# Factoring help, again...

#### Nervous

$$\displaystyle -(x-2)(x^2+2x+5)$$
$$\displaystyle -(x^3+2x^2+5x-2x^2+4x-10)$$
$$\displaystyle -(x^3+x-10)$$
$$\displaystyle -x^3-x+10$$

How do you factor this? I tried adding in a $$\displaystyle 0x^2$$ and grouping, but that didn't work out. The other group can't factor out to $$\displaystyle x-0$$...

#### chiro

MHF Helper
Hey Nervous.

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

#### Nervous

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

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

#### HallsofIvy

MHF Helper
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 $$\displaystyle a^2- b^2= (a- b)(a+ b)$$.

• 1 person

#### Nervous

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

#### Nervous

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

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

#### Nervous

Edit
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.

#### topsquark

Forum Staff
Have you tried the "Rational Root Test?"

-Dan