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

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

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

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

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(x + 4) (x + 2) (x - 1) = 0

x = - 4, x = - 2, x = 1

Solution steps

(x + 4) (x + 2) (x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 4 = 0 or x + 2 = 0 or x - 1 = 0

Solve

x + 4 = 0 : x = - 4

Solve

x + 2 = 0 : x = - 2

Solve

x - 1 = 0 : x = 1

The solutions are

x = - 4, x = - 2, x = 1

3 x^{3} - 12 x^{2} + 9 x - 4 = 0

x, x^{3} + y^{3} = a^{3}

- x^{3} - 1 = 0

(u + 10)^{3} = - (u + 10)^{2}

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

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