What are the solutions to the equation (x-2)(x+1)(x-5)=0 ?

The solutions to the equation (x-2)(x+1)(x-5)=0 are x=2,x=-1,x=5

Find the zeros of (x-2)(x+1)(x-5)=0

The zeros of (x-2)(x+1)(x-5)=0 are x=2,x=-1,x=5

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(x - 2) (x + 1) (x - 5) = 0

x = 2, x = - 1, x = 5

Solution steps

(x - 2) (x + 1) (x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x + 1 = 0 or x - 5 = 0

Solve

x - 2 = 0 : x = 2

Solve

x + 1 = 0 : x = - 1

Solve

x - 5 = 0 : x = 5

The solutions are

x = 2, x = - 1, x = 5

x^{5} + 8 x^{2} = 0

- 27 = x^{3}

- 2 x (x + 2) (x^{2} - 15) = 0

5 x^{3} + 4 x^{2} + 6 x + 2 = 0

x^{3} - 2 x^{2} - 19 x + 20 = 0

What are the solutions to the equation (x-2)(x+1)(x-5)=0 ?
The solutions to the equation (x-2)(x+1)(x-5)=0 are x=2,x=-1,x=5
Find the zeros of (x-2)(x+1)(x-5)=0
The zeros of (x-2)(x+1)(x-5)=0 are x=2,x=-1,x=5