Expanding parenthesis when a negative is involved

In summary: I guess that's another way to look at it :)In summary, the individual terms in the expression (x-2y)^2 - (2x - y)^2 can be solved using the FOIL method or by using the binomial theorem. The negative sign in front of the second set of parentheses can be handled by introducing a new set of parentheses within the expression and solving it separately. This is a basic concept in mathematics and can be easily understood by thinking of -n as +(-1)*n. Additionally, placing multiple parentheses around a single term is always allowed.
  • #1
educatingrob
2
0
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 don't 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 there's a negative, I would be grateful as I can't find an example or comment about how to think of this.

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

but that's 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 what's missing or what rule to follow.

(apologies for the stupid question, thanks for any help)
 
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  • #2
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...
 
  • #3
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
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
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)

This is very elementary stuff... a high school teacher could probably explain it better.
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.
 

Related to Expanding parenthesis when a negative is involved

1. What is the purpose of expanding parentheses when a negative is involved?

Expanding parentheses when a negative is involved helps to simplify algebraic expressions by removing the parentheses and dealing with the negative sign appropriately.

2. How do you expand parentheses when a negative is involved?

To expand parentheses when a negative is involved, you can use the distributive property by multiplying the negative sign to each term inside the parentheses.

3. Can you provide an example of expanding parentheses when a negative is involved?

For example, if we have the expression -3(x + 2), we can distribute the negative sign to both terms inside the parentheses to get -3x - 6.

4. Why is it important to expand parentheses when a negative is involved?

Expanding parentheses when a negative is involved is important because it helps to simplify expressions and make them easier to solve.

5. Are there any special rules to keep in mind when expanding parentheses when a negative is involved?

Yes, it is important to remember that when multiplying a negative to a term inside parentheses, the sign of the term will also change. For example, -3(x + 2) becomes -3x - 6, not -3x + 6.

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