MHB Did I misinterpret the sign attached to a term when factoring?

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The discussion revolves around the correct factoring of the expression xy+x-2y-2. The original attempt resulted in x(y+1)-2(y-1), which was incorrect due to a misunderstanding of sign when factoring out -2. The correct factorization is x(y+1)-2(y+1), leading to (x-2)(y+1). The key takeaway is that when factoring out a negative, the signs within the parentheses must change, highlighting the importance of careful sign management in algebraic manipulation. Understanding these nuances is crucial for accurate factoring in future problems.
find_the_fun
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I wanted to factor [math]xy+x-2y-2[/math]
I got [math]x(y+1)-2(y-1)[/math] and got stuck

I tried somethings out and noticed [math](x-2)(y+1)=xy+x-2y-2[/math] so how come I did get stuck? Did I extract the [math]2[/math] incorrectly with the sign? For example should I have interpreted it at [math]-2y-2[/math] not [math]2y-2[/math]?
 
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find_the_fun said:
I wanted to factor [math]xy+x-2y-2[/math]
I got [math]x(y+1)-2(y-1)[/math] and got stuck

I tried somethings out and noticed [math](x-2)(y+1)=xy+x-2y-2[/math] so how come I did get stuck? Did I extract the [math]2[/math] incorrectly with the sign? For example should I have interpreted it at [math]-2y-2[/math] not [math]2y-2[/math]?

Yes, you simply factored incorrectly.

[math]xy+x-2y-2=x(y+1)-2(y+1)=(x-2)(y+1)[/math]

When you factor a negative factor from a group of terms, the signs of everything within the resulting parentheses must change.
 
MarkFL said:
Yes, you simply factored incorrectly.

[math]xy+x-2y-2=x(y+1)-2(y+1)=(x-2)(y+1)[/math]

When you factor a negative factor from a group of terms, the signs of everything within the resulting parentheses must change.

But [math]2y-2[/math] can be factored to [math]2(y-1)[/math]. Since [math]2y-2[/math] appears in the equation can't we say [math]...-(2y-2)=...-(2(y-1))[/math]?
 
It is $-2y-2$ that appears in the expression...and both terms have $-2$ as a factor. :D
 
find_the_fun said:
But [math]2y-2[/math] can be factored to [math]2(y-1)[/math]. Since [math]2y-2[/math] appears in the equation can't we say [math]...-(2y-2)=...-(2(y-1))[/math]?

That's the point. [math]2y-2[/math] does not appear in the equation.
You can read the equation as:
$$xy+x−2y−2 = xy+x+(−2)y+(−2)$$

As you can see $(−2)y+(−2)$ is definitely different from $2y-2$.

From here we can get to:
$$xy+x+(−2)y+(−2) = x(y+1)+(−2)(y+1)$$

Which can also be written as:
$$x(y+1)−2(y+1)$$
 
I guess I see what I did wrong, I did something like, took a -2 inserted a bracket and pretend it was positive. I try to understand exactly what went wrong in hopes of not making the same mistake again.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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