How do you factor (2x-1)^2 - 9?
I'm taking Calculus 1 right now and I totally get the idea of limits and limit laws, but I haven't had PreCal Algebra in over 8 years. That being said, I'm forgetful on some of the basic algebra stuff. So how do i factor (2x-1)^2 - 9? Also (x-b)^50 - x + b?
Thanks for any help.
Comments
Let u = 2x - 1.
u^2 - 9 =
(u + 3)(u - 3) =
(2x - 1 + 3)(2x - 1 - 3) =
(2x + 2)(2x - 4)
(x - b)^50 - x + b =
(x - b)^50 - (x - b) =
(x - b)((x - b)^49 - 1) =
Factor 2x 1
First Problem:
This is the difference of two squares. Use this formula:
a^2-b^2 = (a+b)(a-b)
(2x-1)^2-9
[(2x-1)+3][(2x-1)-3]
(2x+2)(2x-4)
2(x+1)2(x-2)
4(x+1)(x-2)
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Second Problem:
(x-b)^50-x+b
(x-b)^50-(x-b)---------->Factor out the x-b:
(x-b)[(x-b)^49-1]
(2x-1)^2 - 9 use a² – b² = (a + b)(a – b)
= (2x-1) - 3)(2x-1) + 3)
= (2x - 4)(2x + 2)
= 2(x - 2)(2x + 2)
= 4(x - 2)(x + 1)
(x-b)^50 - x + b
= (x – b)^50 - (x – b)
= (x – b)[(x – b)^49 – 1]
= (x – b)[(x – b)^49/2 – 1][(x – b)^49/2 + 1]
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