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

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

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

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

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

x = - \frac{1}{2}, x = i, x = - i

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x + 1 = 0 or x^{2} + 1 = 0

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

Solve

x^{2} + 1 = 0 : x = i, x = - i

The solutions are

x = - \frac{1}{2}, x = i, x = - i

3 y^{2} \cdot (y - 5) = 0

x, (d^{4} - a^{4}) \cdot y = x^{4}

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

- x^{4} + 2 x^{2} + 8 = 0

x^{4} - 18 x^{3} + 103 x^{2} + 214 x + 192 = 0

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