Find all the real zeros of the following polynomial. \[ f(x)=x^{5}+3 x^{3}-10 x \] Write your answers as exact values. Provide your answer below: \[ x=\square, \square, \square \]

To find the real zeros of the polynomial \( f(x) = x^5 + 3x^3 - 10x \), we can first factor out the common term: \[ f(x) = x(x^4 + 3x^2 - 10) \] Setting \( f(x) = 0 \), we have: \[ x(x^4 + 3x^2 - 10) = 0 \] This gives us one root: \( x = 0 \). Next, we need to solve the quartic equation \( x^4 + 3x^2 - 10 = 0 \). We can make a substitution by letting \( y = x^2 \), transforming the quartic into a quadratic: \[ y^2 + 3y - 10 = 0 \] Now we can apply the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2} \] Calculating the two possible \( y \) values: 1. \( y = \frac{4}{2} = 2 \) 2. \( y = \frac{-10}{2} = -5 \) Since \( y = x^2 \) and \( x^2 = -5 \) does not yield real solutions, we focus on \( y = 2 \): \[ x^2 = 2 \implies x = \pm \sqrt{2} \] Thus, the real zeros of the polynomial are: \[ x = 0, \sqrt{2}, -\sqrt{2} \] So the final answer is: \[ x = 0, \sqrt{2}, -\sqrt{2} \]