

the discriminant is:

$Delta=(-6)^2-4*1*(-8)=36+32=68$

The discriminant is greater than zero, so the equation has 2 distinct real roots:

$x_1=(6-sqrt(68))/2=3-2sqrt(17)$

and

$x_2=(6+sqrt(68))/2=3+2sqrt(17)$

So the complete factorization is:

$P(x)=x^2*(x-3+2sqrt(17))*(x-3-2sqrt(17))$

How do you factor P(x)= x^4 - 6x^3 - 8x^2P(x)= x^4 - 6x^3 - 8x^2?