factoring

• Jan 15th 2009, 08:29 PM
MathMack
factoring
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
• Jan 15th 2009, 08:36 PM
Mush
Quote:

Originally Posted by MathMack
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

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 $\displaystyle -2n(16n^4-8n^2+1)$. This can be further factorised by letting $\displaystyle y=n^2$ and considering the quadratic!

$\displaystyle 16y^2-8y+1 = 0$

By the quadratic equation $\displaystyle y = \frac{1}{4}$ hence $\displaystyle 4y - 1$ is the only factor! Hence $\displaystyle 16y^2-8y+1 = (4y - 1)^2 = (4n^2 - 1)^2$

Hence the entire expression is

$\displaystyle -2n(4n^2 - 1)^2$

2) Use the quadratic equation to solve the equation $\displaystyle x^2+\frac{2}{3}x+\frac{1}{9} = 0$. You should get a single root $\displaystyle \frac{1}{3}$. This means that $\displaystyle (x - \frac{1}{3})$ is a factor. And this the discriminant is zero, in fact TWO of these are factors. Hence the answer is $\displaystyle (x - \frac{1}{3})(x - \frac{1}{3}) =(x - \frac{1}{3}) ^2$

3) Notice that this can be written as $\displaystyle (4m)^2+40mn+(5n)^2 = (4m)^2+4\times 5 \times 2mn+(5n)^2$. This can be written $\displaystyle (4m+5n)^2$
• Jan 15th 2009, 08:48 PM
MathMack
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.
• Jan 15th 2009, 08:56 PM
Mush
Quote:

Originally Posted by MathMack
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.