## Factoring: x - d + (x-d)^2, 36(2x-y)^2 - 25(u-2y)^2,

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lia
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### Factoring: x - d + (x-d)^2, 36(2x-y)^2 - 25(u-2y)^2,

I have some factoring questions that I wasn't sure how to do. The methods we can ue are common factoring, difference of squares, trinomial factoring, grouping, sum and difference of cubes, factor theorem, and extended factor theorem.) I'm not sure how to factor a trinomial though. These questions are to be factors completely as possible. If I could receive help it would be greatly appreciated.

1: x - d + (x-d)^2 I am thinking that this is simplified and can no longer be factored?

2: 36(2x-y)^2 - 25(u-2y)^2
After expanding and simplifying ( I'm not sure if I should expand or not) I end up with 144x^2-144xy-25u^2+100uy-99y^2 and then I factored the 144 out and got:
144(x^2-xy)-25u^2+100uy-99y^2 (not sure if factoring 144 out was correct and not sure what do to after this step)

3: Not sure how to factor this: y^5 - y^4 + y^3 - y^2 + y - 1

4: 9(x + 2y + z)^2 - 16(x - 2y + z) ^2 after expanding and simplifying I end up with -7x^2 + 100xy - 14xz - 28y^2 + 100xy - 7z^2
Then I factored some number out and get -7(x^2 + Z^2) + 100y(x + z) - 14(xz+y^2) ( I am not sure if this is correct and what to do next)

5: 3(x + 2W)^3 - 3p^3r^3 I get 3(x+2w)^3 - (pr) (^2r^2) not sure if this is right.

6: I am to common factor this question but am not sure how to:
-3(x-1)^2 - 12(x-1)^-3

stapel_eliz
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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.
2: 36(2x-y)^2 - 25(u-2y)^2
This is actually just a nasty difference of squares. If they'd given you a2 - b2, 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.
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:

. . . . .y4(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.
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.
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 - p3r3]

. . . . .3[(x + 2W)3 - (pr)3]

If they'd given you a3 - b3, 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.
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...?

Eliz.

lia
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### Re: Factoring: x - d + (x-d)^2, 36(2x-y)^2 - 25(u-2y)^2,

Thanks a lot that was really helpful.

But for number 2 not sure want you meant by "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."

Number 4 I'm not sure what you mean by simplifying inside the factors? Am I suppose to expand and collect like terms then factor again?

For number 5 is this right? :
3[(x + 2W)(x^2 + 2xw +4w^2) - (pr)(p^2+pr+r^2)]

stapel_eliz
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2) You have the formula a2 - b2 = (a - b)(a + b). Instead of "a" and "b" for the difference of squares, you have "12x - 6y" and "5u - 10y". But the pattern is the same: ([first thing] - [second thing])([first thing] + [second thing]).

4) To "simplify", you just combine the "like" terms (the terms containing "y").

5) I'm not sure what you're doing on this exercise...? You don't have (x)3 - (2W)3; you have (x + 2W)3 - (pr)3. Try copying down the differences-of-cubes formula, a3 - b3 = (a - b)(a2 + ab + b2), and then replacing "a" with "x - 2W" and "b" with "pq". See where that leads.

Eliz.

lia
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Okay, thank you.