How to factor? 12x^6-27y^4.
I know to take out the three, but I don't know what to do from there. Any help? Thanks.

- anonymous

- schrodinger

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- anonymous

once you take out the 3, you have the difference of two squares

- anonymous

\[3(4x^6-9y^4)\] the inner piece is like \(a^2-b^2=(a+b)(a-b)\) with \(a=2x^3, b=3y^2\)

- anonymous

Ah, okay, I didn't know the formula. Thank you.

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## More answers

- anonymous

May I ask for the formula if it is a^2+b^2, @satellite73 ?

- jim_thompson5910

`a^2+b^2` is a sum of squares and can't be factored (assuming 'a' and 'b' have nothing in common)

- anonymous

I'm working on the problem now, just gimme a minute and we'll see if I got it right. :-)

- anonymous

I got 2x^5+3y^2)(2x^4-3y^2)

- anonymous

what @jim_thompson5910 said
you can factor the "difference of two squares" but not the "sum of two squares"

- anonymous

hmm no

- anonymous

@satellite73 How did you get 2x^3?

- anonymous

the exponent for the \(x\) term should be \(3\) in both cases

- anonymous

that is a good question

- anonymous

i visualized \(4x^6\) as \((2x^3)^2\)

- anonymous

I thought you just subtract. Since to the squared is to, you'd minus two from 6.

- anonymous

So you divide the exponent?

- jim_thompson5910

satellite73 is using the rule
\[\LARGE (x^a)^b = x^{a*b}\]
notice how the inner exponent 'a' is being multiplied with the outer exponent b

- jim_thompson5910

so with 4x^6, satellite73 broke up the 6 into 3*2 and used the rule above

- anonymous

Does my understanding of it? You divide?

- anonymous

*does it work ,too?

- jim_thompson5910

yeah 6/2 = 3

- jim_thompson5910

and 4/2 = 2

- anonymous

Does that work in general or was I just lucky there?

- jim_thompson5910

well again, notice how 6 factors into 3*2
we want the *2 portion so we can ultimately have something squared
so that's why we divide by 2 to figure out that other factor

- jim_thompson5910

if we had say x^10, we'd break it up like 5*2
how do we find 5? well we can say 10/2 = 5

- anonymous

Ah, okay, I am working on another problem right now. It would mean so much if you could stay for a bit to see if I got it right. :-)

- anonymous

This may take more for more as me as it involves big numbers. Thanks for your patience.

- anonymous

Finally got the factors. 28 and 28.

- anonymous

Hold on a bit.

- anonymous

(4x^2+7)^3

- anonymous

Is that right?

- anonymous

- anonymous

I just realized I didn't give you the problem. XD

- anonymous

16x^4+56x^2+49

- anonymous

your answer is correct

- anonymous

Actually, according to the calculator, it's squared not cubed. :(

- anonymous

oh no it isn't the exponent should be 2

- anonymous

\[16x^4+56x^2+49\] is a "perfect square"

- anonymous

\[(a+b)^2=a^2+2ab+b^2\] in your case \(a=4x^2,b=7\)

- anonymous

Thank you. I'm fine for now, but I may be good. Thanks again for your time! :-)

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