factor difference of two squares.
x^4-16
is this correct?
(x+2)(x-2)

Question

factor difference of two squares.
x^4-16
is this correct?
(x+2)(x-2)

e.mccormick · Answer

Have you ever used substitution to do squares?  Because technically, these are not squares at this point.

anonymous · Answer

No I have not. The question that I have here says, factor each difference of two square.

e.mccormick · Answer

Yes, I know what it says.  But is $x^4$ a square when you see it just like that?

anonymous · Answer

I meant in my book.

e.mccormick · Answer

So did I.  I am familliar with the question.  I have seen it in a few books, or ones almost exactly like it.

Think about this, you know how $(x^m)^n$ works?

anonymous · Answer

yep

e.mccormick · Answer

So, lets say we take $x^4$ and call it $(x^2)^2$, that us valid.  Then we say, Let $u=x^2$.  Now we can subsitute u inside those ( ) and get:
$u^2-16$

That make sense?

anonymous · Answer

yes

e.mccormick · Answer

Now you can factor that as a difference of squares and theen subsitute back the $x^2$

anonymous · Answer

You almost have it right...you're just forgetting to square you x's, which would give you $$(x^2+4)(x^2-4)$$ and then your x=2 and -2, if you needed to find those as well

e.mccormick · Answer

After you backsubstitute, yes, it would look lie what amberjordan2006.  However, the probelm would still not be finished.

anonymous · Answer

|dw:1367641155491:dw| I got this

e.mccormick · Answer

Yes.  And there is $(x^2+4)(x^2-4)\leftarrow$ another difference of squares there.  So you end up following the instructions, via a little substitution so you can use the rule for difference of squares.  Then you have to do the difference of squares one more time and you are done.

anonymous · Answer

(x+2)(x-2) This is what i got using this rule: (A^2-B^2)=(A+B)(A-B)

e.mccormick · Answer

Yes.  So that takes the place of that term.

anonymous · Answer

thanks

e.mccormick · Answer

np.  I am typing up a summary of it, since I had some time.

e.mccormick · Answer

To summarize.  Use $u=x^2$ to make it a difference of squares (and make the teacher happy cause it is not really needed.)  Use the rule $a^2+b^2=(a+b)(a-b)$ and get $(u+4)(u-4)$.  Substitute back the $x^2$ for $u$.

Now it is $(x^2+4)(x^2-4)\leftarrow$ Another difference of squares there.  So you end up following the instructions and doing it again!.

$(x^2+4)\leftarrow$ not a difference of squares or even able to be factored, so done.
$(x^2-4)\leftarrow$ a new, second difference of squares so you need to factor it.

The end result:

$(x^2+4)(x+2)(x-2)$

anonymous · Answer

got it!

e.mccormick · Answer

Yah, not too hard.  You can see how using $(x^m)^n$ you could do it without the substitution, but subs become important in calculus, so it is good to get a little used to them in algebra.

anonymous · Answer

thanks for your help. I'm scare of calculus! ;-)

e.mccormick · Answer

Ha!  Calculus is easy!  90% of calculus is NOT calculus.  It is tons of arithmetic, algebra, geometry, and trigonometry.  So the calculus part is generally short and easy. The pain in the anatomy part is using everything else to get it into the right form where you can do that simple step, and then clean it up after that simple step makes a huge mess out of it.

e.mccormick · Answer

Talk to any professor that teaches calculus.  Most people are not getting the calculus wrong.  They make mistakes in everything else.  Forget to distribute a negative, cancel the wrong thing in a quotent, do trig identities wrong, and so on.

anonymous · Answer

ha! I'll keep that in mind!