Thread: 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

Rearranging






Now using the quadratic formula


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

Originally Posted by yshridhar

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

Let . Then your expressin becomes . The factors are irreducible quadratic factors.