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)

    ________________________________________

    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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