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

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- Dec 7th 2009, 02:00 AMyshridharResolve 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, 05:32 AMI-Think
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 - Dec 8th 2009, 01:15 AMmr fantastic