# Thread: Reducing to lowest terms

1. ## Reducing to lowest terms

Could someone help explain these problems? They've both stumped me.

(z^2-11z+24)/(5z^2-25z-120)

(48-3z^2)/(z^2-7z+12)

Not sure how to use the [itex] coding...

Thanks so much

2. Originally Posted by tony_rose88
Could someone help explain these problems? They've both stumped me.

(z^2-11z+24)/(5z^2-25z-120)

(48-3z^2)/(z^2-7z+12)

Not sure how to use the [itex] coding...

Thanks so much
Hints:
1) $z^2-11z+24 = (z-8)(z-3)$
2) $5z^2-25z-120 = 5(z^2 - 5z - 24) = 5(z-8)(z+3)$
3) $z^2-7z+12 = (z-3)(z-4)$
4) $48-3z^2 = 3(16 - z^2) = 3(4-z)(4+z)$

EDIT:Thank you Danshader. Edited my mistake

3. there is a small mistake there iso o.o" in 3)
$
z^2-7z+12 = (z-5)(z-2)
$

shouldn't it be:
$
z^2-7z+12 = (z-4)(z-3)
$

4. Originally Posted by Isomorphism
Hints:
1) $z^2-11z+24 = (z-8)(z-3)$
2) $5z^2-25z-120 = 5(z^2 - 5z - 24) = 5(z-8)(z+3)$
3) $z^2-7z+12 = (z-3)(z-4)$
4) $48-3z^2 = 3(16 - z^2) = 3(4-z)(4+z)$

EDIT:Thank you Danshader. Edited my mistake

So I've got $(z-8)(z-3)/5(z-8)(z+3)$ on the first one. Cancel out those z-8's cancel out so it leaves me with $(z-3)/(5z+3)$?

And on the second: $3(z-4)(z+4))/(z-3)(z-4)$. Cancel out the z-4's, I get $(3z+4)/(z-3)$

Are those correct?