# factoring (common factoring complex trinomials)

• Nov 5th 2006, 04:57 PM
smallman_11
factoring (common factoring complex trinomials)
so the exuation is :

2x^2+18x+40

I know I have to common factor first, but I don't know exaclly where to go from there. It's grade 10 accademic math..I duuno if that means anything?

so I put it so it was like this:

2(x^2+9x+20)

and I duuno where to go from there?

Any help would be appreciated.
• Nov 5th 2006, 05:05 PM
ThePerfectHacker
Quote:

Originally Posted by smallman_11
so the exuation is :

2x^2+18x+40

I know I have to common factor first, but I don't know exaclly where to go from there. It's grade 10 accademic math..I duuno if that means anything?

so I put it so it was like this:

2(x^2+9x+20)

and I duuno where to go from there?

Any help would be appreciated.

And how does that factor?
$x^2+9x+20=(x+4)(x+5)$
$2(x+4)(x+5)$.
• Nov 5th 2006, 05:19 PM
smallman_11
Thank you
• Nov 6th 2006, 12:02 AM
earboth
Quote:

Originally Posted by smallman_11
...
2(x^2+9x+20)

and I duuno where to go from there?
Any help would be appreciated.

Hello, smallman,

I presume that you are a little bit uncertain how to factorize
$x^2+9x+20$

To do this probelm you must know the binomial formulae:
(a+b)² = a² + 2ab + b² [1]
(a-b)² = a² - 2ab + b² [2]
(a+b)(a-b) = a² - b² [3]

The formula [3] is very important with your problem:
1. Complete square with the first two summands by adding a third summand and subtract this summand immediately:
$x^2+9x+20 = x^2 + 9x +\left( \frac{9}{2} \right)^2-\left( \frac{9}{2} \right)^2 +20$ = $\left( x^2 + 9x +\left( \frac{9}{2} \right)^2 \right)-\left( \frac{9}{2} \right)^2 +20$
2. Unite the first three summands to a binom [1] and the last two summands to a square:
$\left( x +\left( \frac{9}{2} \right) \right)^2-\left( \frac{1}{4} \right)$

3. Now you've got a term similar to the RHS of formula [3]. Use the LHS of this formula:

$\left( x +\left( \frac{9}{2} \right) \right)^2-\left( \frac{1}{2} \right)^2 = \left(x +\left( \frac{9}{2} \right)+\left( \frac{1}{2} \right) \right) \cdot \left(x +\left( \frac{9}{2} \right)-\left( \frac{1}{2} \right) \right)$

4. Finally you get:

$x^2 + 9x +20 = \left(x +\left( \frac{9}{2} \right)+\left( \frac{1}{2} \right) \right) \cdot \left(x +\left( \frac{9}{2} \right)-\left( \frac{1}{2} \right) \right) = (x+5)(x+4)$

That's the TPH's result.

EB