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

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

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

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

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

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

Solution steps

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 1 = 0 : x = - 1

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

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

x^{4} - 4 x^{3} - 3 x^{2} + 16 x - 4 = 0

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

- 5 x^{4} - 32 = - 12

x^{3} - 8 x^{2} + 17 x + 12 = 0

x^{3} - 5 x^{2} - 8 x + 12 = 0

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