Factoring help, again...

Oct 2011
90
2
\(\displaystyle -(x-2)(x^2+2x+5)\)
\(\displaystyle -(x^3+2x^2+5x-2x^2+4x-10)\)
\(\displaystyle -(x^3+x-10)\)
\(\displaystyle -x^3-x+10\)

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
Sep 2012
6,608
1,263
Australia
Hey Nervous.

Try factoring the quadratic by getting the roots through the quadratic formula.
 
Oct 2011
90
2
I thought quadratics were only second degree polynomials...?


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

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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 \(\displaystyle a^2- b^2= (a- b)(a+ b)\).
 
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Oct 2011
90
2
That's a beautiful idea... God, I still have so much to learn...
 
Oct 2011
90
2
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.
 
Oct 2011
90
2
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
Jan 2006
11,602
3,457
Wellsville, NY
Have you tried the "Rational Root Test?"

-Dan