# Resolve into factors

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• Dec 7th 2009, 03:00 AM
yshridhar
Resolve into factors
Resolve into factors :
1) x^2 +2(a^2 + b^2) + 3ax - b(3x + 5a)
2) x^4 + 6x^3 + 4x^2 - 15x + 6
Thanking you
Regards
Sreedhar
• Dec 7th 2009, 06:32 AM
I-Think
Rearranging
$1) x^2 +2(a^2 + b^2) + 3ax - b(3x + 5a)$

$x^2+2a^2+2b^2+3ax-3bx-5ab$

$x^2+3(a-b)x+(2a^2-5ab+2b^2)$

Now using the quadratic formula
$\frac{3(b-a)\pm{\sqrt{9a^2-18ab+9b^2-8a^2+20ab-8b^2}}}{2}$

Should be able to solve from there, solve for x and these are your factors

2) x^4 + 6x^3 + 4x^2 - 15x + 6
This equation is in its simplest form.
Any polynomial whose coefficients are integers will, if it has an integer solution, have an integer solution that is a factor of the coefficient of the zeroth order term i.e. the constant. For example look at the general quartic:

a x^4 + b x^3 + c x^2 + d x + e = 0

where the coefficients a through e are integers. Well if there is an integer solution to this then it will be a factor of e.
No factor of 6 makes this equation, hence it's in its simplest form
• Dec 8th 2009, 02:15 AM
mr fantastic
Quote:

Originally Posted by yshridhar
[snip]
2) x^4 + 6x^3 + 4x^2 - 15x + 6
Thanking you
Regards
Sreedhar

Let $u = x^2 + 3x$. Then your expressin becomes $u^2 - 5u + 6 = (u - 3)(u - 2) = (x^2 + 3x - 3)(x^2 + 3x - 2)$. The factors are irreducible quadratic factors.