# factoring expressions

• Jan 11th 2010, 09:23 PM
jay1
factoring expressions
Please assist me in completely factoring the following expressions: (the numbers are exponents) 1)16t3 - 50t2 + 36t I have 2t (8t+9)(T+2) is this correct???
2) 48u4v4 - 18u2v2 - 3u8v5 I'm stuck on this one
3) 4q2 + 27r ? Thank you!
• Jan 11th 2010, 09:41 PM
bigwave
Quote:

Originally Posted by jay1
Please assist me in completely factoring the following expressions: (the numbers are exponents) 1)16t3 - 50t2 + 36t I have 2t (8t+9)(T+2) is this correct???
2) 48u4v4 - 18u2v2 - 3u8v5 I'm stuck on this one
3) 4q2 + 27r ? Thank you!

2)
\$\displaystyle 48u^4v^4-18u^2v^2-3u^8v^5\$

\$\displaystyle 3u^2v^2(16u^2v^2-6-u^6v^3)\$

are you sure these exponents are correct?
• Jan 12th 2010, 09:06 AM
jay1
factoring expressions
Yes, Big Wave thank you!
Can you assist me with questions 1 and 3? I would appreciate it greatly.
• Jan 12th 2010, 09:27 AM
masters
Quote:

Originally Posted by jay1
Please assist me in completely factoring the following expressions: (the numbers are exponents)

1)\$\displaystyle 16t^3 - 50t^2 + 36t\$

I have \$\displaystyle 2t (8t+9)(t+2)\$ is this correct???
Thank you!

Hi jay1,

Almost. You missed the sign.

\$\displaystyle 16t^3-50t^2+36t=2t(8t-9)(t-2)\$
• Jan 12th 2010, 09:34 AM
masters
Quote:

Originally Posted by jay1
Please assist me in completely factoring the following expressions: (the numbers are exponents)

3) \$\displaystyle 4q^2 + 27r\$ ? Thank you!

jay1,

It would appear that this one is already in its simplest form. Make sure you copied it correctly.
• Jan 12th 2010, 10:28 AM
jay1
Thanks I copied it incorrectly. It should read 4q2 + 27r4. Thanks for your help. Also to factor 2a3 - 128, is this prime?
• Jan 12th 2010, 10:38 AM
masters
Quote:

Originally Posted by jay1
Thanks I copied it incorrectly. It should read 4q2 + 27r4. Thanks for your help. Also to factor 2a3 - 128, is this prime?

Hi jay1,

You still can't do anything with \$\displaystyle 4q^2+27r^4\$

And for \$\displaystyle 2a^3-128\$, first factor out a 2.

\$\displaystyle 2(a^3-64)\$ and recognize that \$\displaystyle a^3-64\$ is the difference of two cubes. You should know how to handle that one, don't you?

\$\displaystyle 2(a^3-4^3)=\$
• Jan 13th 2010, 11:12 AM
jay1
Masters, 2 (a-4)3 ??? (the 3 being an exponent)