Factoring

galactus · Jan 15, 2007, 12:03 PM

Quote:

]How do I factor these problems

X^3-2x^2-4x-8

This one is not easily factorable. It has 2 complex roots.

Are you sure you don't mean $x^{3}-2x^{2}-4x+8$

Otherwise, as you have it, the factorization is(which I pumped out with Maple):

$(x+\frac{14403}{17161}+\frac{23073}{19028}I)\left( x+\frac{14403}{17161}-\frac{23073}{19028}I\right)(x-\frac{63025}{17133})$

Quote:

x^7y^2-4x^5yy^2-21x^3y^2

Is this what you mean:

$x^{7}y^{2}-4x^{5}y^{2}-21x^{3}y^{2}$ ?.

The why I have highlighted. I suppose it's errant?.

If this is correct, then factor out $x^{3} \;\ and \;\ y^{2}$ :

$x^{3}y^{2}(x^{4}-4x^{2}-21)$

Factor in the parentheses by letting $u=x^{2}$

$(u^{2}-4u-21)=(u-7)(u+3)$

Resub:

$x^{3}y^{2}(x^{2}-7)(x^{2}+3)$

Elisha Grey · Jan 17, 2007, 09:17 AM

If you mean x^3 - 2x^2 - 4x + 8, then this equals
X^2(x-2) - 4(x-2) = (x^2 - 4) (x-2) = (x+2)(x-2)^2

Factoring

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