# Expand and simplify

• Mar 9th 2010, 09:11 PM
brumby_3
Expand and simplify
a) (x+1/y)^2
b) (x-(1/3y)+(2/5z))^2
• Mar 9th 2010, 09:19 PM
Prove It
Quote:

Originally Posted by brumby_3
a) (x+1/y)^2
b) (x-(1/3y)+(2/5z))^2

$\displaystyle \left(x + \frac{1}{y}\right)^2 = \left(x + \frac{1}{y}\right)\left(x + \frac{1}{y}\right)$.

FOIL it out.

I can't read the second. Has it got $\displaystyle \frac{1}{3y}$ and $\displaystyle \frac{2}{5z}$, or $\displaystyle \frac{1}{3}y$ and $\displaystyle \frac{2}{5}z$?
• Mar 9th 2010, 09:27 PM
brumby_3
Hi Prove it,
It's the second combination.
So for the first one would it be x^2 + (1/y)x + (1/y)x + (1/y)^2 or is it x^2 + (1/xy) + (1/xy) + (1/y)^2
???
• Mar 9th 2010, 09:37 PM
Prove It
Quote:

Originally Posted by brumby_3
Hi Prove it,
It's the second combination.
So for the first one would it be x^2 + (1/y)x + (1/y)x + (1/y)^2 or is it x^2 + (1/xy) + (1/xy) + (1/y)^2
???

The first combination.

And remember that $\displaystyle x\left(\frac{1}{y}\right) = \frac{x}{y}$.
• Mar 9th 2010, 09:39 PM
brumby_3
Cool, so my final answer for the first one is x^2 + (2x/y) + (1/y)^2
Am I right?
• Mar 9th 2010, 09:57 PM
Prove It
Quote:

Originally Posted by brumby_3
Cool, so my final answer for the first one is x^2 + (2x/y) + (1/y)^2
Am I right?

Yes. But remember that $\displaystyle \left(\frac{1}{y}\right)^2 = \frac{1}{y^2}$.
• Mar 10th 2010, 09:29 AM
Arcane10
Solution
a) (x+1/y)^2

(x+1/y)^2 = (x+1/y)(x+1/y)

(x+1/y)(x+1/y)= x^2+x/y+x/y+1/y

x^2+x/y+x/y+1/y= x^2+2x/y+1/y

• Mar 10th 2010, 04:59 PM
Prove It
Quote:

Originally Posted by Arcane10
a) (x+1/y)^2

(x+1/y)^2 = (x+1/y)(x+1/y)

(x+1/y)(x+1/y)= x^2+x/y+x/y+1/y

x^2+x/y+x/y+1/y= x^2+2x/y+1/y

1. There was no need to post this, as the question had already been solved.

2. If you're going to post a full solution, at least make sure you get it right...

$\displaystyle \left(x + \frac{1}{y}\right)^2 = x^2 + \frac{2x}{y} + \frac{1}{y\color{red}^2}$.
• Mar 12th 2010, 08:05 AM
Arcane10
and it was right
• Mar 12th 2010, 04:18 PM
Prove It
No it wasn't.

Reread my post, you'll notice something you missed.