Factor completely: (6g^2+13gw+6w^2) (81c^6-3g^6) (12u^2-7u-12)
Favorites|Homepage
Subscriptions | sitemap
HOME > Mathematics > Factor completely: (6g^2+13gw+6w^2) (81c^6-3g^6) (12u^2-7u-12)

Factor completely: (6g^2+13gw+6w^2) (81c^6-3g^6) (12u^2-7u-12)

[From: ] [author: ] [Date: 11-05-14] [Hit: ]
use the quadratic formula.-1.2.3.......
I have another three situations that I am having problems with. The answers I came up with for these three problems, they were incorrect.

-
When nothing jumps out at me, I use the quadratic equation.

g = (-13w +/- sqrt((13w)^2 - 4 * 6 * 6w^2)) / (2 * 6)
g = (-13w +/- sqrt(169w^2 - 144w^2)) / 12
g = (-13w +/- sqrt(25w^2)) / 12
g = (-13w +/- 5w) / 12
g = -8w/12 or g = -18w/12
g = -2w/3 or g = -3w/2

(3g + 2w)(2g + 3w)

***

81c^6 - 3g^6

3 * (27c^6 - g^6)

Whenever you see a problem like this, see if you can get it in terms of a^2 - b^2 or a^3 - b^3

3 * ((3c^2)^3 - (g^2)^3)

3 * (3c^2 - g^2) * ((3c^2)^2 + (3c^2) * (g^2) + (g^2)^2)

3 * (3c^2 - g^2) * (9c^4 + 3c^2 g^2 + g^4)

***
The two minus signs says we're looking for something like

(? - ?)(? + ?)

12 = 1 * 12 or 2 * 6 or 3 * 4

I sort of realized that 4^2 = 16 and 3^2 = 9, and -16 + 9 = -7.

(4u + 3)(3u - 4)

If you can't work it out quickly, use the quadratic formula.

-
1. (3 x+2 y) (2 x+3 y)
2. 3 (3 c^2-y^2) (9 c^4+3 c^2 y^2+y^4)
3. (3 x-4) (4 x+3)
1
keywords: completely,13,gw,Factor,12,81,Factor completely: (6g^2+13gw+6w^2) (81c^6-3g^6) (12u^2-7u-12)
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .