Expanding parenthesis when a negative is involved

  • Thread starter Thread starter educatingrob
  • Start date Start date
  • Tags Tags
    Expanding Negative
educatingrob
Messages
2
Reaction score
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)
 
Mathematics news on Phys.org
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...
 
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.
 
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
 
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.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Thread 'Imaginary pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Replies
3
Views
504
Replies
4
Views
1K
Replies
9
Views
3K
Replies
3
Views
2K
Replies
6
Views
2K
Replies
2
Views
984
Back
Top