What are the solutions to the equation 0=12x^3-48x^2+36x ?

The solutions to the equation 0=12x^3-48x^2+36x are x=0,x=1,x=3

Find the zeros of 0=12x^3-48x^2+36x

The zeros of 0=12x^3-48x^2+36x are x=0,x=1,x=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0 = 12 x^{3} - 48 x^{2} + 36 x

x = 0, x = 1, x = 3

Solution steps

0 = 12 x^{3} - 48 x^{2} + 36 x

Switch sides

12 x^{3} - 48 x^{2} + 36 x = 0

Factor

12 x^{3} - 48 x^{2} + 36 x : 12 x (x - 1) (x - 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 1 = 0 or x - 3 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 3 = 0 : x = 3

The solutions are

x = 0, x = 1, x = 3

x^{4} + 8 x^{3} + 14 x^{2} - 2 x + 1 = 0

m^{3} + m^{2} + 4 m + 4 = 0

4 (7) = 1 + 64 (2 - n)^{3}

x^{4} - 5 x^{3} + 5 x^{2} + 5 x - 6 = 0

x, 3 y^{2} = (2 x + 3)^{3}

What are the solutions to the equation 0=12x^3-48x^2+36x ?
The solutions to the equation 0=12x^3-48x^2+36x are x=0,x=1,x=3
Find the zeros of 0=12x^3-48x^2+36x
The zeros of 0=12x^3-48x^2+36x are x=0,x=1,x=3