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{eq}f(x) = 2x^3-3x^2-50x-26{/eq}.

Find all complex zeros. (a) {eq}x ={/eq} (b) {eq}f(x) ={/eq}

Question:

{eq}f(x) = 2x^3-3x^2-50x-26{/eq}.

Find all complex zeros. (a) {eq}x ={/eq} (b) {eq}f(x) ={/eq}

Cubic Functions:

Recall the the fundamental theorem of algebra ensures us that all cubic functions have exactly three roots, where we either have three real roots, or one real and two complex roots (where the complex roots are complex conjugates of one another). Since we have the technology, the best way to get a leg up on these functions is to take a look at what they look like.

Answer and Explanation: 1

This whole "(a) x= (b) f(x) =" business is silly. We know we are looking for the zeroes, so we want to know the {eq}x {/eq} values when {eq}f (x) = 0 {/eq}. So obviously {eq}f (x) = 0 {/eq}. If the point was to be needlessly confusing, then job well done.

Our function is

{eq}\begin{align*} y &= 2x^{3}-3x^{2}-50x-26 \end{align*} {/eq}

Let's take a look at it.

Ugh, this is messed up; we are being hoodwinked. Clearly, the graph passes through {eq}y = 0 {/eq} at three real points (we use our trace feature to see that they are approximately -3.97, -0.54, and 6.02). Thus there are no complex zeroes.


Learn more about this topic:

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Cubic Function: Definition, Formula & Examples

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Chapter 8 / Lesson 14
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What is a cubic function? See examples of cubic functions and learn how to graph cubic functions. Learn the equation and properties of a standard cubic function.


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