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$101$ is a root of $f(x) = x^2 + 10,201$ . Find the other roots of $f(x)$ . $\newline$ Write your answer as a list of simplified values separated by commas, if there is more than one value.

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Complex conjugate theorem

Full solution

Q.

101

is a root of

f(x) = x^2 + 10,201

. Find the other roots of

f(x)

.

\newline

Write your answer as a list of simplified values separated by commas, if there is more than one value.

Polynomial Degree: $f(x) = x^2 + 10,201$ has a degree of $2$ , so it has $2$ roots.
Conjugate Root Theorem: If $101$ is a root, then by the conjugate root theorem, the other root is $-101$ since the polynomial has real coefficients.
Check Roots: Check the roots by plugging them into the polynomial: $\newline$ $f(101) = 101^2 + 10,201 = 10,201 + 10,201 = 20,402$ , which is not $0$ . Oops, there's a mistake here.

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