SOLUTION: If {{{(x^2 + 3x + 6)(x^2 + ax + b) = x^4 + mx^2 + n}}} for integers a, b, m and n, what is the product of m and n?

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Question 1030211: If %28x%5E2+%2B+3x+%2B+6%29%28x%5E2+%2B+ax+%2B+b%29+=+x%5E4+%2B+mx%5E2+%2B+n
for integers a, b, m and n, what is the product of m and n?
Found 2 solutions by robertb, Edwin McCravy:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
after expansion. This is supposed to be equal to +x%5E4+%2B+mx%5E2+%2B+n
By equating coefficients, we obtain...
==> a+3 = 0,
m = b+3a+6,
3b+6a = 0, or b+2a = 0 after reduction,
6b = n
==> a = -3,
b = 6,
m = 6 - 9 +6 = 3, and
n = 6*6 = 36
==> mn = 3*36 = highlight%28108%29.

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
If %28x%5E2+%2B+3x+%2B+6%29%28x%5E2+%2B+ax+%2B+b%29+=+x%5E4+%2B+mx%5E2+%2B+n
for integers a, b, m and n, what is the product of m and n?
Since this true for all x,

%28x%5E2+%2B+3x+%2B+6%29%28x%5E2+%2B+ax+%2B+b%29+=+x%5E4+%2B+mx%5E2+%2B+n

then it's true when x=0



6b=n.  Substitute 6b for n

%28x%5E2+%2B+3x+%2B+6%29%28x%5E2+%2B+ax+%2B+b%29+=+x%5E4+%2B+mx%5E2+%2B+6b

It's also true when x=1:



10%281%2Ba%2Bb%29=1%2Bm%2B6b

10%2B10a%2B10b=1%2Bm%2B6b

10a%2B4b-m=-9

It's also true when x=-1:



4%281-a%2Bb%29=1%2Bm%2B6b

4-4a%2B4b=1%2Bm%2B6b

-4a-2b-m=-3

Multiply through by -1:

4a%2B2b%2Bm=3

It's also true when x=2:



16%284%2B2a%2Bb%29=16%2B4m%2B6b

Divide through by 2

8%284%2B2a%2Bb%29=8%2B2m%2B3b

32%2B16a%2B8b=8%2B2m%2B3b

16a%2B5b-2m=-24

Solve the system:

10a%2B4b-m=-9
4a%2B2b%2Bm=3
16a%2B5b-2m=-24

Then take the value for b, substitute it in 6b=n
to find n.  Take the value for m, multiply it by
the value you get for n and get mn.  You finish.

Edwin