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

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

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

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

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(x - 2)^{2} (x - 3) (x + 1) = 0

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

Solution steps

(x - 2)^{2} (x - 3) (x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

x - 3 = 0 : x = 3

Solve

x + 1 = 0 : x = - 1

The solutions are

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

4 x^{5} - 25 x^{3} = 0

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

3 x^{3} + 8 x^{2} - 1 = 0

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

(1 - r)^{3} = 3 - 3 r

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