What is x^4+2x^3-4x^2-8x<= 0 ?

The solution to x^4+2x^3-4x^2-8x<= 0 is x=-2\lor 0<= x<= 2

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x^4+2x^3-4x^2-8x<= 0

x^{4} + 2 x^{3} - 4 x^{2} - 8 x \leq 0

x = - 2 or 0 \leq x \leq 2

Interval Notation

x = - 2 \cup [0, 2]

Solution steps

x^{4} + 2 x^{3} - 4 x^{2} - 8 x \leq 0

Factor

x^{4} + 2 x^{3} - 4 x^{2} - 8 x : x (x + 2)^{2} (x - 2)

x (x + 2)^{2} (x - 2) \leq 0

Identify the intervals

x = - 2 or 0 \leq x \leq 2

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

- 2 \leq - 2 x + 4 < 12

- 2 + (- 7)

\frac{1}{2} x^{2} + \frac{1}{6} x - 1 = 0

\frac{2}{5} + \frac{4}{3} \times \frac{3}{5} - (\frac{1}{4})^{2}

What is x^4+2x^3-4x^2-8x<= 0 ?
The solution to x^4+2x^3-4x^2-8x<= 0 is x=-2\lor 0<= x<= 2