What are the solutions to the equation 12x^3-48x^2+12x+72=0 ?

The solutions to the equation 12x^3-48x^2+12x+72=0 are x=-1,x=2,x=3

Find the zeros of 12x^3-48x^2+12x+72=0

The zeros of 12x^3-48x^2+12x+72=0 are x=-1,x=2,x=3

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12 x^{3} - 48 x^{2} + 12 x + 72 = 0

x = - 1, x = 2, x = 3

Solution steps

12 x^{3} - 48 x^{2} + 12 x + 72 = 0

Factor

12 x^{3} - 48 x^{2} + 12 x + 72 : 12 (x + 1) (x - 2) (x - 3)

12 (x + 1) (x - 2) (x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 2 = 0 or x - 3 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

Solve

x - 3 = 0 : x = 3

The solutions are

x = - 1, x = 2, x = 3

x^{6} - 9 x^{2} - 8 x^{4} = 0

2 x^{3} - 50 x = 0

a, 54 = a^{4}

p (2 p - 5)^{2} - 3 (2 p - 5) = 0

x^{3} - 6 x^{2} - 15 x + 40 = 0

What are the solutions to the equation 12x^3-48x^2+12x+72=0 ?
The solutions to the equation 12x^3-48x^2+12x+72=0 are x=-1,x=2,x=3
Find the zeros of 12x^3-48x^2+12x+72=0
The zeros of 12x^3-48x^2+12x+72=0 are x=-1,x=2,x=3