3x^2(4x-12)^2 + x^3(2)(4x-12)(4) HALP

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The discussion focuses on factoring the expression 3x^2(4x-12)^2 + x^3(2)(4x-12)(4). The initial approach involves factoring out x^2(4x-12) and simplifying to arrive at 16x^2(x-3)(5x-9). Participants confirm the correctness of this factoring process and discuss the importance of recognizing factors. A new question is posed about factoring 4a^2c^2 - (a^2 - b^2 + c^2)^2, with suggestions to apply the difference of squares method and look for simplifications. The conversation emphasizes the relevance of these techniques in calculus and real-world applications.
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3x^2(4x-12)^2 + x^3(2)(4x-12)(4)
Factor this expression completely. This type of question occurs in calculus in using the "product rule".

Attempt:
I first factor out a x^2 (4x-12) factor out the 4 from this, then have 4x^2 (X-3)
The left overs are: 3(4x-12) + x(2)(4)
I factor out a 4 from the left overs, I now have. 16x^2(x-3) 3(x-3) + x(2)
Then I simplify: 16x^2 (x-3) 3(x-3) +2x = 16x^2 (x-3) (5x-9)


Which is the right answer, but I may have been biased into finding this as I did know the answer before hand. So is my reasoning sound? Maybe you could go through this yourself and show me your process, practice does not hurt :)
 
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When you factor out x^2 (4x-12) you have:
[x^2 (4x-12)]*[3(4x-12) + 8x] ... which is what you got. Then you factor out the 4:
[16 x^2 (x-13)]*[3(x-3)+2x] => [16 x^2 (x-3)]*[3x - 9 +2x] => [16 x^2 (x-3)]*[5x-9]
And then discard the extra brackets.

Like you, I did not notice the extra factor of 4 in (4x-12) on the first pass.
 
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Thanks for the reply. I have another question...

Factor this completely.

4a^2c^2 - (a^2 - b^2 + c^2)^2

I am so stumped on this one.

Thanks again for the previous answer.
 
Square the second term ... you should get six terms. One of these six can be combined with the first term of the original expression.

Then look for simplifications ...
 
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Hierophant said:
Thanks for the reply. I have another question...

Factor this completely.

4a^2c^2 - (a^2 - b^2 + c^2)^2

I am so stumped on this one.

Thanks again for the previous answer.

You often get things like this in exercises and even sometimes in real science problems. This is a "difference of two squares". You've probably seen the factorising of that. After using that there might be some further simplification possible.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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