Re: Factoring help, again...

Hey Nervous.

Try factoring the quadratic by getting the roots through the quadratic formula.

Re: Factoring help, again...

I thought quadratics were only second degree polynomials...?

EDIT: Ohhhhh I get it, you meant the first trinomial... OK...

Re: Factoring help, again...

Why in the world would you first **multiply** and then factor? You start with $\displaystyle -(x- 2)(x^2+ 2x+ 5)$ so obviously, **one** factor is x- 2. In order to factor $\displaystyle x^2+ 2x+ 5$ into factors with **integer** coefficients, I would note that 5 factors only as 1(5) but neither 5-1 nor 5+ 1 is equal to 2 so we **can't** factor with integer coefficients. Not every quadratic can be factored with integer (or even real) coefficients.

So instead, **complete** the square! $\displaystyle x^2+ 2x+ 1= (x+ 1)^2$ so that $\displaystyle x^2+ 2x+ 5= x^2+ 2x+ 1+ 4= (x+1)^2+ 4= (x+1)^2- (-4)= (x+ 1)^2- (2i)^2$. Now use the identity $\displaystyle a^2- b^2= (a- b)(a+ b)$.

Re: Factoring help, again...

That's a beautiful idea... God, I still have so much to learn...

Re: Factoring help, again...

Another one:

$\displaystyle 2x^3+11x^2-12x-36$

I tried grouping, I tried factoring out an x and using quadratic formula on $\displaystyle x^+11x-12$ I don't know how to do this one.

Re: Factoring help, again...

Edit

Quote:

Originally Posted by

**nervous** another one:

$\displaystyle 2x^3+11x^2-12x-36$

i tried grouping, i tried factoring out an x and using quadratic formula on $\displaystyle x^2+11x-12$ i don't know how to do this one.

Re: Factoring help, again...

Have you tried the "Rational Root Test?"

-Dan