# Expanding parenthesis when a negative is involved

1. Oct 27, 2012

### educatingrob

Im completing Engineering Maths cover to cover in an attempt to get more familiar with maths as I finished my education many years ago without really understanding many basic maths concepts. This problem is at the back of the introduction to algebra.

(x-2y)^2 - (2x - y)^2

Now I can expand fairly easily:

x(x-2y) - 2y(x-2y) - (2x)(2x-y) - (-y)(2x-y)

x^2 - 2yx - 2yx + 4y^2 - 4x^2 ... Hmm

I know the rest as its in the book but I don't know the correct method for determining:
- (-y)(2x-y)

because of the negative at the front. without it, its easy:
-y * 2x = -2xy
-y * -y = y^2

thus:
-2xy + y^2

but then you have ... - -2xy + y^2

minus minus???

What I dont understand is how the negative before the brackets affects the result. When expanding is it:
- (-y)(2x-y)
-y * 2x = -2xy * -1 = 2xy
-y * -y = y^2 * -1 = -y^2

or maybe the - term belongs to all the last bit?

- (-2xy + y^2)
so its
-1 * (-2xy + y^2)
= +2xy - y^2

If someone could explain the rule dictated to expand when theres a negative, I would be grateful as I cant find an example or comment about how to think of this.

Of course theres the other FILO way (a+b)(c+d) = ac+bc+bd+ad

but thats just confusing me more WRT expanding the two elements of (x-2y)^2 - (2x - y)^2 because of the negative.

My maths is riddled with these inconsistencies where I just used to guess without understanding whats missing or what rule to follow.

(apologies for the stupid question, thanks for any help)

2. Oct 27, 2012

This is very elementary stuff... a high school teacher could probably explain it better. You can always replace a - by a multiplication with (-1). So $$- (-y)(2x-y) = (-1)\cdot (-y) \cdot (2x-y)$$ Now there are many ways to calculate the results. The easiest is this one: $$(-1)\cdot ((-1) \cdot y) \cdot (2x - y) = (-1)\cdot (-1) \cdot y \cdot (2x - y)= y \cdot (2x - y) = 2xy - y^2$$ because $$(-1)\cdot(-1)=1$$ Btw $$a - b = a + (-1)\cdot b$$ and your equation would usually be solved using the binomial theorem...

3. Oct 27, 2012

### haruspex

Looks to me that you got the right answer by both methods above.
You can think of -n as +(-1)*n. So - (-y)(2x-y) = +(-1) (-y)(2x-y). Multiplying the first two terms together gives +y(2x-y). Or multiplying the last two terms first gives +(-1)*(-2xy+y^2) = 2xy-y^2.

4. Oct 27, 2012

### arildno

The simplest, and safest way is to introduce a new parenthesis between the expression to be expanded, and then solve that paranthesis later. You can ALWAYS place as many parentheses you want about a single term.

So:
-(2x-y)^2=-((2x-y)^2). Now, internally, there is no dangerous minus sign, so you can proceed
-((2x-y)^2)=-(2x(2x-y)-y(2x-y))=
-(4x^2-2xy-2xy+y^2)=-(4x^2-4xy+y^2)=-4x^2+4xy-y^2

5. Oct 27, 2012

### educatingrob

Annoyingly Id just typed out a long response and the session timed out when I went to post it :grumpy:

Anyway,

I managed to factorize
(x-2y)^2-(2x-y)^2
to
3(y^2-x^2)
via, as you posted introducing (-1), more parenthesis.

i.e.
(x)(x-2y)(-2y)(x-2y)(-1)(2x)(2x-y)(-1)(-y)(2x-y)

The last time I was around high school teachers was 20 years ago and for whatever reason, I didnt get it then ;) Seems elementary now though.

Thanks all :)

Just looking at the binomial theorem.