# Last two, I promise (for tonite)

• Sep 13th 2006, 08:40 PM
Suwanee
Last two, I promise (for tonite)
I'm not sure if I know how to write these out, but here goes: (multiply and simplify)

k^2+11k+28 X k^2+8k
k^2+12k+32 X k^2+4k-21

m^2-8m + 15
m-5 m-5

That is m^2-8m over m-5 PLUS
15 over m-5
• Sep 13th 2006, 08:47 PM
Jameson
The first one (k^2+11k+28)(k^2+8k)=k^4+8k^3+11k^3+88k^2+28k^2+22 4k=k^4+19k^3+116k^2+224k

That looks like it to me unless I messed up my mental math. Do you see the method?
• Sep 13th 2006, 09:05 PM
Suwanee
Quote:

Originally Posted by Jameson
The first one (k^2+11k+28)(k^2+8k)=k^4+8k^3+11k^3+88k^2+28k^2+22 4k=k^4+19k^3+116k^2+224k

That looks like it to me unless I messed up my mental math. Do you see the method?

Not really, but I'm a bit dense tonite. My answer was nowhere near that.

I think I will go beddy-bye. It is 12:03 a.m. here. My eyes are soooo tired:eek:

Thanks. Bye.
• Sep 14th 2006, 05:19 AM
topsquark
Quote:

Originally Posted by Suwanee
m^2-8m + 15
m-5 m-5

Both fractions have the same denominator, so you can just add the numerators:
(m^2 - 8m + 15)/(m - 5)

Now, the numerator factors: m^2 - 8m + 15 = (m-5)(m-3), so we get
[(m-5)(m-3)]/(m-5) = m - 3 because the m-5 terms cancel.

-Dan