lia wrote:1: x - d + (x-d)^2 I am thinking that this is simplified and can no longer be factored?
There is a common factor, but it's big and lumpy:. . . . .
(x - d) + (x - d)2
= (x - d) + (x - d)(x - d)
If they'd given you x + x2
, you'd have noted the common factor of x and factored
as x(1 + x). This time, the common factor is x - d, but the process is the same.
lia wrote:2: 36(2x-y)^2 - 25(u-2y)^2
This is actually just a nasty difference of squares
. If they'd given you a2
, you'd have factored as (a + b)(a - b). This is bigger, but the same:. . . . .
[6(2x - y)]2
- [5(u - 2y)]2. . . . .
[12x - 6y]2
- [5u - 10y]2
Apply the same formula to 12x - 6y and 5u - 10y as you would to a and b. Make sure to simplify (by adding y-terms) inside the factors.
lia wrote:3: Not sure how to factor this: y^5 - y^4 + y^3 - y^2 + y - 1
They may have mentioned a formula related to dividing by y - 1...? If they didn't give you a formula, then I don't see how you're supposed to "know" how to do this on your own, but:. . . . .
(y - 1) + y2
(y - 1) + 1(y - 1)
Or, if you've learned how to do polynomial long division, you can divide the original polynomial by y - 1.
lia wrote:4: 9(x + 2y + z)^2 - 16(x - 2y + z) ^2
This is like (2) above: a difference of squares.. . . . .
[3(x +2y + z)]2
- [4(x - 2y + z)]2. . . . .
[3x + 6y + 3z]2
- [4x - 8y + 4z]2
You'll be able to simplify inside the factors.
lia wrote:5: 3(x + 2W)^3 - 3p^3r^3
You've got a common factor of 3, which leaves behind a difference of cubes:. . . . .
3[(x + 2W)3
]. . . . .
3[(x + 2W)3
If they'd given you a3
, you'd have factored as (a - b)(a2
+ ab + b2
). Apply the formula using x + 2W and pq instead of a and b. There won't be any simplifying that can be done inside the factors this time, though.
lia wrote:6: I am to common factor this question but am not sure how to:
-3(x-1)^2 - 12(x-1)^-3
I'm not sure what is meant by "common factoring" something, and the negative exponent on the x - 1 in the second term would seem to make this impossible...?