Factoring cubic polynomial by grouping?

[jwpaine]

Hello, it's been a while sense I did this, and I seem to have forgotten a bit of it over the summer:

Can I factor x^3 + 9x^2 + 26x + 24 by grouping, just as it is?

If I do: (x^3 + 9x^2) + (26x + 24) the first binomial can be factored out to x^2(x+9) but if I factor the other binomial, they won't match:

My three binomial factors should come out to (x+2)(x+3)(x+4). Can I factor this by grouping without having to preform a rational roots test, and divide off the polynomial down until I have my three binomial factors? I'm trying to refresh myself a bit.

Thanks,
John.

 

[galactus]

Yes, JW, I beleive you can with some ingenuity.

Rewrite as:

\(\displaystyle \L\\x^{3}+7x^{2}+2x^{2}+12x+14x+24\)

\(\displaystyle \L\\x(x^{2}+7x+12)+2(x^{2}+7x+12)\)

\(\displaystyle \L\\(x+2)(x^{2}+7x+12)\)

\(\displaystyle \L\\(x+2)(x+3)(x+4)\)

There ya' go, JW

 

[jwpaine]

OK, so you broke up the 9x^2 and the 26x into separate terms, and then switches around the placement of the 12x and then 2x^2 so that you could group them..... tricky....

Thanks.