What are the solutions to the equation (2x^2-1)(x+2)(x^2-2x)(x^3+5)=0 ?

The solutions to the equation (2x^2-1)(x+2)(x^2-2x)(x^3+5)=0 are

Find the zeros of (2x^2-1)(x+2)(x^2-2x)(x^3+5)=0

The zeros of (2x^2-1)(x+2)(x^2-2x)(x^3+5)=0 are

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(2 x^{2} - 1) (x + 2) (x^{2} - 2 x) (x^{3} + 5) = 0

Solution steps

(2 x^{2} - 1) (x + 2) (x^{2} - 2 x) (x^{3} + 5) = 0

Factor

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

Solve

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

Solve

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

Solve

The solutions are

(3 x^{2} + 2 x + 4)^{2} = (x^{2} - 4)^{2}

x^{3} - 10 x^{2} - 784 = 0

x^{4} = 3184

x^{4} - 6 x^{2} + x + 1 = 0

x^{3} - 6 x^{2} + x + 2 = 0

What are the solutions to the equation (2x^2-1)(x+2)(x^2-2x)(x^3+5)=0 ?
The solutions to the equation (2x^2-1)(x+2)(x^2-2x)(x^3+5)=0 are
Find the zeros of (2x^2-1)(x+2)(x^2-2x)(x^3+5)=0
The zeros of (2x^2-1)(x+2)(x^2-2x)(x^3+5)=0 are