Algebra 2
posted by Luke .
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

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
(x3)(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. 
Thank you. So can that be factored anymore?

and would I do this oen the same way?
7. 2x^4 + 16x 
yes, I gave you the formula and the hint.
64 + a^3
= (4+a)(16  4a + a^2) 
What about the other one?? the same way??

Thanks reiny for your help.

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