What are the solutions to the equation 3x^3-4x^2-27x+36=0 ?

The solutions to the equation 3x^3-4x^2-27x+36=0 are x= 4/3 ,x=-3,x=3

Find the zeros of 3x^3-4x^2-27x+36=0

The zeros of 3x^3-4x^2-27x+36=0 are x= 4/3 ,x=-3,x=3

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3 x^{3} - 4 x^{2} - 27 x + 36 = 0

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

Decimal Notation

x = 1.33333 \dots, x = - 3, x = 3

Solution steps

3 x^{3} - 4 x^{2} - 27 x + 36 = 0

Factor

3 x^{3} - 4 x^{2} - 27 x + 36 : (3 x - 4) (x + 3) (x - 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

x + 3 = 0 : x = - 3

Solve

x - 3 = 0 : x = 3

The solutions are

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

(x + 7)^{3} (x - 2 i) (x + 2 i) = 0

4 x^{3} + 7 x^{2} - 12 x - 21 = 0

x^{4} - 5 x^{2} + 6 x - 4 = 0

5 x^{4} = 80

n^{3} - 3 n = 40

What are the solutions to the equation 3x^3-4x^2-27x+36=0 ?
The solutions to the equation 3x^3-4x^2-27x+36=0 are x= 4/3 ,x=-3,x=3
Find the zeros of 3x^3-4x^2-27x+36=0
The zeros of 3x^3-4x^2-27x+36=0 are x= 4/3 ,x=-3,x=3