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Precalculus : Pre-Calculus

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Example Questions

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Find all the real and complex zeroes of the following equation: $\displaystyle 2x^5+x^4-2x-1$

Possible Answers:

$x=1, \frac{-1}{2}$ $\displaystyle x=1, \frac{-1}{2}$

$x=1, -1, \frac{-1}{2}, \pm i$ $\displaystyle x=1, -1, \frac{-1}{2}, \pm i$

$x=1, -1, \frac{-1}{2}, i$ $\displaystyle x=1, -1, \frac{-1}{2}, i$

$x=1,\frac{-1}{2}, i$ $\displaystyle x=1,\frac{-1}{2}, i$

$x=1, \frac{-1}{2}, \pm i$ $\displaystyle x=1, \frac{-1}{2}, \pm i$

Correct answer:

$x=1, -1, \frac{-1}{2}, \pm i$ $\displaystyle x=1, -1, \frac{-1}{2}, \pm i$

Explanation:

First, factorize the equation using grouping of common terms:

$2x^5+x^4-2x-1\rightarrow x^4(2x+1)-1(2x+1)\rightarrow (x^4-1)(2x+1)\rightarrow (x+1)(x-1)(x^2+1)(2x+1)$ $\displaystyle 2x^5+x^4-2x-1\rightarrow x^4(2x+1)-1(2x+1)\rightarrow (x^4-1)(2x+1)\rightarrow (x+1)(x-1)(x^2+1)(2x+1)$

Next, setting each expression in parenthesis equal to zero yields the answers.

$(x+1)=0\rightarrow x=-1$ $\displaystyle (x+1)=0\rightarrow x=-1$ $(x-1)=0\rightarrow x=1$ $\displaystyle (x-1)=0\rightarrow x=1$ $(x^2+1)=0\rightarrow x=\pm i$ $\displaystyle (x^2+1)=0\rightarrow x=\pm i$

$(2x+1)=0\rightarrow x=\frac{-1}{2}$ $\displaystyle (2x+1)=0\rightarrow x=\frac{-1}{2}$

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Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Find all the zeroes of the following equation and their multiplicity: $\displaystyle g(t)=t^5-6t^3+9t$

Possible Answers:

$t=0, \pm \sqrt{3i}$ $\displaystyle t=0, \pm \sqrt{3i}$ (multiplicity of 2 on 0, multiplicity of 1 on $\pm \sqrt{3i}$ $\displaystyle \pm \sqrt{3i}$

$t=0, \pm \sqrt{3i}$ $\displaystyle t=0, \pm \sqrt{3i}$ (multiplicity of 1 on 0, multiplicity of 2 on $\pm \sqrt{3i}$ $\displaystyle \pm \sqrt{3i}$

$t=0, \pm \sqrt{3}$ $\displaystyle t=0, \pm \sqrt{3}$ (multiplicity of 1 on 0, multiplicity of 2 on $\pm \sqrt{3}$ $\displaystyle \pm \sqrt{3}$

$t=0, \pm \sqrt{3}$ $\displaystyle t=0, \pm \sqrt{3}$ (multiplicity of 2 on 0, multiplicity of 1 on $\pm \sqrt{3}$ $\displaystyle \pm \sqrt{3}$

Correct answer:

$t=0, \pm \sqrt{3}$ $\displaystyle t=0, \pm \sqrt{3}$ (multiplicity of 1 on 0, multiplicity of 2 on $\pm \sqrt{3}$ $\displaystyle \pm \sqrt{3}$

Explanation:

First, pull out the common t and then factorize using quadratic factoring rules:

This equation has solutions at two values: when t=0 $\displaystyle t=0$ and when

$(t^2-3)^2=0\rightarrow t=\pm \sqrt{3}$ $\displaystyle (t^2-3)^2=0\rightarrow t=\pm \sqrt{3}$ Therefore, But since the degree on the former equation is one and the degree on the latter equation is two, the multiplicities are 1 and 2 respectively.

