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)
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
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