What are the solutions to the equation x^4-8x^3+23x^2-28x+12=0 ?

The solutions to the equation x^4-8x^3+23x^2-28x+12=0 are x=1,x=2,x=3

Find the zeros of x^4-8x^3+23x^2-28x+12=0

The zeros of x^4-8x^3+23x^2-28x+12=0 are x=1,x=2,x=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{4} - 8 x^{3} + 23 x^{2} - 28 x + 12 = 0

x = 1, x = 2, x = 3

Solution steps

x^{4} - 8 x^{3} + 23 x^{2} - 28 x + 12 = 0

Factor

x^{4} - 8 x^{3} + 23 x^{2} - 28 x + 12 : (x - 1) (x - 2)^{2} (x - 3)

(x - 1) (x - 2)^{2} (x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x - 2 = 0 or x - 3 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 2 = 0 : x = 2

Solve

x - 3 = 0 : x = 3

The solutions are

x = 1, x = 2, x = 3

6 x^{4} + x^{3} - 22 x^{2} - 11 x + 6 = 0

(x^{2} + 3 x)^{2} - 2 (x^{2} + 3 x) - 8 = 0

216 x = x^{4}

g (n) = n^{3} + 4 n, g (4)

4 x^{4} = 10

What are the solutions to the equation x^4-8x^3+23x^2-28x+12=0 ?
The solutions to the equation x^4-8x^3+23x^2-28x+12=0 are x=1,x=2,x=3
Find the zeros of x^4-8x^3+23x^2-28x+12=0
The zeros of x^4-8x^3+23x^2-28x+12=0 are x=1,x=2,x=3