1. ## 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

2. Rearranging
$\displaystyle 1) x^2 +2(a^2 + b^2) + 3ax - b(3x + 5a)$

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

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

$\displaystyle \frac{3(b-a)\pm{\sqrt{9a^2-18ab+9b^2-8a^2+20ab-8b^2}}}{2}$

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

3. Originally Posted by yshridhar
[snip]
2) x^4 + 6x^3 + 4x^2 - 15x + 6
Thanking you
Regards
Sreedhar
Let $\displaystyle u = x^2 + 3x$. Then your expressin becomes $\displaystyle u^2 - 5u + 6 = (u - 3)(u - 2) = (x^2 + 3x - 3)(x^2 + 3x - 2)$. The factors are irreducible quadratic factors.

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