anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
once you take out the 3, you have the difference of two squares
anonymous
  • 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
  • anonymous
Ah, okay, I didn't know the formula. Thank you.

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anonymous
  • anonymous
May I ask for the formula if it is a^2+b^2, @satellite73 ?
jim_thompson5910
  • 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
  • anonymous
I'm working on the problem now, just gimme a minute and we'll see if I got it right. :-)
anonymous
  • anonymous
I got 2x^5+3y^2)(2x^4-3y^2)
anonymous
  • anonymous
what @jim_thompson5910 said you can factor the "difference of two squares" but not the "sum of two squares"
anonymous
  • anonymous
hmm no
anonymous
  • anonymous
@satellite73 How did you get 2x^3?
anonymous
  • anonymous
the exponent for the \(x\) term should be \(3\) in both cases
anonymous
  • anonymous
that is a good question
anonymous
  • anonymous
i visualized \(4x^6\) as \((2x^3)^2\)
anonymous
  • anonymous
I thought you just subtract. Since to the squared is to, you'd minus two from 6.
anonymous
  • anonymous
So you divide the exponent?
jim_thompson5910
  • 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
  • jim_thompson5910
so with 4x^6, satellite73 broke up the 6 into 3*2 and used the rule above
anonymous
  • anonymous
Does my understanding of it? You divide?
anonymous
  • anonymous
*does it work ,too?
jim_thompson5910
  • jim_thompson5910
yeah 6/2 = 3
jim_thompson5910
  • jim_thompson5910
and 4/2 = 2
anonymous
  • anonymous
Does that work in general or was I just lucky there?
jim_thompson5910
  • 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
  • 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
  • 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
  • anonymous
This may take more for more as me as it involves big numbers. Thanks for your patience.
anonymous
  • anonymous
Finally got the factors. 28 and 28.
anonymous
  • anonymous
Hold on a bit.
anonymous
  • anonymous
(4x^2+7)^3
anonymous
  • anonymous
Is that right?
anonymous
  • anonymous
@satellite73 ?
anonymous
  • anonymous
I just realized I didn't give you the problem. XD
anonymous
  • anonymous
16x^4+56x^2+49
anonymous
  • anonymous
your answer is correct
anonymous
  • anonymous
Actually, according to the calculator, it's squared not cubed. :(
anonymous
  • anonymous
oh no it isn't the exponent should be 2
anonymous
  • anonymous
\[16x^4+56x^2+49\] is a "perfect square"
anonymous
  • anonymous
\[(a+b)^2=a^2+2ab+b^2\] in your case \(a=4x^2,b=7\)
anonymous
  • anonymous
Thank you. I'm fine for now, but I may be good. Thanks again for your time! :-)

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