# Thread: Factoring the difference of two perfect squares

1. ## Factoring the difference of two perfect squares

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

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

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

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

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

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

2. $\displaystyle (x^{2})^{2}-(y^{3})^{2}$

There, now it is two perfect squares.

$\displaystyle (x^{2}-y^{3})(x^{2}+y^{3})$

3. 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 $\displaystyle (x^{2})^{2}-(y^{3})^{2}$ are both perfect squares if you wouldst mind elaborating.

4. 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 $\displaystyle (x^{2})^{2}-(y^{3})^{2}$ are both perfect squares if you wouldst mind elaborating.
There was no error.
$\displaystyle x^4 - y^6 = (x^2)^2 - (y^3)^2$

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

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

-Dan

5. 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 $\displaystyle x^4 - y^6$ has nothing to do with factoring the difference of two perfect squares?

IE
$\displaystyle 25z^2 - 81y^2$?

6. Originally Posted by cmf0106
So the problem at hand $\displaystyle x^4 - y^6$ has nothing to do with factoring the difference of two perfect squares?
It does. What I am saying is that $\displaystyle y^6 = (y^3)^2$ is a perfect square: the square of $\displaystyle y^3$.

-Dan

7. Originally Posted by cmf0106
So the problem at hand $\displaystyle x^4 - y^6$ has nothing to do with factoring the difference of two perfect squares?

$\displaystyle x^4 - y^6$

Both of these are perfect squares

Here is a numerical expample to help clarify.

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

So we can write both of the above as perfect squares

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

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

I hope this helps.