Factor:

a)64x^3+8 b)72x^3-108x^2-140x c)x^3+3x^2-6x-8

I've become so depended on my calculator that I actually forget how to factor...help! :(

Thank you

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- Jan 5th 2008, 12:02 PMjohettFactoring
Factor:

a)64x^3+8 b)72x^3-108x^2-140x c)x^3+3x^2-6x-8

I've become so depended on my calculator that I actually forget how to factor...help! :(

Thank you - Jan 5th 2008, 12:22 PMgalactus
The first one is the sum of two cubes. Remember the factoring for the sum of two cubes?. $\displaystyle (x+y)(x^{2}-xy+y^{2})$

You have $\displaystyle 64x^{3}+8 = (4x)^{3}+(2)^{3}$

Now, use the formula.

For the second one. You can factor out 4x:

$\displaystyle 4x(18x^{2}-27x-35)$

Now, factor the quadratic. What two numbers when multiplied equal -630 (because (-35)(18) = -630) and when added equal -27.

Hmmmm, let's see......how about -42 and 15.

$\displaystyle 18x^{2}+15x-42x-35$

Group: $\displaystyle (18x^{2}+15x)-(42x+35)$

Factor out 3x in the first one and -7 out of the second:

$\displaystyle 3x(6x+5)-7(6x+5)$

$\displaystyle (3x-7)(6x+5)$

Don't forget the 4x we factored out in the beginning.

$\displaystyle 4x(3x-7)(6x+5)$

See how it's done now?.

Oh yeah, BTW, there is a way to remember the two different factorings for the cubes.

$\displaystyle SOAP: \;\ \;\ \boxed{S}ame, \;\ \boxed{O}pposite, \;\ \boxed{A}lways \;\ \boxed{P}ositive$

$\displaystyle x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$

$\displaystyle x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$

Note the signs in the factoring on the right side of the equals sign.

Then the first sign is the Same, then Opposite, then the last one is Always Positive. See what I mean?.