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
Rearranging
$\displaystyle 1) x^2 +2(a^2 + b^2) + 3ax - b(3x + 5a)$
$\displaystyle x^2+2a^2+2b^2+3ax-3bx-5ab$
$\displaystyle x^2+3(a-b)x+(2a^2-5ab+2b^2)$
Now using the quadratic formula
$\displaystyle \frac{3(b-a)\pm{\sqrt{9a^2-18ab+9b^2-8a^2+20ab-8b^2}}}{2}$
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