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

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

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

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

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

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

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 3 = 0 : x = 3

Solve

x + 1 = 0 : x = - 1

Solve

x + 2 = 0 : x = - 2

The solutions are

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

- x^{3} - 8 x^{2} + 25 = 0

x^{4} - 10 x^{2} - 3 x + 25 = 0

x^{4} - 11 x^{2} + 24 = 0

x^{4} = \frac{1}{80}

x^{3} - 10 x^{2} = 0

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