What are the solutions to the equation (x+1)(x-2)(x+3)=0 ?

The solutions to the equation (x+1)(x-2)(x+3)=0 are x=-1,x=2,x=-3

Find the zeros of (x+1)(x-2)(x+3)=0

The zeros of (x+1)(x-2)(x+3)=0 are x=-1,x=2,x=-3

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(x + 1) (x - 2) (x + 3) = 0

x = - 1, x = 2, x = - 3

Solution steps

(x + 1) (x - 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

x^{2} (x + 1) - 64 (x + 1) = 0

x^{3} + 6 x^{2} = 16 x

x^{3} = 100

x^{3} = \frac{8}{27}

x^{3} + 4 x^{2} + 13 x = 0

What are the solutions to the equation (x+1)(x-2)(x+3)=0 ?
The solutions to the equation (x+1)(x-2)(x+3)=0 are x=-1,x=2,x=-3
Find the zeros of (x+1)(x-2)(x+3)=0
The zeros of (x+1)(x-2)(x+3)=0 are x=-1,x=2,x=-3