# Thread: help on some algebra problems plz!

1. ## help on some algebra problems plz!

Simplify .

----------------------------------
If 4ax^2 = 3y . . .then A/Y = ?
1) 12x^2
2) (4x^2)/3
3) 3/(4x^2)
4)12
5) 4x^2
-------------------------------------
Which of the following is NOT a factor of ?

2. Originally Posted by rafaeli
Simplify .

Use these properties:

For real numbers $a, b, m, n$:

$a^ma^n = a^{m + n}$

$\frac{a^m}{a^n} = a^{m - n}$

$a^{m/n} = \sqrt[n]{a^m}$

$(a^m)^n = a^{mn}$

$(ab)^m = a^mb^m$

$\left(\frac ab\right)^m = \frac{a^m}{b^m}$

I'll do the first one for you:

$\sqrt[3]{x^2y^6} + \sqrt{x^{4/3}y^4}$

$=\left(x^2y^6\right)^{1/3} + \left(x^{4/3}y^4\right)^{1/2}$

$=x^{2/3}y^{6/3} + x^{4/6}y^{4/2}$

$=x^{2/3}y^2 + x^{2/3}y^2$

$=2x^{2/3}y^2 = 2y^2\sqrt[3]{x^2}$

Originally Posted by rafaeli
If 4ax^2 = 3y . . .then A/Y = ?
1) 12x^2
2) (4x^2)/3
3) 3/(4x^2)
4)12
5) 4x^2
I assume you are looking for $\frac ay$? Note that variables written in a different case (e.g. $x$ and $X$) are usually considered distinct.

For this one, just solve for $\frac ay$:

$4ax^2 = 3y\Rightarrow \frac{4ax^2}y = 3$

and continue from there.

Originally Posted by rafaeli
Which of the following is NOT a factor of ?
Factor the expression first: what is the greatest common factor of 162 and 2? The answer should be fairly obvious: it's 2. So we have

$162x^4 - 2y^8 = 2\left(81x^4 - y^8\right)$

Now notice that the second factor is a difference of squares, because $81x^4 - y^8 = \left(9x^2\right)^2 - \left(y^4\right)^2$. You can factor it using the identity

$a^b - b^2 = (a + b)(a - b)$.

Then you should end up with another difference of squares which you can further factor.

Once the expression is completely factored, it should be easy to see which of those choices is not a factor.