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Mathematics 6 Online
OpenStudy (anonymous):

b2(p + 3) + q(p + 3) In factored form... I'm still working on it the way she taught me...

OpenStudy (bibby):

\(ab+ac=a(b+c)\)

OpenStudy (anonymous):

that absolutely does nothing to help me. I don't understand why but I think the answer is (b^2*p)(q+3)

Directrix (directrix):

b^2(p + 3) + q(p + 3) - Factoring by Grouping The terms b^2 *(p + 3) and q*(p + 3) have a common factor of (p + 3).

OpenStudy (anonymous):

That wasn't the right answer.

Directrix (directrix):

Factor out (p+3) in this way: b^2 *(p + 3) and q*(p + 3) = (p + 3)* ( b^2 + q)

Directrix (directrix):

By the commutative property, that answer can also be written as: (b^2 + q) ( p + 3)

OpenStudy (anonymous):

I was talking about my guess.

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