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Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

This one is a bit of a journey. The expressions for the first two zeroes are easily calculated, (x+2) $\displaystyle (x+2)$ and (x-5) $\displaystyle (x-5)$ respectively. The last expression must be broken up into two equations:

which are then set equal to zero to yield the expressions $(x+2-i\sqrt{2})$ $\displaystyle (x+2-i\sqrt{2})$ and $(x+2+i\sqrt{2})$ $\displaystyle (x+2+i\sqrt{2})$

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

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Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

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

Possible Answers:

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

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

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

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

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

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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Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Find all the real and complex zeros of the following equation: $2x^{5}+x^{4}-2x-1$ $\displaystyle 2x^{5}+x^{4}-2x-1$

Possible Answers:

$x=1,-1,\frac{-1}{2},i$ $\displaystyle x=1,-1,\frac{-1}{2},i$

$x=1,\frac{-1}{2},i$ $\displaystyle x=1,\frac{-1}{2},i$

$x=1, \frac{-1}{2},\pm i$ $\displaystyle x=1, \frac{-1}{2},\pm i$

$x=1,\frac{-1}{2}$ $\displaystyle x=1,\frac{-1}{2}$

$x=1, -1, \frac{-1}{2},\pm i$ $\displaystyle x=1, -1, \frac{-1}{2},\pm i$

Correct answer:

$x=1, -1, \frac{-1}{2},\pm i$ $\displaystyle x=1, -1, \frac{-1}{2},\pm i$

Explanation:

First, factorize the equation using grouping of common terms:

$2x^{5}+x^{4}-2x-1\rightarrow x^{4}(2x+1)-1(2x+1)\rightarrow (x^{4}-1)(2x+1)\rightarrow (x+1)(x-1)(x^{2}+1)(2x+1)$ $\displaystyle 2x^{5}+x^{4}-2x-1\rightarrow x^{4}(2x+1)-1(2x+1)\rightarrow (x^{4}-1)(2x+1)\rightarrow (x+1)(x-1)(x^{2}+1)(2x+1)$

Next, setting each expression in parentheses equal to zero yields the answers.

$(x+1)=0\rightarrow x=-1(x-1)=0\rightarrow x=1(x^{2}+1)=0\rightarrow x=\pm i{(2x+1)=0\rightarrow x=\frac{-1}{2}}$ $\displaystyle (x+1)=0\rightarrow x=-1(x-1)=0\rightarrow x=1(x^{2}+1)=0\rightarrow x=\pm i{(2x+1)=0\rightarrow x=\frac{-1}{2}}$

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Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Find all the zeroes of the following equation and their multiplicity: $g(t)=t^{5}-6t^{3}+9t$ $\displaystyle g(t)=t^{5}-6t^{3}+9t$

Possible Answers:

$t=0,\pm \sqrt{3}$ $\displaystyle t=0,\pm \sqrt{3}$ (multiplicity of 2 on 0, multiplicity of 1 on $\pm \sqrt{3}$ $\displaystyle \pm \sqrt{3}$)

$t=0,\pm \sqrt{3}$ $\displaystyle t=0,\pm \sqrt{3}$ (multiplicity of 1 on 0, multiplicity of 2 on $\pm \sqrt{3}$ $\displaystyle \pm \sqrt{3}$)

$t=0,\pm \sqrt{3i}$ $\displaystyle t=0,\pm \sqrt{3i}$ (multiplicity of 2 on 0, multiplicity of 1 on $\pm \sqrt{3i}$ $\displaystyle \pm \sqrt{3i}$)

$t=0,\pm \sqrt{3i}$ $\displaystyle t=0,\pm \sqrt{3i}$ (multiplicity of 1 on 0, multiplicity of 2 on $\pm \sqrt{3i}$ $\displaystyle \pm \sqrt{3i}$)

Correct answer:

$t=0,\pm \sqrt{3}$ $\displaystyle t=0,\pm \sqrt{3}$ (multiplicity of 1 on 0, multiplicity of 2 on $\pm \sqrt{3}$ $\displaystyle \pm \sqrt{3}$)

Explanation:

