SOLUTION: How do I factor this? {{{m4 -38m^2n^2 + 72n^4}}} The answer is {{{(m-6n)(m+6n)(m^2-2n^2)}}} I'm confused on how they got this answer. The 38 is not a typo, either.

Algebra ->  Expressions-with-variables -> SOLUTION: How do I factor this? {{{m4 -38m^2n^2 + 72n^4}}} The answer is {{{(m-6n)(m+6n)(m^2-2n^2)}}} I'm confused on how they got this answer. The 38 is not a typo, either.       Log On


   

Question 903823: How do I factor this?
m4+-38m%5E2n%5E2+%2B+72n%5E4
The answer is %28m-6n%29%28m%2B6n%29%28m%5E2-2n%5E2%29
I'm confused on how they got this answer. The 38 is not a typo, either.
Found 3 solutions by jim_thompson5910, josgarithmetic, MathTherapy:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
m%5E4-38m%5E2n%5E2%2B72n%5E4

m%5E4-2m%5E2n%5E2-36m%5E2n%5E2%2B72n%5E4 Break up -38m%5E2n%5E2 into -2m%5E2n%5E2-36m%5E2n%5E2 (see note below)

%28m%5E4-2m%5E2n%5E2%29%2B%28-36m%5E2n%5E2%2B72n%5E4%29 Pair up the terms

m%5E2%28m%5E2-2n%5E2%29%2B%28-36m%5E2n%5E2%2B72n%5E4%29 Factor m%5E2 from the first group

m%5E2%28m%5E2-2n%5E2%29-36n%5E2%28m%5E2-2n%5E2%29 Factor -36n%5E2 from the second group

%28m%5E2-36n%5E2%29%28m%5E2-2n%5E2%29 Factor out the overall GCF

%28m-6n%29%28m%2B6n%29%28m%5E2-2n%5E2%29 Factor m%5E2-36n%5E2 (difference of squares rule)


Note: the last coefficient is 72. We need to find factors of 72 that add to -38. Put another way, we need to find two numbers that multiply to 72 AND add to -38. Those two numbers are -2 and -36. That explains how I broke up -38m%5E2n%5E2 into -2m%5E2n%5E2-36m%5E2n%5E2 (notice the coefficients)

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Not remembering any instruction during Introductory Algebra from so long ago that dealt with factoring degree 4 trinomial, eventually further instruction combined from Introductory and College Algebra contained enough knowledge to use polynomial division for trying to handle something like you have; including Rational Roots Theorem, although you are not interested in roots here.

There is m^4 and n^4, the n^4 term having coefficient 72. Factors of 72 are 2, 3, 6, 12, 24,36.

You would want to test DIVISORS of m-2n, m%2B2n, m-3n, m%2B3n, m-6n, m%2B6n, ... possibly others. Your DIVIDEND must be formed as m%5E4%2B0m%5E3%2An%2B%28-38%29m%5E2n%5E2%2B0mn%5E3%2B72n%5E4;

After finding the quotient with remainder of zero, you next use whatever skills you have for factoring, or you might try another polynomial division on the quotient, choosing a divisor that you believe is worth testing.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

How do I factor this?
m4+-38m%5E2n%5E2+%2B+72n%5E4
The answer is %28m-6n%29%28m%2B6n%29%28m%5E2-2n%5E2%29
I'm confused on how they got this answer. The 38 is not a typo, either. 


m%5E4+-+38m%5E2n%5E2+%2B+72n%5E4+

m%5E4+-+38m%5E2n%5E2+%2B+72n%5E4+

The factors of 72 that sum to 72 (a * c), and differ by 38 (b), are 36 and 2

Therefore, m%5E4+-+38m%5E2n%5E2+%2B+72n%5E4 becomes:

m%5E4+-+36m%5E2n%5E2+-+2m%5E2n%5E2+%2B+72n%5E4

m%5E2%28m%5E2+-+36n%5E2%29+-+2n%5E2%28m%5E2+-+36n%5E2%29 -------- Factoring out GCF from each pair of binomials

%28m%5E2+-+2n%5E2%29%28m%5E2+-+36n%5E2%29

highlight_green%28highlight_green%28%28m%5E2+-+2n%5E2%29%28m+-+6n%29%28m+%2B+6n%29%29%29 -------- Factoring %28m%5E2+-+36n%5E2%29 further