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

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

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

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

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

x = - 2 i, x = 2 i, x = 5

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 2 i = 0 : x = - 2 i

Solve

x - 2 i = 0 : x = 2 i

Solve

x - 5 = 0 : x = 5

The solutions are

x = - 2 i, x = 2 i, x = 5

5 x^{3} + 13 x^{2} - 56 x + 20 = 0

4 x^{3} + 42 x^{2} + 240 x + 800 = 0

x^{(4)} - 29 x^{(2)} + 100 = 0

- 2 x^{3} - 4 x^{2} + 10 x + 12 = 0

x, x^{4} - 11 x^{2} + 10 = 0

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