First, pull out the common t and then factorize using quadratic factoring rules: $t^{5}-6t^{3}+9t\rightarrow t(t^{4}-6t^{2}+9)\rightarrow t(t^{2}-3)^{2}$ $\displaystyle t^{5}-6t^{3}+9t\rightarrow t(t^{4}-6t^{2}+9)\rightarrow t(t^{2}-3)^{2}$

This equation has a solution as two values: when t=0 $\displaystyle t=0$, and when $\left ( t^{2}-3 \right )^{2}=0\rightarrow t=\pm \sqrt{3}$ $\displaystyle \left ( t^{2}-3 \right )^{2}=0\rightarrow t=\pm \sqrt{3}$. Therefore, But since the degree on the former equation is one and the degree on the latter equation is two, the multiplicities are 1 and 2 respectively.

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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 #431 : 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:

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

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

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

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

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

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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Example Question #1 : Find The Sum And Product Of The Zeros Of A Polynomial

Given $\displaystyle y=5x^2+14x-3$, determine the sum and product of the zeros respectively.

Possible Answers:

$\frac{7}{5},-\frac{21}{5}$ $\displaystyle \frac{7}{5},-\frac{21}{5}$

$\frac{14}{5},-\frac{3}{5}$ $\displaystyle \frac{14}{5},-\frac{3}{5}$

$\frac{1}{5},-3$ $\displaystyle \frac{1}{5},-3$

$\frac{16}{5}, -\frac{3}{5}$ $\displaystyle \frac{16}{5}, -\frac{3}{5}$

$-\frac{14}{5},-\frac{3}{5}$ $\displaystyle -\frac{14}{5},-\frac{3}{5}$

Correct answer:

$-\frac{14}{5},-\frac{3}{5}$ $\displaystyle -\frac{14}{5},-\frac{3}{5}$

Explanation:

To determine zeros of $\displaystyle y=5x^2+14x-3$, factorize the polynomial.

$\displaystyle 5x^2+14x-3=(5x-1)(x+3)$

Set each of the factorized components equal to zero and solve for $\displaystyle x$.

5x-1=0 $\displaystyle 5x-1=0$

$x=\frac{1}{5}$ $\displaystyle x=\frac{1}{5}$

x+3=0 $\displaystyle x+3=0$

x=-3 $\displaystyle x=-3$

The sum of the roots:

$\frac{1}{5}+(-3)= \frac{1}{5}-\frac{15}{5}= -\frac{14}{5}$ $\displaystyle \frac{1}{5}+(-3)= \frac{1}{5}-\frac{15}{5}= -\frac{14}{5}$

The product of the roots:

$(\frac{1}{5})(-3)= -\frac{3}{5}$ $\displaystyle (\frac{1}{5})(-3)= -\frac{3}{5}$

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Example Question #431 : Pre Calculus

Please choose the best answer from the following choices.

Find the sum and product of the zeros of the following polynomial:
$\displaystyle x^2-5x+6$

Possible Answers:

Sum=1 $\displaystyle Sum=1$

$\displaystyle Product=3/2$

Sum=5 $\displaystyle Sum=5$

$\displaystyle Product=6$

Sum=1 $\displaystyle Sum=1$

$\displaystyle Product=5$

Sum=-5 $\displaystyle Sum=-5$

$\displaystyle Product=6$

Correct answer:

Sum=5 $\displaystyle Sum=5$

$\displaystyle Product=6$

Explanation:

To find the zeros you have to factor the polynomial.

This is easily factorable and you will get (x-3) $\displaystyle (x-3)$ and (x-2) $\displaystyle (x-2)$.

Next, set both of these equal to zero. x-3=0 $\displaystyle x-3=0$ and x-2=0 $\displaystyle x-2=0$.

Isolate the x's and you will get x=3 $\displaystyle x=3$ and x=2 $\displaystyle x=2$.

The sum will be $\displaystyle 5$ since you add the two together, and the product will be $\displaystyle 6$ because you multiply the two together.

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #431 : Pre Calculus

Example Question #431 : Pre Calculus

Example Question #1 : Find The Sum And Product Of The Zeros Of A Polynomial

Example Question #431 : Pre Calculus

All Precalculus Resources

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