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April 1, 2015

April 1, 2015

Posted by **Mindy** on Sunday, December 1, 2013 at 7:14pm.

A.) 8a^4 b^5 + 3a^3 b^5 - 2a^3 b^4

B.) 8a^4 b^3 + 6a^3 b^5 - 2a^3 b^4

C.) 8a^4 b^3 + 6a^3 b^5 + 2a^3 b^4

D.) 8a^4 b^5 + 3a^3 b^5 + 2a^3 b^4

I'm guessing its A

- Math -
**Jai**, Sunday, December 1, 2013 at 7:34pm2a^2 b^3 (4a^2 + 3ab -ab)

Distribute to each:

2a^2 b^3 * 4a^2 = 8a^4 b^3

2a^2 b^3 * 3ab = 6a^3 b^4

2a^2 b^3 * (-ab) = -2a^3 b^4

Adding them,

8a^4 b^3 + 6a^3 b^4 -2a^3 b^4

= 8a^4 b^3 + 4a^3 b^4

Actually either there's no answer in the choices, or there is a typo. :3

- TYPO -
**Damon**, Sunday, December 1, 2013 at 7:36pm2a^2 b^3 (4a^2 + 3ab -ab)

Nah, no b^5 term results

Perhaps you mean

2a^2 b^3 (4a^2 + 3ab^2 -ab)

which is

8 a^4 b^3 + 6 a^3 b^5 - 2 a^3 b^4

which is B

- Math -
**Terri**, Sunday, December 1, 2013 at 7:38pmNone of those b/c the 2nd b is b^4

It is

8a^4b^3 + 6a^3b^4 - 2^3b^4

- Math -
**Mindy**, Sunday, December 1, 2013 at 8:04pmThank you Damon the answer was B

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