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

Question

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

and factor completely

64 + a^3

Answer 1

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.

Answer 2

Thank you. So can that be factored anymore?

Answer 3

and would I do this oen the same way?

7. 2x^4 + 16x

Answer 4

yes, I gave you the formula and the hint.

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

Answer 5

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

Answer 6

Thanks reiny for your help.

Answer 7

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

Answer 8

Thanks