Precalculus (algebra review)

[aimforthehead]

I have about 100 problems with a lot of factoring. (Difference of squares, sum of cubes, difference of cubes, etc.) But I need help on a few.

First is x^3+3x^2+3x+1.

What kind of problem is this...?

I did some steps that I thought appropriate and ended up with...

(3x+(x^2-x+1))+(x+1)

If anyone can double check this that'd be great

Edited August 28, 2011 by aimforthehead

[Cap'n Refsmmat]

[math](3x+(x^2-x+1))+(x+1) = 3x + x^2 + 2[/math]

 

So no, that's not the same as what you started with. If there's no multiplication, how can you get an x³?

 

In this case you have the cube of a binomial. This site gives some helpful tips on factoring it:

 

http://www.sosmath.com/algebra/factor/fac05/fac05.html

[aimforthehead]

Hmm okay, I've done a few wrong then...Thanks.

Also.

64(x-a)^3-x+a

I know difference of cubes, but the x+a is not cubed, so I am lost...

Edited August 28, 2011 by aimforthehead

[Cap'n Refsmmat]

I don't think that one actually factors very well, the way you've written it. I mean, you could maybe do this:

 

[math](x-a)(64(x-a)^2 - 1)[/math]

 

but that doesn't really simplify it any.

[DrRocket]

Hmm okay, I've done a few wrong then...Thanks.

Also.

64(x-a)^3-x+a

I know difference of cubes, but the x+a is not cubed, so I am lost...

 

[math]x+a = \ (^3\sqrt {x+a})^3[/math]

[aimforthehead]

Thanks for the time guys.

I got...

(4(x+a)- (^3sqrt {x+a}) (16(x+a)^2+4(x+a) (^3sqrt {x+a})+ (^3sqrt {x+a})^2)

 

with a=4(x+a) and b= (^3sqrt {x+a})

Edited August 28, 2011 by aimforthehead