Expanding parenthesis when a negative is involved

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)

0xDEADBEEF

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 [tex]- (-y)(2x-y) = (-1)\cdot (-y) \cdot (2x-y) [/tex] Now there are many ways to calculate the results. The easiest is this one: [tex](-1)\cdot ((-1) \cdot y) \cdot (2x - y) = (-1)\cdot (-1) \cdot y \cdot (2x - y)= y \cdot (2x - y) = 2xy - y^2[/tex] because [tex](-1)\cdot(-1)=1[/tex] Btw [tex]a - b = a + (-1)\cdot b[/tex] and your equation would usually be solved using the binomial theorem...

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.

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

educatingrob

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

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.