What are the solutions to the equation -3x(2x-5)(x^2+25)=0 ?

The solutions to the equation -3x(2x-5)(x^2+25)=0 are x=0,x= 5/2 ,x=5i,x=-5i

Find the zeros of -3x(2x-5)(x^2+25)=0

The zeros of -3x(2x-5)(x^2+25)=0 are x=0,x= 5/2 ,x=5i,x=-5i

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- 3 x (2 x - 5) (x^{2} + 25) = 0

x = 0, x = \frac{5}{2}, x = 5 i, x = - 5 i

Solution steps

- 3 x (2 x - 5) (x^{2} + 25) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 2 x - 5 = 0 or x^{2} + 25 = 0

Solve

2 x - 5 = 0 : x = \frac{5}{2}

Solve

x^{2} + 25 = 0 : x = 5 i, x = - 5 i

The solutions are

x = 0, x = \frac{5}{2}, x = 5 i, x = - 5 i

3 x^{2} = x^{4}

x^{6} - 7 x^{3} = 8

t^{4} = 3

2 x^{3} - 18 x = 0

z^{3} + 216 = 0

What are the solutions to the equation -3x(2x-5)(x^2+25)=0 ?
The solutions to the equation -3x(2x-5)(x^2+25)=0 are x=0,x= 5/2 ,x=5i,x=-5i
Find the zeros of -3x(2x-5)(x^2+25)=0
The zeros of -3x(2x-5)(x^2+25)=0 are x=0,x= 5/2 ,x=5i,x=-5i