Question

Factoring/grouping question?

x^2+xy+5x+5y
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x+5

So far I have (x^2+xy)+(5x+5y)
__________________
x+5

Based on a guide I found it says that now I have to find the GCF between the two groups, but what would it be? x? If so, then how do I do it?

Answer 1

NUMERATOR:
\[x^2 + xy + 5x + 5y\]
\[x(x+y) + 5(x+y)\] [Now take it common]
Like so..
\[(x+y)(x+5)\]
DENOMINATOR:
\[(x+5)\]
So Finally!! :D
\[(x+y)(x+5)/(x+5) = (x+y)\]
Getting me? :D

Answer 2

I feel like I'm almost getting it! I see how you got x(x+y)+5(x+y) but how did you get just (x+y)(x+5)? I dont understand that part? D:

Answer 3

Good question! :)
It is just like this!
ab + cb? = b(a+c)
So
x(x+y) + 5(x+y)? = (x+5)(x+y)
Now getting it? :)

Answer 4

It is just like using the Distributive property but in an opposite way! :D

Answer 5

ahh I'm sorry I still don't get it >.< I kinda see what you mean though. What I have written down on my paper is x(x+y)+5(x+y) meeeh

Answer 6

Great! :D
Now look it is just like
Alright
Do this question!
Break
a(b+c) using distributive property! :D

Answer 7

No problem! :D

Answer 8

It would be ab+ac

Answer 9

Good now!
If I tell you to factorize this how will you do that?

Answer 10

a(b+c)?

Answer 11

\(\bf x^2+xy+5x+5y \implies x(x+y) + 5(x+y)\\
x\color{red}{(x+y)} + 5\color{red}{(x+y)} \textit{ common factor to } (x+5)\color{red}{(x+y)}\)

Answer 12

see the common factor?

Answer 13

I do!

Answer 14

Exactly now take that COMMON
And write like a(b+c)
Tke the (x+y) as an 'a'
So x(x+y) + 5(x+y) = xa + 5a = a(x+5) or (x+y)(x+5)!! :D
Get it now? :D

Answer 15

Oh! It all clicked! The answer was right! thanks for your time :D

Answer 16

If I fully simplified it, it would be just x+y right?

Answer 17

:)
Yes! :D