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

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Answer 1 · 2018-04-30T08:42:26

See explanation.

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

$P (x) = x^{2} \cdot (x^{2} - 6 x - 8)$ $P(x)=x^2*(x^2-6x-8)$

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

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

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

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)=x4−6x3−8x2P(x)= x^4 - 6x^3 - 8x^2?