I have three I'm having problems with:

1.) 16n^3 - 32n^5 - 2n

2.) x^2 - 2/3x + 1/9

3.) 16m^2 + 40mn +25n^2

Printable View

- January 15th 2009, 09:29 PMMathMackfactoring
I have three I'm having problems with:

1.) 16n^3 - 32n^5 - 2n

2.) x^2 - 2/3x + 1/9

3.) 16m^2 + 40mn +25n^2 - January 15th 2009, 09:36 PMMush
1) Each term has two things in common! They're all divisible by -2, and they're all divisible by n. Hence they're all divisible by -2n. So that's your factor!

This gives . This can be further factorised by letting and considering the quadratic!

By the quadratic equation hence is the only factor! Hence

Hence the entire expression is

2) Use the quadratic equation to solve the equation . You should get a single root . This means that is a factor. And this the discriminant is zero, in fact TWO of these are factors. Hence the answer is

3) Notice that this can be written as . This can be written - January 15th 2009, 09:48 PMMathMack
on number 1, I understand that I take out the two and go from there, but I'm confused how to factor it after that.

- January 15th 2009, 09:56 PMMush