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

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

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

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

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

x = 0, x = - 1, x = 1

Solution steps

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

Move

(x + 1) (x - 4)

to the left side

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 1 = 0 or x - 1 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

The solutions are

x = 0, x = - 1, x = 1

z^{5} - 2 = 0

x^{3} - 5 x^{2} - 9 x + 45 = 0

- x^{3} + 6 x^{2} + 135 x + 400 = 0

z^{4} = - 81

x^{4} - 7 x^{2} + 10 = 0

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