factoring

**MathMack** · January 15th 2009, 08:29 PM

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

**Mush** · January 15th 2009, 08:36 PM

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

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

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

Hence the entire expression is

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

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

3) Notice that this can be written as $(4m)^2+40mn+(5n)^2 = (4m)^2+4\times 5 \times 2mn+(5n)^2$ . This can be written $(4m+5n)^2$

**MathMack** · January 15th 2009, 08:48 PM

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.

**Mush** · January 15th 2009, 08:56 PM

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.

Please review my editted post!

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