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Mathematics 14 Online
OpenStudy (anonymous):

Factor by Grouping: 4z² + 12z - az - 3a

Parth (parthkohli):

In factoring by grouping, you factor the first two terms and then the last two.

OpenStudy (anonymous):

So then the first two terms would come out to 4z(z + 3) ?

Parth (parthkohli):

Yes, and the last two?

OpenStudy (anonymous):

-a(z + 3)

Parth (parthkohli):

Okay, so you have \(4z(z + 3) - a(z + 3)\). Can you factor that further?

OpenStudy (anonymous):

Nope, not that I can see. Unless I can do something with both (z+3)'s...

Parth (parthkohli):

Yes, you can!

Parth (parthkohli):

For example, you have \(ax + bx\). You may factor it so it becomes \((a + b)x\). The same way, you can always apply the trick to this one ;)

OpenStudy (anonymous):

then in this case, I would replace x with (z + 3)? (4z - a)(z + 3)?

Parth (parthkohli):

Yes!

OpenStudy (anonymous):

Much obliged!

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