Fundamental Theorem of Algebra

Problem 264: Give an expression for $f(x)$ with leading coefficient 4, of degree 3, and having roots $x = 9, x = -5 - 3i, x = -5 + 3i$ .

Solution: Using the Fundamental Theorem of Algebra, we can set up a polynomial with the given roots. That is,

$\begin{align*} f(x) & = a(x - \text{root})(x - \text{root})(x - \text{root}) \\ & = 4(x - 9)(x-[-5-3i])(x-[-5+3i]) \\ & = 4(x - 9)(x + 5 + 3i)(x + 5 - 3i) \\ & = 4(x - 9)(x^2 + 10x + 25 - 9(-1)) \\ & = 4(x - 9)(x^2 + 10x + 34) \\ & = 4(x^3 + 10x^2 + 34x - 9x^2 - 90x - 306) \\ & = 4(x^3 + x^2 - 56x - 306) \\ & = 4x^3 + 4x^2 - 224x - 1224. \end{align*}$

Thus the expression for $f(x)$ is given by

$4x^3 + 4x^2 - 224x - 1224.$

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