WEBVTT

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Here is the problem: x to the power 5 minus x equals 0.

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A polynomial equation of degree 5 in x.

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Start with the equation.

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Move every term to one side.

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Test candidates from the rational root theorem: x equals 0 is a root. Synthetic division by (x minus 0) leaves a polynomial of degree 4.

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Test candidates from the rational root theorem: x equals negative 1 is a root. Synthetic division by (x plus 1) leaves a polynomial of degree 3.

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Test candidates from the rational root theorem: x equals 1 is a root. Synthetic division by (x minus 1) leaves a polynomial of degree 2.

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Solve the remaining quadratic.

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Compute the discriminant.

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delta less than 0, so there are no real roots; the two roots are complex conjugates.

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Apply the quadratic formula with the square root of negative 1 equals i.

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Simplify.

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Collect the real solutions.

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Now look at the graph of the left side minus the right side.

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It crosses zero exactly at the solutions we found.

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So the answer is x equals negative 1, x equals 0, x equals 1; complex: plus or minus i.

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A degree-5 polynomial has exactly 5 roots counted with multiplicity (some may be complex). Exact rational roots are found first, then the rest are solved exactly or approximated.

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x equals negative 1: left side equals 0, right side equals 0, which checks out.

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x equals 0: left side equals 0, right side equals 0, which checks out.

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x equals 1: left side equals 0, right side equals 0, which checks out.