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))#