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

2qr2(99q2-92q+9)

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

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

The expression as I wrote it is fully factored.

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.

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

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.

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.

Thank you so much for your help! :)

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