Solve (8x^3 - 2x^23) - (4x^36x 8)

Question

Simplify the expression

8 x^{3} - 6 x^{2} - 192 x^{4}

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2 x^{2} (4 x - 3 - 96 x^{2})

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x_{1} = \frac{1}{48} - \frac{71}{48} i, x_{2} = \frac{1}{48} + \frac{71}{48} i, x_{3} = 0

Alternative Form

x_{1} \approx 0.0208 \dot{3} - 0.175545 i, x_{2} \approx 0.0208 \dot{3} + 0.175545 i, x_{3} = 0

Evaluate

(8 x^{3} - 2 x^{2} \times 3) - (4 x^{3} \times 6 x \times 8)

To find the roots of the expression,set the expression equal to 0

(8 x^{3} - 2 x^{2} \times 3) - (4 x^{3} \times 6 x \times 8) = 0

Multiply the terms

(8 x^{3} - 6 x^{2}) - (4 x^{3} \times 6 x \times 8) = 0

Remove the parentheses

8 x^{3} - 6 x^{2} - (4 x^{3} \times 6 x \times 8) = 0

Multiply

More Steps

8 x^{3} - 6 x^{2} - 192 x^{4} = 0

Factor the expression

2 x^{2} (4 x - 3 - 96 x^{2}) = 0

Divide both sides

x^{2} (4 x - 3 - 96 x^{2}) = 0

Separate the equation into 2 possible cases

x^{2} = 0 4 x - 3 - 96 x^{2} = 0

The only way a power can be 0 is when the base equals 0

x = 0 4 x - 3 - 96 x^{2} = 0

Solve the equation

More Steps

x = 0 x = \frac{1}{48} + \frac{71}{48} i x = \frac{1}{48} - \frac{71}{48} i

Solution

x_{1} = \frac{1}{48} - \frac{71}{48} i, x_{2} = \frac{1}{48} + \frac{71}{48} i, x_{3} = 0

Alternative Form

x_{1} \approx 0.0208 \dot{3} - 0.175545 i, x_{2} \approx 0.0208 \dot{3} + 0.175545 i, x_{3} = 0

Show Solution