# Thread: Factor the expression

1. ## Factor the expression

First question:

From the book... whomever wrote this doesn't know HOW to explain things.

Solve:
$p(m-n)-q(n-m)$

$p(m-n)-q(-1)(-n+m)$
From the second two terms, pull out a ‘-1’. You can double check that this is correct by distributing the “-1” back in.

$p(m-n)-q(-1)(m-n)$
Inside the parenthesis on the far right, rearrange terms.

$p(m-n)+q(m-n)$
Now, in the second group of terms, we have a “-1” times a “-1” which gives a “+1”.
$(m-n)(p+q)$

Can someone explain how the answer is obtained better? I'm fumbling on factoring with -1 on these types of equations.

-----------------------------------------
Second question:

Solve:
$3x(c-3d)+6y(c-3d)$

$3(c-3d)(x+2y)$

If you apply the order of distribution it comes out to $3c-9d...$

Why can't the answer be
$(c-3d)(x+2y)3$
or
$3(x+2y)(c-3d)$

-----------------------------------------

Anyone have a better explanation? I'm starting to cram for a COMPASS test this coming Friday and still need to refresh my math skills from algebra, adv. algebra, geo, and trig.

2. It can be either of those. I would hesitate to write the first one, though, because a standard convention is to write constants first. Also, you might get someone thinking that the 3 is an exponent, which is not the case.

3. Originally Posted by pychon
Solve:
$3x(c-3d)+6y(c-3d)$
$3(c-3d)(x+2y)$
If you apply the order of distribution it comes out to $3c-9d...$
Why can't the answer be
$(c-3d)(x+2y)3$
or
$3(x+2y)(c-3d)$
Usually when you multiply a number by a variable or expression you write the number first, so that there is no ambiguity. I could look at
$(c-3d)(x+2y)3$
and thought that you forgot to make the 3 an exponent, like this:
$(c-3d)(x+2y)^3$
... which is certainly wrong. So keep the 3 in front.

This:
$3(x+2y)(c-3d)$
is just fine. Commutative property of multiplication says that the order you multiply doesn't matter. So it doesn't matter if you wrote $3(c-3d)(x+2y)$ or $3(x+2y)(c-3d)$ as your answer.

EDIT: Ack! Too slow!

4. Ok, so is this legal to check the answer?

$3(c-3d)(x+2y)$

distribute

$3x-9d+3x+6y
$

grrr... that doesn't look right.

5. Originally Posted by pychon
From the book... whomever wrote this doesn't know HOW to explain things.

Solve:
$p(m-n)-q(n-m)$

$p(m-n)-q(-1)(-n+m)$

$p(m-n)-q(-1)(m-n)$

$p(m-n)+q(m-n)$

$(m-n)(p+q)$

-----------------------------------------

Solve:
$3x(c-3d)+6y(c-3d)$

$3(c-3d)(x+2y)$

If you apply the order of distribution it comes out to $3c-9d...$

Why can't the answer be
$(c-3d)(x+2y)3$
or
$3(x+2y)(c-3d)$

-----------------------------------------

Anyone have a better explanation?
They are all the same really!

$3x(c-3d)+6y(c-3d)$ is (c-3d) times both 3x and 6y

so $(c-3d)(3x+6y)=(c-3d)3(x+2y)=(3c-9d)(x+2y)$

All give the exact same result if you pick values for x, y, c, d.

Also.

$(m-n)=-(n-m)$

because $1=(-1)(-1)$

$m-n=(-1)(-1)(m-n)=(-1)((-1)m-(-1)n)=(-1)(-m--n)=(-1)(n-m)$

It's simplest to realise you only need to switch the terms and change the sign.
You are really multiplying by -1 twice.

6. I don't think you distributed correctly. You lost a c in there, among other things. Double-check your distribution, and go one step at a time.

7. Originally Posted by pychon
Ok, so is this legal to check the answer?

$3(c-3d)(x+2y)$

distribute

$3x-9d+3x+6y
$

grrr... that doesn't look right.
Assuming you meant

$3c-9d+3x+6y$

you multiplied out (distributed) $3(c-3d+x+2y)$

8. Ok, so is this legal to check the answer?

$3(c-3d)(x+2y)$

distribute

$3c-9d+3x+6y$

$3c-9d+3x+6y$

3 is common...

$c-3d+x+2y$

so, now.. I'm failing to see where this can validate (c-3d)(x+2y) since (*)(*) implies multiplication.

9. No.

Originally Posted by pychon
$3(c-3d)(x+2y)$
distribute
$3c-9d+3x+6y$
First, you between the 3d and x you changed from a multiplication to an addition. Can't do that.

You can only distribute the 3 over EITHER the (c - 3d) factor OR the (x + 2y) factor, NOT both. If you try to distribute over both, you would be multiplying by 3 TWICE. Since in the original problem I see a 3x and a 6y, that tells me that I should distribute the 3 over the (x + 2y) factor:
$3(c-3d)(x+2y)$
$(c-3d)(3)(x+2y)$
$(c-3d)(3x+6y)$

Now distribute the (c - 3d) over the (3x + 6y):
$(c-3d)(3x) + (c-3d)(6y)$
$3x(c-3d) + 6y(c-3d)$
...which what we started with.

10. Originally Posted by pychon
Ok, so is this legal to check the answer?

$3(c-3d)(x+2y)$

distribute

$3c-9d+3x+6y$

$3c-9d+3x+6y$

3 is common...

$c-3d+x+2y$

so, now.. I'm failing to see where this can validate (c-3d)(x+2y) since (*)(*) implies multiplication.
Hi pychon,

think in terms of numbers

3(4)5=60 as it's either 3(20) or (12)5, however it's also 4(3)5=4(15)

You can multiply them in any order you like.
3(4)5=3(4)+3(4)+3(4)+3(4)+3(4) or 4(5)+4(5)+4(5) or 3(5)+3(5)+3(5)+3(5).

Alternatively 3(6-2)(8-3) must also equal 60.
We can "distribute" that..

3[(6-2)(8-3)]=3[6(8)+6(-3)-2(8)-2(-3)]=3[48-18-16+6]=3[30-16+6]=3[36-16]=3[20]=60

That's the idea, then the algebra is easy from there.

11. Originally Posted by pychon
Ok, so is this legal to check the answer?

$3(c-3d)(x+2y)$

distribute

$3x-9d+3x+6y
$

grrr... that doesn't look right.
No, that isn't right because you have changed the product to a sum.

You can distribute the 3 into the first term only: 3(c-3d)(x+2y)= (3c- 3d)(x+ 2y) and now separate that 3c- 3d into parts:
3c(x+ 2y)- 3d(x+ 2y)= 3cx+ 6cy- 3dx+ 6dy.