SOLUTION: How would you factor: 3x^3+192 2x^3-11x^2+12x+9 9x^2-3x-2 and 2x^5/4+x^3/4-15x^1/4 p.s. the 5/4, 3/4 and 1/4 in the above equation are exponents (2x raised to the 5/4)

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: How would you factor: 3x^3+192 2x^3-11x^2+12x+9 9x^2-3x-2 and 2x^5/4+x^3/4-15x^1/4 p.s. the 5/4, 3/4 and 1/4 in the above equation are exponents (2x raised to the 5/4)      Log On


   

Question 205385: How would you factor:
3x^3+192
2x^3-11x^2+12x+9
9x^2-3x-2
and
2x^5/4+x^3/4-15x^1/4
p.s. the 5/4, 3/4 and 1/4 in the above equation are exponents (2x raised to the 5/4)
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
3x^3 + 192
Factor out 3
3(x^3 + 64)
The sum of cubes can be factored to:
3(x + 4)(x^2 - 4x + 16)
:
:
2x^3 - 11x^2 + 12x + 9
Use long division or synthetic division (divide by 3):
(x-3)(2x^2 - 5x - 3)
Factor again
(x-3)(2x + 1)(x - 3)
:
;
9x^2 - 3x - 2
(3x - 2)(3x + 1)
:
:
and
2x^(5/4) + x^(3/4) - 15x^(1/4)
:
Factor out x^(1/4)
x^(1/4)[2x^(4/4) + x^(2/4) - 15]
:
which is
x^(1/4) [2x + x^(1/2) - 15]
:
Factors to
x^(1/4) [2x^(1/2) - 5][x^(1/2) + 3]