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Posted by on Monday, March 9, 2009 at 7:30pm.

I would really appreciate it someone helped me with this information.

1. Solve the inequality: |3 - 3x/4| >= 9
Solve for x
2.(x - 2)(x + 1) = 4

and factor completely

64 + a^3

  • Algebra 2 - , Monday, March 9, 2009 at 7:46pm

    1. 3 - 3x/4 ≥ 9 OR -3 + 3x/4 ≥ 9

    12 - 3x ≥ 36 OR -12 + 3x ≥ 36
    -3x ≥ 24 OR 3x ≥ 48
    x ≤ -8 or x ≥ 16

    2. expand and rearrange to a quadratic
    x^2 - x - 6 = 0
    (x-3)(x+2) = 0
    x = 3 or x = -2

    3. You should know a formula for factoring the sum of cubes:
    A^3 + B^3 = (A+B)(A^2 - AB + B^2)

    notice that
    64 + a^3
    = 4^3 + a^3

    take it from there.

  • Algebra 2 - , Monday, March 9, 2009 at 7:51pm

    Thank you. So can that be factored anymore?

  • Algebra 2 - , Monday, March 9, 2009 at 7:53pm

    and would I do this oen the same way?

    7. 2x^4 + 16x

  • Algebra 2 - , Monday, March 9, 2009 at 7:53pm

    yes, I gave you the formula and the hint.

    64 + a^3
    = (4+a)(16 - 4a + a^2)

  • Algebra 2 - , Monday, March 9, 2009 at 7:55pm

    What about the other one?? the same way??

  • Algebra 2 - , Monday, March 9, 2009 at 7:58pm

    Thanks reiny for your help.

  • Algebra 2 - , Monday, March 9, 2009 at 7:58pm

    yes, take out a common factor of 2x first to get
    2x(x^3 + 8) and recognize 8 as 2^3

  • Algebra 2 - , Monday, March 9, 2009 at 8:02pm

    Thanks

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