Simplifying an expression

**Wraith** · December 6th 2010, 03:14 PM

Hello,
I would like to know how to simplify this expression step-by-step:
Wolfram Alpha

If I try and do it by hand on paper, I get results like this:
http://www.upload.ee/image/969390/clipboard_upped.png
(I forgot to add bottom of fraction at the end, because top was already wrong, but the point of the problem remains...)

**I-Think** · December 6th 2010, 03:55 PM

Don't expand out until you have factorized as fully as possible.

In this case, we have
$\frac{(m+n)(m^2-n^2)-(m-n)(m^2+mn+n^2)}{(m+n)(m-n)}$

Note $m^2-n^2=(m+n)(m-n)$

That leaves us with
$\frac{(m+n)(m-n)(m+n)-(m-n)(m^2+mn+n^2)}{(m+n)(m-n)}$

$\frac{(m+n)^2-(m^2+mn+n^2)}{(m+n)}$

Answer should be visible now

**skeeter** · December 6th 2010, 03:56 PM

$\frac{m^2-n^2}{m-n} - \frac{m^3-n^3}{m^2-n^2}<br />$

common denominator ...

$\frac{(m+n)(m^2-n^2)}{(m+n)(m-n)} - \frac{m^3-n^3}{m^2-n^2}<br />$

$\frac{(m+n)(m^2-n^2)-(m^3-n^3)}{(m+n)(m-n)}<br />$

factor the difference of squares and cubes ...

$\frac{(m+n)(m+n)(m-n)-(m-n)(m^2+mn+n^2)}{(m+n)(m-n)}<br />$

divide out the common factor (m-n) ...

$\frac{(m+n)(m+n)-(m^2+mn+n^2)}{m+n}$

expand (m+n)(m+n) ...

$\frac{(m^2+2mn+n^2)-(m^2+mn+n^2)}{m+n}$

combine like terms ...

$\frac{mn}{m+n}$

**Wraith** · December 7th 2010, 10:27 AM

Thank you, I understand that exercise now

But, I came over this new one:

What is the common denominator here? I could use ${(x+2)(x-2)$ as a common denominator on first two blocks, but the last one has $(x-2)^2}$ under it, how do I deal with that?

This is where I am stuck..

**Soroban** · December 7th 2010, 10:53 AM

Hello, Wraith!

$\text{Simplify: }\;\dfrac{(m+n)(m^2-n^2) - (m-n)(m^2+mn + n^2)}{(m+n)(m-n)}$

We have: . $\dfrac{(m^3 - mn^2 + m^2n - n^3) - (m^3 - n^3)}{(m+n)(m-n)}$

. . . . . . $=\;\dfrac{m^3 - mn^2 + m^2n - n^3 - m^3 + n^3}{(m+n)(m-n)}$

. . . . . . $=\;\dfrac{m^2n - mn^2}{(m+n)(m-n)}$

. . . . . . $=\;\dfrac{mn(m-n)}{(m+n)(m-n)}$

. . . . . . $=\;\dfrac{mn}{m+n}$

**skeeter** · December 7th 2010, 02:03 PM

Originally Posted by Wraith

Thank you, I understand that exercise now

But, I came over this new one:

What is the common denominator here? I could use ${(x+2)(x-2)$ as a common denominator on first two blocks, but the last one has $(x-2)^2}$ under it, how do I deal with that?

This is where I am stuck..

note that the common denominator is $(x+2)(x-2)^2$

Thread: Simplifying an expression

LinkBack

Thread Tools

Display

Simplifying an expression

Similar Math Help Forum Discussions

Help simplifying expression

simplifying the expression

I need help simplifying this expression...

Once again... simplifying yet another expression

Simplifying an expression

Search Tags