Call Now to Set Up Tutoring:

(888) 888-0446

College Algebra : Zeros/Roots of a Polynomial

Study concepts, example questions & explanations for College Algebra

Create An Account

All College Algebra Resources

5 Diagnostic Tests 84 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #31 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=306x - 270x^{2} - 72\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

x=-180,-432

x=180,432

$x=-\frac{1}{3},-\frac{4}{5}$

$x=\frac{1}{3},\frac{4}{5}$

Correct answer:

$x=\frac{1}{3},\frac{4}{5}$

Explanation:

$\begin{align*}&\text{The function }f(x)=306x - 270x^{2} - 72\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-18\cdot (3x - 1)\cdot (5x - 4)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(306)\pm\sqrt{(306)^2-4(-270)(-72)}}{2(-270)}=\frac{-306\pm126}{-540}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=\frac{1}{3}\text{ and }\frac{4}{5}.\end{align*}$

Report an Error

Example Question #225 : College Algebra

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=232x + 120x^{2} - 208\\&\text{has a zero value.}\end{align*}$

Possible Answers:

$x=-\frac{2}{3},\frac{13}{5}$

x=160,-624

$x=\frac{2}{3},-\frac{13}{5}$

x=-160,624

Correct answer:

$x=\frac{2}{3},-\frac{13}{5}$

Explanation:

$\begin{align*}&\text{Given }f(x)=232x + 120x^{2} - 208\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&8\cdot (5x + 13)\cdot (3x - 2)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(232)\pm\sqrt{(232)^2-4(120)(-208)}}{2(120)}=\frac{-232\pm392}{240}\\&\text{Either way, we find that the function crosses the x-axis at} \ x=\frac{2}{3}\text{ and }-\frac{13}{5}.\end{align*}$

Report an Error

Example Question #32 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the root(s) the function: }f(x)=71x - 153x^{2} - 6\end{align*}$

Possible Answers:

$x=-\frac{1}{9},-\frac{6}{17}$

x=-34,-108

$x=\frac{1}{9},\frac{6}{17}$

x=34,108

Correct answer:

$x=\frac{1}{9},\frac{6}{17}$

Explanation:

$\begin{align*}&\text{The roots of the function }f(x)=71x - 153x^{2} - 6\text{ are the x values}\\&\text{that give it a value of zero. To find these x values}\\&\text{we can either factor the equation:}\\&-(9x - 1)\cdot (17x - 6)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(71)\pm\sqrt{(71)^2-4(-153)(-6)}}{2(-153)}=\frac{-71\pm37}{-306}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=\frac{1}{9}\text{ and }\frac{6}{17}.\end{align*}$

Report an Error

Example Question #33 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=224 - 580x^{2} - 328x\\&\text{has a zero value.}\end{align*}$

Possible Answers:

x=-1120,464

$x=-\frac{2}{5},\frac{28}{29}$

$x=\frac{2}{5},-\frac{28}{29}$

x=1120,-464

Correct answer:

$x=\frac{2}{5},-\frac{28}{29}$

Explanation:

$\begin{align*}&\text{Given }f(x)=224 - 580x^{2} - 328x\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&-4\cdot (29x + 28)\cdot (5x - 2)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-328)\pm\sqrt{(-328)^2-4(-580)(224)}}{2(-580)}=\frac{328\pm792}{-1160}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=\frac{2}{5}\text{ and }-\frac{28}{29}.\end{align*}$

Report an Error

Example Question #231 : College Algebra

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=- 176x - 28x^{2} - 48\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=6,\frac{2}{7}$

x=336,16

$x=-6,-\frac{2}{7}$

x=-336,-16

Correct answer:

$x=-6,-\frac{2}{7}$

Explanation:

$\begin{align*}&\text{The function }f(x)=- 176x - 28x^{2} - 48\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-4\cdot (x + 6)\cdot (7x + 2)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-176)\pm\sqrt{(-176)^2-4(-28)(-48)}}{2(-28)}=\frac{176\pm160}{-56}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=-6\text{ and }-\frac{2}{7}.\end{align*}$

Report an Error

Example Question #232 : College Algebra

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=208x^{2} - 6x - 154\\&\text{has a zero value.}\end{align*}$

Possible Answers:

x=-364,352

$x=\frac{7}{8},-\frac{11}{13}$

x=364,-352

$x=-\frac{7}{8},\frac{11}{13}$

Correct answer:

$x=\frac{7}{8},-\frac{11}{13}$

Explanation:

$\begin{align*}&\text{Given }f(x)=208x^{2} - 6x - 154\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&2\cdot (13x + 11)\cdot (8x - 7)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-6)\pm\sqrt{(-6)^2-4(208)(-154)}}{2(208)}=\frac{6\pm358}{416}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=\frac{7}{8}\text{ and }-\frac{11}{13}.\end{align*}$

Report an Error

Example Question #51 : Polynomial Functions

Consider the polynomial

$p(x) = x^{5}+ 5x^{3}- 7x +1$

Which of the following is true of the rational zeroes of ?

