What are the solutions to the equation (x^2-1)(x-4)-(x+1)(-x)=0 ?

The solutions to the equation (x^2-1)(x-4)-(x+1)(-x)=0 are x=2,x=-1

Find the zeros of (x^2-1)(x-4)-(x+1)(-x)=0

The zeros of (x^2-1)(x-4)-(x+1)(-x)=0 are x=2,x=-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^{2} - 1) (x - 4) - (x + 1) (- x) = 0

x = 2, x = - 1

Solution steps

(x^{2} - 1) (x - 4) - (x + 1) (- x) = 0

Factor

(x^{2} - 1) (x - 4) - (x + 1) (- x) : (x - 2)^{2} (x + 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x + 1 = 0

Solve

x - 2 = 0 : x = 2

Solve

x + 1 = 0 : x = - 1

The solutions are

x = 2, x = - 1

x^{3} + 3 x^{2} + 4 x + 1 = 0

x^{3} - 10 x^{2} + 37 x - 52 = 0

1000 x^{3} - 300 x^{2} + 1.5 = 0

x^{2} = 6 x^{4}

- x^{3} + 41 x + 24 = 0

What are the solutions to the equation (x^2-1)(x-4)-(x+1)(-x)=0 ?
The solutions to the equation (x^2-1)(x-4)-(x+1)(-x)=0 are x=2,x=-1
Find the zeros of (x^2-1)(x-4)-(x+1)(-x)=0
The zeros of (x^2-1)(x-4)-(x+1)(-x)=0 are x=2,x=-1