1. ## Factoring by grouping...

I'm refreshing my math and a bit confused with the following expression and answer.

Factoring:

Solve: $\displaystyle (x+y)2+(x+y)b$
Answer: $\displaystyle (x+y)(2+b)$
or
Answer: $\displaystyle x+y(2+b)$

Solve: $\displaystyle 3(r-2s)-x(r-2s)$
Answer: $\displaystyle (r-2s)(3-x)$
or
Answer: $\displaystyle r-2s(3-x)$

If I were to try and check the answer given $\displaystyle (r-2s)(3-x)$... I've learned I need to distribute the problem to obtain the answer. In this case $\displaystyle r(3), r(-x), -2s(3), -2s(-x)$, which would give me a totally different answer of $\displaystyle 3r-rx-6s+2sx$, that is not an answer or how could one even derive $\displaystyle 3(r-2s)-x(r-2s)$.

If I were to check the answer with my solution $\displaystyle r-2s(3-x)$ ... $\displaystyle r-2s(3), r-2s(-x)$ ... I would then get $\displaystyle 3(r-2s)-x(r-2s)$

I'm lost...

Edit:

Going to basic math... mutliply/divide, add, subtract... anything in parenthesis is always solved first... so how can one justify using parenthesis in the first set of numbers in this answer!? "()" doesn't mean quantity... or does it...

2. first $\displaystyle (x+y)(2+b)$

second $\displaystyle (r-2s)(3-x)$

when u solve as u said... u'll get

$\displaystyle (r-2s)(3-x) = 3r -xr -6s +2xs = 3r - 6s -xr +2xs = 3(r-2s) -x ( r -2s)$

where do u get lost ?

3. Thanks, but doesn't explain/asnwer my question... you just restated what I asked.

Anyone?

4. pychon. You are facing the problem that in practice factoring is much harder than expanding.

$\displaystyle (x+y)c = xc+yc$ by the Distributive property.

Now, if you take $\displaystyle c = (2+b)$, you get $\displaystyle (x+y)c = xc+yc = x(2+b)+y(2+b)$

When you look at this:$\displaystyle (r-2s)(3-x) = 3r -xr -6s +2xs = 3r - 6s -xr +2xs$ you need to notice the following: $\displaystyle 3r - 6s -xr +2xs = 3(r)-3(2s)-x(r)+x(2s)$. Then you need to recognize that the distributive property can be used in reverse. $\displaystyle 3(r)-3(2s) = 3(r-2s)$ and $\displaystyle 3(r)-3(2s) = 3(r-2s)$ and $\displaystyle -x(r)+x(2s) = x(-r+2s) = (-x)(r-2s)$. So, $\displaystyle 3(r)-3(2s)-x(r)+x(2s) =3(r-2s) +(-x)(r-2s) = 3(r-2s)-x(r-2s)$

If you do it in reverse, you need to recognize the distributive property whenever you see it.

5. Originally Posted by pychon
I'm refreshing my math and a bit confused with the following expression and answer.

Factoring:

Solve: $\displaystyle (x+y)2+(x+y)b$
Answer: $\displaystyle (x+y)(2+b)$
or
Answer: $\displaystyle x+y(2+b)$

Solve: $\displaystyle 3(r-2s)-x(r-2s)$
Answer: $\displaystyle (r-2s)(3-x)$
or
Answer: $\displaystyle r-2s(3-x)$

there's no way to both of ur answer to be correct.... that's first of all
second, when u do factoring what do u mean with this :

Originally Posted by pychon
... which would give me a totally different answer of...
Originally Posted by pychon
anything in parenthesis is always solved first...
there's no way that u get totally different result...

P.S. look in definition of real numbers ... and that five axioms for Addition , five for Multiplying , one AM (that says that multiplying is distributable to addition) ... there's those five for relations u don't need it for this, but look at it anyway

6. I don't know if it's the good answer but I'll try :P:

I think that the answer is:

Solve: $\displaystyle (x+y)2+(x+y)b$
A: $\displaystyle (x+y)^2(2+b)$

I think it's the correct answer, correct me if 'm wrong.

7. Originally Posted by Wissem
I don't know if it's the good answer but I'll try :P:

I think that the answer is:

Solve: $\displaystyle (x+y)2+(x+y)b$
A: $\displaystyle (x+y)^2(2+b)$

I think it's the correct answer, correct me if 'm wrong.
No, there is no square there.

