Issue on expanding -(X-2)^2

[drjsellis]

HI, I am teaching ,myself algebra and eventually calculus.

I am doing great with expanding, but I have an issue with one questions; which is basically finding the area of a shaded area, when a cube inside is taking up some space.

So, the large rectangle had the formula (x+7) * (3x-y+4)
Which I was able to expand to:
3x^2-xy+4x+21x-7y+28

If the square inside has the area (x-2)² then the formula would be:

3x^2-xy+4x+21x-7y+28 -(x-2)^2

its the -(x-2)^2 that is causing me an issue.

I can not for the life in me understand why -X^2+2x+2x-4 is the correct answer for -(x-2)^2 expansion. As in my head I am doing this

-(x-2)^2 = -1*(x-2)* -1*(x-2) = -(-x+2) * (-x+2) which I expand as X^2 -4x +4

So I guess what I should be doing is expanding -(x-2)*(x-2) which could give

My question is; how does -(x-2)^2 become -X^2+4x-4; where is my thought process betraying me?

With KR,

James

 

[firemath]

It is because of the troublesome little negative sign on the outside of your parentheses.

You must multiply like this:
(x-2)(x-2)
x^2-2x+4-2x

Now this is where the negative sign is added. Not before.
-(x^2-2x+4-2x)

If you move a negative sign into parentheses, you change the sign of all the values inside to the opposite sign.
-x^2+2x-4+2x

 

[drjsellis]

Ah thank you so much; so always you multiply before switching the signs over.

The rule of thumb with parentheses is that if there is a negative sign, all in the bracket have their signs revered; but on expanding, one does this after expansion.

so

-(x+3)^2 would be :

(x+3).(x+3) = +x^2+3x+3x+9 which then equals -X^2-6x-9

That was very helpful thanks

 

[Dr.Peterson]

Another way to think of it is that you are expanding this:

3x^2-xy+4x+21x-7y+28 - (x-2)^2​

You are subtracting (x-2)^2, which expands to x^2 - 4x + 4, so you have this:

3x^2-xy+4x+21x-7y+28 - (x^2 - 4x + 4)​

That means you are subtracting each term inside the parentheses, which changes their signs:

3x^2-xy+4x+21x-7y+28 - x^2 + 4x - 4​

Now you just combine like terms.

In terms of the negative, the main idea is the order of operations: the squaring is done (and therefore expanded) before either negation or subtraction. That's why -(x-2)^2 does not mean (-(x-2))^2 = (-x + 2)^2. The negation can't be done before squaring, because that would change the meaning.

 

[Steven G]

You can always think of -(x-2)^2 as 0 - (x-2)^2.

Or as Dr Peterson said (but paraphrased), you should know what you are subtracting BEFORE you subtract. From 3x^2-xy+4x+21x-7y+28 you are subtracting (x-2)^2 so you better know what it equals before you do the subtracting. Now (x-2)^2 = x^2 - 4x + 4.

Now you have to compute (3x^2-xy+4x+21x-7y+28) - ( x^2 - 4x + 4)

 

[pka]

HI, I am teaching ,myself algebra and eventually calculus.
its the -(x-2)^2 that is causing me an issue.

Frankly I prefer to answer what is asked: \(\displaystyle -(x-2)^2=~?\).
I assume that we are dealing with real numbers, \(\displaystyle \mathbb{R}\).
So do you understand that for every real number \(\displaystyle x\) it is a fact that \(\displaystyle -x^2\le 0~?\)
Thus \(\displaystyle -(x-2)^2=-(x^2-4x+4)=-x^2+4x-4\).

Using the expression \(\displaystyle -x^2+4x-4\) you should convince yourself that no matter for what value of \(\displaystyle x\) you pick the expression is not positive.
One way is to look at this graph.