a) You made a mistake. This is how you should have started:
b) Good work. Now you can factor the difference of squares:
c) Again, you made a mistake. This is how you should have started:
This one is similar to (b).
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?
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
.
The reason this is equal to
is that the negative sign is "distributed" across all the terms in the sum
.
If this is difficult to see, consider that
is the same as
and use the distributive law to distribute the -1 across all the terms.
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?