can someone factor these problems until they cant be factored anymore thanks:)

a^3+a^2b-ab^2-b^3

x^4-10x^2+9

x^4-24x^2+144

a^3+2a^2-5a-10

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- Nov 30th 2006, 10:31 PMelfen_liedneed help on some factor problems
can someone factor these problems until they cant be factored anymore thanks:)

a^3+a^2b-ab^2-b^3

x^4-10x^2+9

x^4-24x^2+144

a^3+2a^2-5a-10 - Nov 30th 2006, 10:52 PMCaptainBlack
- Nov 30th 2006, 11:01 PMCaptainBlack
Clearly if $\displaystyle x=1$ this is zero so $\displaystyle x-1$ is a factor. Long division gives:

$\displaystyle

x^4-10x^2+9=(x-1)(x^3+x^2-9x-9)

$

and if $\displaystyle x=-1$ the second term on the right is zero, so $\displaystyle x+1$ is a factor

of that term, and you should be able to complete this yourself from this

point.

RonL - Nov 30th 2006, 11:29 PMelfen_lied
- Nov 30th 2006, 11:32 PMCaptainBlack
- Nov 30th 2006, 11:46 PMelfen_lied
- Nov 30th 2006, 11:58 PMCaptainBlack
- Dec 1st 2006, 12:06 AMelfen_lied
- Dec 1st 2006, 03:47 AMtopsquark
Perhaps a slightly different approach would work better for you.

$\displaystyle x^4-10x^2+9$

Notice how we only have even powers of x listed here. So to momentarily simplify the expression I'm going to define $\displaystyle y = x^2$:

$\displaystyle x^4-10x^2+9 = y^2 - 10y + 9$

This is a simple quadratic, which you should know how to factor.

I get that

$\displaystyle y^2 - 10y + 9 = (y - 1)(y - 9)$

Now we put the x's back in:

$\displaystyle x^4-10x^2+9 = y^2 - 10y + 9 = (y - 1)(y - 9) = (x^2 - 1)(x^2 - 9)$

$\displaystyle x^4-10x^2+9 = (x^2 - 1)(x^2 - 9)$

Now. Notice that each of these factors is of the form: $\displaystyle x^2 - a^2 = (x + a)(x - a)$, so

$\displaystyle x^4-10x^2+9 = (x^2 - 1)(x^2 - 9) = (x + 1)(x - 1)(x + 3)(x - 3)$

The same approach will also work on your third problem.

-Dan