Factoring the difference of two perfect squares

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June 1st 2008, 05:15 PM
cmf0106

Factoring the difference of two perfect squares

The author starts by giving the example $25z^2 - 81y^2$ , which equals $(5z + 9y) (5z - 9y)$ .

Both 81, and 25 are perfect squares so this makes sense.

$\sqrt{25} = 5*5, \sqrt{9} = 3*3$

Then the author uses this example factoring $x^4 - y^6$
Which equals $(x^2 + y^3)(x^2 - y^3)$ .

But this does not make sense to me
$\sqrt{4} = 2 * 2, \sqrt{6} = 2.45...$

Can some one please help clarify what I am doing wrong and how the author is pulling y^3 out of y^6 in the second example?

Many thanks
June 1st 2008, 05:18 PM
galactus

$(x^{2})^{2}-(y^{3})^{2}$

There, now it is two perfect squares.

$(x^{2}-y^{3})(x^{2}+y^{3})$
June 1st 2008, 05:21 PM
cmf0106

Galactus forgive me for being math dense, but are you rearranging the problem because the original problem had an error?

Also I do not see how $<br /> (x^{2})^{2}-(y^{3})^{2}<br />$ are both perfect squares if you wouldst mind elaborating.
June 1st 2008, 06:09 PM
topsquark

Quote:

Originally Posted by cmf0106

Galactus forgive me for being math dense, but are you rearranging the problem because the original problem had an error?

Also I do not see how $<br /> (x^{2})^{2}-(y^{3})^{2}<br />$ are both perfect squares if you wouldst mind elaborating.

There was no error.
$x^4 - y^6 = (x^2)^2 - (y^3)^2$

This follows from the general rule
$(z^p)^q = z^{pq}$

Perhaps one more clarification: We are not taking the square root of the exponent, we are dividing it by 2.

-Dan
June 1st 2008, 06:33 PM
cmf0106

Quote:

Originally Posted by topsquark

Perhaps one more clarification: We are not taking the square root of the exponent, we are dividing it by 2.

-Dan

So the problem at hand $x^4 - y^6$ has nothing to do with factoring the difference of two perfect squares?

IE
$25z^2 - 81y^2$ ?
June 1st 2008, 06:39 PM
topsquark

Quote:

Originally Posted by cmf0106

So the problem at hand $x^4 - y^6$ has nothing to do with factoring the difference of two perfect squares?

It does. What I am saying is that $y^6 = (y^3)^2$ is a perfect square: the square of $y^3$ .

-Dan
June 1st 2008, 06:40 PM
TheEmptySet

Quote:

Originally Posted by cmf0106

So the problem at hand $x^4 - y^6$ has nothing to do with factoring the difference of two perfect squares?

$x^4 - y^6$

Both of these are perfect squares

Here is a numerical expample to help clarify.

lets say that x=2 then $x^4=16$
but $x^2=4$ and $(x^2)^2=x^4=16=4^2$

So we can write both of the above as perfect squares

$(x^2)^2 \\\ (y^3)^2$

Remember that if you have $(x^a)^b=x^{a \cdot b}$

I hope this helps.