8. $\displaystyle (x+y)2 + (x+y)b$
$\displaystyle = (x+y)(2+b)$

9. Originally Posted by pychon
I'm refreshing my math and a bit confused with the following expression and answer.

Factoring:

Solve: $\displaystyle (x+y)2+(x+y)b$
Answer: $\displaystyle (x+y)(2+b)$
or
Answer: $\displaystyle x+y(2+b)$

Solve: $\displaystyle 3(r-2s)-x(r-2s)$
Answer: $\displaystyle (r-2s)(3-x)$
or
Answer: $\displaystyle r-2s(3-x)$

If I were to try and check the answer given $\displaystyle (r-2s)(3-x)$... I've learned I need to distribute the problem to obtain the answer. In this case $\displaystyle r(3), r(-x), -2s(3), -2s(-x)$, which would give me a totally different answer of $\displaystyle 3r-rx-6s+2sx$, that is not an answer or how could one even derive $\displaystyle 3(r-2s)-x(r-2s)$.

If I were to check the answer with my solution $\displaystyle r-2s(3-x)$ ... $\displaystyle r-2s(3), r-2s(-x)$ ... I would then get $\displaystyle 3(r-2s)-x(r-2s)$

I'm lost...

Edit:

Going to basic math... mutliply/divide, add, subtract... anything in parenthesis is always solved first... so how can one justify using parenthesis in the first set of numbers in this answer!? "()" doesn't mean quantity... or does it...
You correctly expanded $\displaystyle (r-2s)(3-x)$

The brackets placed side by side means "both components in one set of brackets multiplied by both components in the other".

You're almost there.

Now, you could write $\displaystyle x+y(2+b)$ as $\displaystyle x+(y)(2+b)$

Therefore the two components in the rightmost bracket are only multiplied by "y"
this time. x is then added to that.

So, if you expand that you will see the difference.

Basically $\displaystyle (x+y)(2+b)$ means both of x and y multiplied by both of 2 and b.

For example.. (6)(7)=42 but it's also (4+2)(4+3) which is 4(4+3)+2(4+3) which is 4(4)+4(3)+2(4)+2(3)=16+12+8+6=42.
That's known as distributing, but it's basis is quite straightforward.

$\displaystyle (x+y)2+(x+y)b$ means (x+y) multiplied by 2 and (x+y) multiplied by b,

so it is (x+y) multiplied by both 2 and b which is (x+y) times (2+b).

Getting back to where you were.
Can you now see whether or not

$\displaystyle (r-2s)(3-x)$

and

$\displaystyle r(3-x)-2s(3-x)$

are the same thing or different ?

10. when there are some common element or any common thing with some numbers u can always pull that out of then and put it out of brackets ...
if u look this :

$\displaystyle 2x^2 + 4x^4 = 2(x^2+2x^4)$

$\displaystyle 2x^2 + 4x^4= 2x(x+2x^3)$

$\displaystyle 2x^2 + 4x^4 = 2x^2(1+2x^2)$

$\displaystyle 2x^2 + 4x^4 = x(2x+4x^3)$

or any another... they are all correct but it depends on u how do u need them to be...

11. Archie, thanks what I was looking for. The first answer to each equation to solve is what was given by the book... the second is what I came up with because I'm not grasping/understand how the equation can be checked; pulling out the correct terms after distribution to check if the answer is correct then placing them in the correct order. I can factor the problem, but having problems checking... I hate brick walls!

Still a bit fuzzy. It's been over 10 years since I've done algebra, geometry and trig... was a B-B+ student, was so easy then :|

12. Originally Posted by pychon
Archie, thanks what I was looking for. The first answer to each equation to solve is what was given by the book... the second is what I came up with because I'm not grasping/understand how the equation can be checked; pulling out the correct terms after distribution to check if the answer is correct then placing them in the correct order. I can factor the problem, but having problems checking... I hate brick walls!

Still a bit fuzzy. It's been over 10 years since I've done algebra, geometry and trig... was a B-B+ student, was so easy then :|
Hi pychon,

I guess it's easiest to consider these from here...

$\displaystyle (x+y)(2+b)$ is both x and y multiplied by both 2 and b,

hence we must multiply 2 by both x and y and also multiply b by both x and y.
So it's x by 2.... and x by b..... and y by 2..... and y by b.
Everything in the left bracket by everything in the right bracket.

Therefore a first step brings us to

$\displaystyle (x+y)(2+b)=x(2+b)+y(2+b)$

Then the last step is

$\displaystyle (x+y)(2+b)=x(2+b)+y(2+b)=2x+bx+2y+by$

If you want to go backwards from there, you just spot the terms that are written more than once.

$\displaystyle 2x+bx+2y+by=x(2+b)+y(2+b)$

But then we can see that (2+b) is a common factor,
so (2+b) is multiplied by both x and y,
which is just another way to say 2 is multiplied by both x and y
and also b is multiplied by both x and y.

In

$\displaystyle x+y(2+b)$

only the y is multiplied by 2 and b, x is only multiplied by 1,
so that works out as

$\displaystyle x+y(2+b)=x+2y+by$