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Precalculus : Find Complex Zeros of a Polynomial Using the Fundamental Theorem of Algebra

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Example Question #431 : Pre Calculus

Find a fourth-degree polynomial whose zeroes are -2,5 $\displaystyle -2,5$, and $-2\pm \sqrt{2i}$ $\displaystyle -2\pm \sqrt{2i}$

Possible Answers:

$x^{4}+x^{3}+16x^{2}+58x-60$ $\displaystyle x^{4}+x^{3}+16x^{2}+58x-60$

$x^{4}-x^{3}-16x^{2}-58x-60$ $\displaystyle x^{4}-x^{3}-16x^{2}-58x-60$

$x^{3}-16x^{2}-58x-60$ $\displaystyle x^{3}-16x^{2}-58x-60$

$x^{4}-16x^{2}-58x-60$ $\displaystyle x^{4}-16x^{2}-58x-60$

$x^{4}+x^{3}-16x^{2}-58x-60$ $\displaystyle x^{4}+x^{3}-16x^{2}-58x-60$

Correct answer:

$x^{4}+x^{3}-16x^{2}-58x-60$ $\displaystyle x^{4}+x^{3}-16x^{2}-58x-60$

Explanation:

This one is a bit of a journey. The expressions for the first two zeroes are easily calculated, $\left ( x+2 \right )$ $\displaystyle \left ( x+2 \right )$ and $\left ( x-5 \right )$ $\displaystyle \left ( x-5 \right )$ respectively. The last expression must be broken up into two equations: $x=-2-i\sqrt{2},x=-2+i\sqrt{2}$ $\displaystyle x=-2-i\sqrt{2},x=-2+i\sqrt{2}$ which are then set equal to zero to yield the expressions $\left ( x+2-i\sqrt{2} \right )$ $\displaystyle \left ( x+2-i\sqrt{2} \right )$ and $\left ( x+2+i\sqrt{2} \right )$ $\displaystyle \left ( x+2+i\sqrt{2} \right )$

Finally, we multiply together all of the parenthesized expressions, which multiplies out to $x^{4}+x^{3}-16x^{2}-58x-60=0$ $\displaystyle x^{4}+x^{3}-16x^{2}-58x-60=0$

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Example Question #432 : Pre Calculus

The third-degree polynomial expression $f(x)=x^{3}-x^{2}+4x-4$ $\displaystyle f(x)=x^{3}-x^{2}+4x-4$ has a real zero at x=1 $\displaystyle x=1$. Find all of the complex zeroes.

Possible Answers:

$x=2\pm \sqrt{2i}$ $\displaystyle x=2\pm \sqrt{2i}$

$x=\pm 2i$ $\displaystyle x=\pm 2i$

$x=2\pm 2i$ $\displaystyle x=2\pm 2i$

$x=\pm \sqrt{2}$ $\displaystyle x=\pm \sqrt{2}$

$x=\pm 4i$ $\displaystyle x=\pm 4i$

Correct answer:

$x=\pm 2i$ $\displaystyle x=\pm 2i$

Explanation:

First, factor the expression by grouping: $x^{3}-x^{2}+4x-4\rightarrow x^{2}(x-1)+4(x-1)\rightarrow (x^{2}+4)\cdot (x-1)$ $\displaystyle x^{3}-x^{2}+4x-4\rightarrow x^{2}(x-1)+4(x-1)\rightarrow (x^{2}+4)\cdot (x-1)$

To find the complex zeroes, set the term $(x^{2}+4)$ $\displaystyle (x^{2}+4)$ equal to zero: $x^{2}=-4\rightarrow x=\pm \sqrt{-2}=\pm 2i$ $\displaystyle x^{2}=-4\rightarrow x=\pm \sqrt{-2}=\pm 2i$

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Precalculus : Find Complex Zeros of a Polynomial Using the Fundamental Theorem of Algebra

Study concepts, example questions & explanations for Precalculus

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Example Question #431 : Pre Calculus

Example Question #432 : Pre Calculus

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