Posted by **Luke** 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 -
**Reiny**, 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 -
**Luke**, Monday, March 9, 2009 at 7:51pm
Thank you. So can that be factored anymore?

- Algebra 2 -
**Luke**, Monday, March 9, 2009 at 7:53pm
and would I do this oen the same way?

7. 2x^4 + 16x

- Algebra 2 -
**Reiny**, 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 -
**Mike**, Monday, March 9, 2009 at 7:55pm
What about the other one?? the same way??

- Algebra 2 -
**Luke**, Monday, March 9, 2009 at 7:58pm
Thanks reiny for your help.

- Algebra 2 -
**Reiny**, 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 -
**Luke**, Monday, March 9, 2009 at 8:02pm
Thanks

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