# Thread: FACTORING! plz help <3 EXAM TOMORROW!

1. ## FACTORING! plz help <3 EXAM TOMORROW!

2. Originally Posted by Noelle

1. $20x^9y^9 - 8x^6y^7 + 36x^3y^5$

from here, the lowest power of x is 3, while of y is 5.. also, the coefficients have 4 as their highest common factor..

so you would have $20x^9y^9 - 8x^6y^7 + 36x^3y^5 = 4x^3y^5 \left( 5x^6y^4 - 2x^3y^2 + 9 \right)$

if you set $u = x^3y^2$, then you will have $u^2 = (x^3y^2)^2 = x^6y^4$, so that the one in the parenthesis would become $5u^2 - 2u + 9$, which is not factorable..
hence,

$20x^9y^9 - 8x^6y^7 + 36x^3y^5 = 4x^3y^5 \left( 5x^6y^4 - 2x^3y^2 + 9 \right)$

2. $15z^2 + 14z - 8$

notice that the factors of 15 are 1, 3, 5, 15 (positive and negative) and the factors of 8 are 1,2,4,8 (positive and negative)..

so what combinations of the factors of 15 and 8 that would give 14?? it is (5)(4) + (3)(-2) = 20 - 6..

(use the what so called FOIL method..)

thus you can factor $15z^2 + 14z - 8$ to $\left( 5z - 2 \right) \left( 3z + 4 \right)$..

3. do same as number two..

4. $\frac{k^2 +11k + 24}{k^2 + 13k + 40} \cdot \frac{k^2 + 5k}{k^2 - 3k - 18}$

so if you do FOIL method to every part, you should come up with

$\frac{(k+3)(k+8)}{(k+5)(k+8)} \cdot \frac{k(k+5)}{(k-6)(k+3)}$

form here, you see that there are k+8, k+3, k+5 in both numerator and denominator, so that they would cancel out (assuming that they are not zzeros)

so, after cancellation, only factors left are k in the numerator and k-6 in the denominator.. therefore

$\frac{k^2 +11k + 24}{k^2 + 13k + 40} \cdot \frac{k^2 + 5k}{k^2 - 3k - 18} = \frac{k}{k-6}$