1. ## Stumbling! Help!!

Need help on how to approach these specific probs.

expand:
(2x^2-1)^2

Factor completely (factor common first)
y^4–6y^3+9y^2

factor/simplify lowest terms
9x^2-1 * x^2-1
__x-1 ___6x^2-2x

1 - 1
x_ x+2

2. Originally Posted by dude

expand:
(2x^2-1)^2
On this one just re-write it as $\displaystyle (2x^2-1)(2x^2-1)$ and foil it out (use the distributive law). After foiling and collecting like terms you should get $\displaystyle 4x^4-4x^2+1$

3. expand:
(2x^2-1)^2

=(2x^2 -1)(2x^2 -1)
=4x^4-2x^2-2x^2+1
=4x^4 -4x^2 +1

Factor completely (factor common first)
y^4–6y^3+9y^2

=y^2(y^2 -6y +9)
=y^2(y-3)(y-3)
or y^2(y-3)^2

factor/simplify lowest terms
9x^2-1 * x^2-1
______ ______
x-1 6x^2-2x

=(9x^2-1)/(x-1)*(x^2-1)/(6x^2-2x)
=(3x-1)(3x+1)/(x-1)*((x+1)x-1)/2x(3x-1)
=(3x+1)(x+1)/2x

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

4. Originally Posted by dude

Factor completely (factor common first)
y^4–6y^3+9y^2
On this one try to find factors that each term have in common i.e. is there some way that you can re-write each term such that they each have a similar item?

for example I can re-write the original statement as
$\displaystyle y^2(y^2)-y^2(6y^1)+y^2(9)$ notice that if you multiply each of them back together it would make the same statement that we started with, and even more, notice that they each have a $\displaystyle y^2$ in common that we can factor out.

so if we factor out the $\displaystyle y^2$ we get $\displaystyle y^2(y^2-6y+9)$ Once again notice that if we multiply everything back out we will get the same as we started with.

Now we can factor what we have left to $\displaystyle y^2(y-3)(y-3)$ That last factoring is a little tricky, it is like undoing foiling.
so we should finish with $\displaystyle y^2(y-3)^2$

Hope this helps a little.