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Exercises Polynomials and factoring

Example 7 from Algebra and Trigonometry, 5.5 Zeros of Polynomial Functions

Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros

Find a fourth degree polynomial with real coefficients that has zeros of –3, 2, $i,$ such that $f (−2) = 100.$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Because $x = i$ is a zero, by the Complex Conjugate Theorem $x = - i$ is also a zero. The polynomial must have factors of $(x + 3), (x - 2), (x - i),$ and $(x + i) .$ Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let’s begin by multiplying these factors.

$\begin{matrix} f (x) & = & a (x + 3) (x - 2) (x - i) (x + i) \\ f (x) & = & a (x^{2} + x - 6) (x^{2} + 1) \\ f (x) & = & a (x^{4} + x^{3} - 5 x^{2} + x - 6) \end{matrix}$
We need to find a to ensure $f (- 2) = 100.$ Substitute $x = - 2$ and $f (2) = 100$ into $f (x) .$

$\begin{matrix} 100 & = & a ({(−2)}^{4} + {(−2)}^{3} - 5 {(−2)}^{2} + (−2) - 6) \\ 100 & = & a (−20) \\ −5 & = & a \end{matrix}$

So the polynomial function is

$f (x) = −5 (x^{4} + x^{3} - 5 x^{2} + x - 6)$
or

$f (x) = - 5 x^{4} - 5 x^{3} + 25 x^{2} - 5 x + 30$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Find a third degree polynomial with real coefficients that has zeros of 5 and $- 2 i$ such that $f (1) = 10.$

How did it go?

PreviousFinding the Zeros of a Polynomial Function with Complex Zeros Next Using Descartes’ Rule of Signs

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 5.5, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.