# Math Help - Group Factoring

1. ## Group Factoring

I have three questions involving group factoring, though I'm not exactly sure how to do them. I've done this so far.

a)
ac + bc - ad - bd
(ac + bc) - (ad - bd)
c(a + b) - d(a - b)
(a + b)(c - d)(-1)

Is the negative supposed to be there?

b)

x^2 + 2x + 1 - y^2
(x^2 + 2x + 1) - y^2
(x + 1)(x + 1) - y^2
(x + 1)^2 - y^2

What do I do from here?

c)
x^2 - y^2 - 10y - 25
x^2 - (y^2 - 10y - 25)
x^2 - (y + 5)(y - 5)

Same problem as in b).

Is what I have even correct so far?

2. a) You made a mistake. This is how you should have started:

$ac + bc - ad - bd$
$(ac + bc) - (ad + bd)$

b) Good work. Now you can factor the difference of squares:

$(x + 1 - y)(x + 1 + y)$
$(x - y + 1)(x + y + 1)$

c) Again, you made a mistake. This is how you should have started:

$x^2 - y^2 - 10y - 25$
$x^2 - (y^2 + 10y + 25)$

This one is similar to (b).

3. a)
ac + bc - ad - bd
(ac + bc) - (ad + bd)
c(a + b) - d(a + b)
(a + b)(c - d)

b)

x^2 + 2x + 1 - y^2
(x^2 + 2x + 1) - y^2
(x + 1)(x + 1) - y^2
(x + 1)^2 - y^2
(x + 1 - y)(x + 1 + y)
(x - y + 1)(x + y +1)

c)
x^2 - y^2 - 10y - 25
x^2 - (y^2 + 10y + 25)
x^2 - (y + 5)(y + 5)
x^2 - (y + 5)^2
(x + y + 5)(x - y - 5)

Not sure about the last line in c). The negative (y + 5)^2 is a little confusing. Got the other two questions. Though, why do the numbers in brackets change to positive?

4. Originally Posted by shadow6
a)
ac + bc - ad - bd
(ac + bc) - (ad + bd)
c(a + b) - d(a + b)
(a + b)(c - d)

b)
x^2 + 2x + 1 - y^2
(x^2 + 2x + 1) - y^2
(x + 1)(x + 1) - y^2
(x + 1)^2 - y^2
(x + 1 - y)(x + 1 + y)
(x - y + 1)(x + y +1)

c)
x^2 - y^2 - 10y - 25
x^2 - (y^2 + 10y + 25)
x^2 - (y + 5)(y + 5)
x^2 - (y + 5)^2
(x + y + 5)(x - y - 5)

Not sure about the last line in c). The negative (y + 5)^2 is a little confusing. Got the other two questions. Though, why do the numbers in brackets change to positive?
You are correct about (c). To answer your question, take a look at

$x^2 - y^2 - 10y - 25$.

The reason this is equal to

$x^2 - (y^2 + 10y + 25)$

is that the negative sign is "distributed" across all the terms in the sum

$(y^2 + 10y + 25)$.

If this is difficult to see, consider that

$x^2 - (y^2 + 10y + 25)$ is the same as

$x^2 + (-1)(y^2 + 10y + 25)$

and use the distributive law to distribute the -1 across all the terms.

5. Got it now, thanks.

6. Got lots more factoring questions today. I managed to complete all of them but 2... The two expressions that need factoring are:

1) 3x^2-27(2-x)^2

This one I tried expanding the (2-x)^2, then multiplying the expanded result by 27. I tried this... and I kept getting a much different answer than what was in my textbook. The back of the text book says that the expression's factored form will be -12(2x - 3)(x - 3), though I have absolutely no idea how they got this answer. I've even tried doing this backwards (expanding the answer to get to the question), and that isn't even successful.

2) 2m^2 + 10m + 10n -2n^2

This one I've also attempted several tries (in terms of rearranging and grouping, and even factoring out the 2), however I end up getting lost near the end. Got this so far:

2(m^2 + 5m + 5n - n^2)
2(m^2 - n^2 + 5m +5n)
2((m^2 - n^2) + (5m+5n))
2((m + n)(m + n) + 5(m + n))
2((m + n) + 5)
...?

Any suggestions?