Question 1030211
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?
<pre>
Since this true for all x,

{{{(x^2 + 3x + 6)(x^2 + ax + b) = x^4 + mx^2 + n}}}

then it's true when x=0

{{{((0)^2 + 3(0) + 6)((0)^2 + a(0) + b) = (0)^4 + m(0)^2 + n}}}

{{{6b=n}}}.  Substitute 6b for n

{{{(x^2 + 3x + 6)(x^2 + ax + b) = x^4 + mx^2 + 6b}}}

It's also true when x=1:

{{{((1)^2 + 3(1) + 6)((1)^2 + a(1) + b) = (1)^4 + m(1)^2 + 6b}}}

{{{10(1+a+b)=1+m+6b}}}

{{{10+10a+10b=1+m+6b}}}

{{{10a+4b-m=-9}}}

It's also true when x=-1:

{{{((-1)^2 + 3(-1) + 6)((-1)^2 + a(-1) + b) = (-1)^4 + m(-1)^2 + 6b}}}

{{{4(1-a+b)=1+m+6b}}}

{{{4-4a+4b=1+m+6b}}}

{{{-4a-2b-m=-3}}}

Multiply through by -1:

{{{4a+2b+m=3}}}

It's also true when x=2:

{{{((2)^2 + 3(2) + 6)((2)^2 + a(2) + b) = (2)^4 + m(2)^2 + 6b}}}

{{{16(4+2a+b)=16+4m+6b}}}

Divide through by 2

{{{8(4+2a+b)=8+2m+3b}}}

{{{32+16a+8b=8+2m+3b}}}

{{{16a+5b-2m=-24}}}

Solve the system:

{{{10a+4b-m=-9}}}
{{{4a+2b+m=3}}}
{{{16a+5b-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</pre>