What are the solutions to the equation 3x^3+x^2-20x+12=0 ?

The solutions to the equation 3x^3+x^2-20x+12=0 are x=2,x= 2/3 ,x=-3

Find the zeros of 3x^3+x^2-20x+12=0

The zeros of 3x^3+x^2-20x+12=0 are x=2,x= 2/3 ,x=-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 x^{3} + x^{2} - 20 x + 12 = 0

x = 2, x = \frac{2}{3}, x = - 3

Decimal Notation

x = 2, x = 0.66666 \dots, x = - 3

Solution steps

3 x^{3} + x^{2} - 20 x + 12 = 0

Factor

3 x^{3} + x^{2} - 20 x + 12 : (x - 2) (3 x - 2) (x + 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

3 x - 2 = 0 : x = \frac{2}{3}

Solve

x + 3 = 0 : x = - 3

The solutions are

x = 2, x = \frac{2}{3}, x = - 3

4 (x^{6} + 5) = 2 x + 10

y, u = y^{3}

(1 + x)^{5} = 4

\frac{8}{125} = x^{3}

x^{3} - 3 x - 2.1 = 0

What are the solutions to the equation 3x^3+x^2-20x+12=0 ?
The solutions to the equation 3x^3+x^2-20x+12=0 are x=2,x= 2/3 ,x=-3
Find the zeros of 3x^3+x^2-20x+12=0
The zeros of 3x^3+x^2-20x+12=0 are x=2,x= 2/3 ,x=-3