We are given the function $f(x) = x^5 + x^3 - 42x$. To find all real zeros, we need to solve for $x$ such that $f(x) = 0$.

Step 1: Factor the polynomial

First, we can factor out the common factor of $x$: $f(x) = x(x^4 + x^2 - 42) = 0$ This gives us two cases to solve:

1. $x = 0$
2. $x^4 + x^2 - 42 = 0$

Step 2: Solve $x^4 + x^2 - 42 = 0$

Let’s make a substitution: let $u = x^2$. This simplifies the equation to: $u^2 + u - 42 = 0$ This is a quadratic equation. We can solve it using the quadratic formula: $u = \frac{-1 \pm \sqrt{1^2 - 4(1)(-42)}}{2(1)}$ $u = \frac{-1 \pm \sqrt{1 + 168}}{2}$ $u = \frac{-1 \pm \sqrt{169}}{2}$ $u = \frac{-1 \pm 13}{2}$ Thus, the solutions for $u$ are: $u = \frac{-1 + 13}{2} = 6 \quad \text{or} \quad u = \frac{-1 - 13}{2} = -7$

Step 3: Solve for $x$

Recall that $u = x^2$, so we have two cases:

1. $x^2 = 6$ gives $x = \pm \sqrt{6}$
2. $x^2 = -7$ has no real solutions since the square of a real number cannot be negative.

Step 4: List all real solutions

The real solutions are: $x = 0, \pm \sqrt{6}$

Thus, the real zeros of the function are: $x = 0, \sqrt{6}, -\sqrt{6}$

Would you like any more details or further clarifications?

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Here are 5 related questions you might find helpful:

1. How can you verify if a function is factored correctly?
2. What is the quadratic formula and how is it applied?
3. How do you handle equations involving higher powers like $x^5$?
4. What are imaginary solutions, and when do they appear in polynomials?
5. What are some methods to approximate irrational solutions like $\sqrt{6}$?

Tip: Always check for common factors first when factoring polynomials!