# Factoring Questions <3 pplz help exam tomorrow :(

• Dec 7th 2007, 08:55 PM
Noelle
Factoring Questions <3 pplz help exam tomorrow :(
i've been cramming all day... im so tired! i still have to learn so much more :(

http://i83.photobucket.com/albums/j3.../factoring.jpg
• Dec 7th 2007, 10:39 PM
red_dog
1) $\displaystyle 20x^9y^9-8x^6y^7+36x^3y^5=4x^3y^5(5x^6y^4-2x^3y^2+9)$

2), 3), 4):
$\displaystyle aX^2+bX+c=a(X-x_1)(X-x_2)$
where $\displaystyle x_1, \ x_2$ are the roots of the quadratic $\displaystyle ax^2+bx+c=0$.

I'll do it for 2):
$\displaystyle 15z^2+14z-8=0\Rightarrow z_1=-\frac{4}{3}, \ z_2=\frac{2}{5}$
Then $\displaystyle 15z^2+14z-8=15\left(z+\frac{4}{3}\right)\left(z-\frac{2}{5}\right)=\left[3\left(z+\frac{4}{3}\right)\right]\left[5\left(z-\frac{2}{5}\right)\right]=(3z+4)(5z-2)$

Now you can do the others.
• Dec 7th 2007, 11:11 PM
angel.white
Quote:

Originally Posted by red_dog
1) $\displaystyle 20x^9y^9-8x^6y^7+36x^3y^5=4x^3y^5(5x^6y^4-2x^3y^2+9)$

2), 3), 4):
$\displaystyle aX^2+bX+c=a(X-x_1)(X-x_2)$
where $\displaystyle x_1, \ x_2$ are the roots of the quadratic $\displaystyle ax^2+bx+c=0$.

I'll do it for 2):
$\displaystyle 15z^2+14z-8=0\Rightarrow z_1=-\frac{4}{3}, \ z_2=\frac{2}{5}$
Then $\displaystyle 15z^2+14z-8=15\left(z+\frac{4}{3}\right)\left(z-\frac{2}{5}\right)=\left[3\left(z+\frac{4}{3}\right)\right]\left[5\left(z-\frac{2}{5}\right)\right]=(3z+4)(5z-2)$

Now you can do the others.

Wow, I never knew that, that is neat!

(always seemed like there ought to be a way to use the roots in factoring, but I never actually knew how)

Another attempt is to note that $\displaystyle 20-6=14$ :eek:
\displaystyle \begin{aligned} 15z^2 + 14z - 8 &= 15z^2 + (20z - 6z) - 8\\ &= \left( {15z^2 + 20z} \right) - (6z + 8)\\ &= 5z(3z + 4) - 2(3z + 4)\\ &= (3z + 4)(5z - 2). \end{aligned}