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

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

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

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

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

x = - 1, x = 2

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 2 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

The solutions are

x = - 1, x = 2

π \cdot r^{4} = 4.86 \cdot 1 0^{- 7}

x^{4} - 265 x^{2} + 1 = 0

x^{4} x - 8 = 2 x - 4

x (x^{2} + x - 6) = 0

- 2 = 2 x^{5} + x^{3} + 1

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