What are the solutions to the equation 3x^3-4x^2-48x+64=0 ?

The solutions to the equation 3x^3-4x^2-48x+64=0 are x= 4/3 ,x=-4,x=4

Find the zeros of 3x^3-4x^2-48x+64=0

The zeros of 3x^3-4x^2-48x+64=0 are x= 4/3 ,x=-4,x=4

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3 x^{3} - 4 x^{2} - 48 x + 64 = 0

x = \frac{4}{3}, x = - 4, x = 4

Decimal Notation

x = 1.33333 \dots, x = - 4, x = 4

Solution steps

3 x^{3} - 4 x^{2} - 48 x + 64 = 0

Factor

3 x^{3} - 4 x^{2} - 48 x + 64 : (3 x - 4) (x + 4) (x - 4)

(3 x - 4) (x + 4) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x - 4 = 0 or x + 4 = 0 or x - 4 = 0

Solve

3 x - 4 = 0 : x = \frac{4}{3}

Solve

x + 4 = 0 : x = - 4

Solve

x - 4 = 0 : x = 4

The solutions are

x = \frac{4}{3}, x = - 4, x = 4

10000 = 1 x^{10}

5 x^{4} - 40 x = 0

555 = 5 x^{3}

(2 x + 1)^{4} + 4 (2 x + 1)^{2} = 12

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

What are the solutions to the equation 3x^3-4x^2-48x+64=0 ?
The solutions to the equation 3x^3-4x^2-48x+64=0 are x= 4/3 ,x=-4,x=4
Find the zeros of 3x^3-4x^2-48x+64=0
The zeros of 3x^3-4x^2-48x+64=0 are x= 4/3 ,x=-4,x=4