I am so stuck on this factoring thing. how do you solve:

1)15a(squared)b-10ab(squared)
2)x(squared)-8x-8
3) 6p(squared)-17p-45
4)2r(cubed)+250
5)x(to the 6th power)-64
6)x(squared)-xy+2x-2y
7)x(to 4th power)-1

I will do some, then you try.

1. 15a^2 b-10ab^2 factor out 5 ab

5ab(3a-2b)

4. 2r^3+250
2(r^3+125) you have a formula for factoring sum of cubes..
2(r+5)(r^2+5r+25) check that.

To solve these factoring problems, you'll need to understand and apply some common factoring techniques. Here's a step-by-step explanation for each problem:

1) 15a(squared)b - 10ab(squared)
To factor this expression, you can see that both terms have common factors of 5 and ab. Therefore, you can factor out 5ab:
15a(squared)b - 10ab(squared) = 5ab(3a - 2b)

2) x(squared) - 8x - 8
To factor a quadratic expression of the form ax(squared) + bx + c, you need to find two numbers whose product is ac (in this case, -8) and whose sum is b (-8).
Since -4 * 2 = -8 and -4 + 2 = -2, you can factor the expression as follows:
x(squared) - 8x - 8 = (x - 4)(x + 2)

3) 6p(squared) - 17p - 45
For this quadratic expression, you need to find two numbers whose product is ac (in this case, -270) and whose sum is b (-17).
Since -27 * 10 = -270 and -27 + 10 = -17, you can factor it out like this:
6p(squared) - 17p - 45 = (2p - 9)(3p + 5)

4) 2r(cubed) + 250
This expression is in the form ax(cubed) + b. To factor it, you can find the greatest common factor (GCF) of the terms.
The GCF of 2r(cubed) and 250 is 2. By factoring it out, you get:
2r(cubed) + 250 = 2(r(cubed) + 125)

5) x(to the 6th power) - 64
This expression is in the form a(sixth power) - b(sixth power). It is a difference of squares because 64 is a perfect square.
The expression can be factored using the difference of squares formula:
x(to the 6th power) - 64 = (x(to the 3rd power) - 8)(x(to the 3rd power) + 8)

6) x(squared) - xy + 2x - 2y
This expression doesn't seem to have any common factors to factor out. It might be rewritten or rearranged but not factored using the traditional factoring method.

7) x(to the 4th power) - 1
This is also a difference of squares. You can notice that x(to the 4th power) can be written as (x(to the 2nd power))^2.
Therefore, it can be factored as follows:
x(to the 4th power) - 1 = (x(to the 2nd power) - 1)(x(to the 2nd power) + 1)

Remember, factoring often requires practice and familiarity with different techniques. Hence, it is crucial to keep practicing and exploring different examples to become more comfortable with factoring.