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
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
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