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

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- Feb 23rd 2008, 12:23 PM #1

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- Feb 23rd 2008, 02:43 PM #2

- Feb 23rd 2008, 02:52 PM #3

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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)

- Feb 23rd 2008, 02:53 PM #4
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.