Question

198q3r2 – 184q2r2 + 18qr2

Can someone please help me factor this expression?

Answer 1

2qr2(99q2-92q+9)

Answer 2

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

Answer 3

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

Answer 4

The expression as I wrote it is fully factored.

Answer 5

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.

Answer 6

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

Answer 7

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.

Answer 8

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.

Answer 9

Thank you so much for your help! :)