What are the solutions to the equation roots x^3-54x^2+51x+106 ?

The solutions to the equation roots x^3-54x^2+51x+106 are x=-1,x=2,x=53

Find the zeros of roots x^3-54x^2+51x+106

The zeros of roots x^3-54x^2+51x+106 are x=-1,x=2,x=53

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

roots

x^{3} - 54 x^{2} + 51 x + 106

x = - 1, x = 2, x = 53

Solution steps

x^{3} - 54 x^{2} + 51 x + 106

The roots are the intercepts with the x-axis

(y = 0)

x^{3} - 54 x^{2} + 51 x + 106 = 0

Factor

x^{3} - 54 x^{2} + 51 x + 106 : (x + 1) (x - 2) (x - 53)

(x + 1) (x - 2) (x - 53) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 2 = 0 or x - 53 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

Solve

x - 53 = 0 : x = 53

The solutions are

x = - 1, x = 2, x = 53

n^{3} - 6 n^{2} - n + 6 = 0

\frac{x ^{3}}{9} - 1 = 2

x^{4} - 16 x^{2} + 75 = 2 x^{2} - 6

m^{3} - 4 m^{2} - 5 m = 0

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

What are the solutions to the equation roots x^3-54x^2+51x+106 ?
The solutions to the equation roots x^3-54x^2+51x+106 are x=-1,x=2,x=53
Find the zeros of roots x^3-54x^2+51x+106
The zeros of roots x^3-54x^2+51x+106 are x=-1,x=2,x=53