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Mathematics
OpenStudy (anonymous):

198q3r2 – 184q2r2 + 18qr2 Can someone please help me factor this expression?

OpenStudy (anonymous):

2qr2(99q2-92q+9)

OpenStudy (anonymous):

You can factor that further into \[ 2qr^2(9q-1)(11q-9) \]

OpenStudy (anonymous):

? I guess I'm not understanding what to do next.... ?

OpenStudy (anonymous):

The expression as I wrote it is fully factored.

OpenStudy (anonymous):

oh ok. Well I appreciate it. BUT do you mind explaining, I'd like to know how to do it as well so I know how to do other questions like this.

OpenStudy (anonymous):

you take out the 2qr2, then since the part in the brackets is a quadratic expression, you can factorise further as demonstrated above

OpenStudy (anonymous):

First step is to look for components that are present in every term. In this case we found that to be true for 2, q, and \(r^2\), so we factored those three all out to get \(2qr^2(99q^2-92q+9).\) From there, we notice that \(99q^2-92q+9\) is a quadratic, so we look for a way to split it into binomials. You're looking for numbers that multiply to 99 and numbers that multiply to 9, and a way of combining them such that you get a middle term of -92. I found that \((9q-1)(11q-9)\) worked.

OpenStudy (anonymous):

Best way to practice factoring of quadratics is just to do a lot of them, they are easy to check because you can just multiply them back out to make sure that you did it right.

OpenStudy (anonymous):

Thank you so much for your help! :)

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