Hint: Think "Rational Zeroes Theorem".

Possible Answers:

is the only rational zero.

1 is the only rational zero.

and 1 are the only rational zeroes.

Neither nor 1 is a rational zero, but there is at least one rational zero.

There are no rational zeroes.

Correct answer:

1 is the only rational zero.

Explanation:

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set $\left \{ -1, 1\right \}$ .

1 is a zero of if and only if p(1) =0 . An easy test for this is to add the coefficients and determine whether their sum, which is p(1) , is 0:

$p(x) = x^{5}+ 5x^{3}- 7x +1$

p(1) = 1 + 5-7+1

p(1)= 0

1 is a zero.

is a zero of if and only if p(-1) =0 . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is p(1) , is 0:

$p(x) = x^{5}+ 5x^{3}- 7x +1$

p(-1) = -1-5+7+1

$p(-1) = 2 \ne 0$

is not a zero.

1 is the only rational zero.

Report an Error

Example Question #52 : Polynomial Functions

The polynomial

$P(x)= 2x^{5}+16x^{4}-40x^{3}-3x^{2}-24x+60$

has a zero on one of the following intervals. Which one?

Possible Answers:

(1.2, 1.3)

( 1.1, 1.2 )

(1.3, 1.4)

(1.0, 1.1)

(1.4, 1.5)

Correct answer:

( 1.1, 1.2 )

Explanation:

The Intermediate value Theorem states that if P(x) is a continuous function, a< b , and P(a) and P(b) are of unlike sign, it must hold that P(c)= 0 for some $c \in [a,b]$ . Equivalently, if P(x) has values on the endpoints of an interval that differ in sign, then P(x) has a zero on that interval. Since all polynomial functions are continuous, we may apply this theorem by evaluating $P(x)= 2x^{5}+16x^{4}-40x^{3}-3x^{2}-24x+60$

for x = 1.0, 1.1, 1.2, 1.3, 1.4, 1.5 , and finding two consecutive values such that P(x) differs in sign.

By substitution, we can find that:

P(1.0)= 11

P(1.1)= 3.37662

P(1.2) = -4.08576

P(1.3) = -11.02654

P(1.4)=-17.01792

P(1.5) = -21.5625

The change in sign occurs on the interval ( 1.1, 1.2 ) ; therefore, a zero of is located on this interval.

Report an Error

Example Question #53 : Polynomial Functions

Consider the polynomial

$x^{5}+ 8 x^{3}- 42x + 17$ .

The Rational Zeroes Theorem allows us to reduce the possible rational zeroes of this polynomial to a set comprising how many elements?

Possible Answers:

Correct answer:

Explanation:

By the Rational Zeroes Theorem, any rational zero of a polynomial with integer coefficients must be equal to a factor of the constant divided by a factor of the leading coefficient, taking both positive and negative numbers into account.

The constant 17, being prime, has only two factors, 1 and 17; the leading coefficient is 1, which only has 1 as a factor. Thus, the only possible rational zeroes of the given polynomial are given in the set

$\left \{ \pm 1, \pm 17 \right \}$ ,

a set of four elements. This makes 4 the correct choice.

Report an Error

Example Question #233 : College Algebra

Consider the polynomial

$x^{5}+ 8 x^{3}- 42x + 17$ .

Going by Descartes' Rule of Signs alone, how many positive real zeroes could the polynomial have?

Possible Answers:

Exactly three

One or three

Exactly two

None or two

Exactly one

Correct answer:

None or two

Explanation:

By Descartes' Rule of Signs, the number of positive real zeroes of a polynomial p(x) with real coefficients is, at maximum, the number of its sign changes, but equal to some number of the same odd-even parity. This number can be seen to be two:

$x^{5}+ 8 x^{3}\textbf{{\color{Red} - }}42x \textbf{{\color{Red} +}} 17$

This means that this polynomial must have either zero or two such zeroes.

Report an Error

View College Algebra Tutors

Ger
Certified Tutor

Pennsylvania State University-Penn State Lehigh Valley, Bachelor of Science, Psychology.

View College Algebra Tutors

Nicholas
Certified Tutor

Stony Brook University, Bachelor of Science, Mechanical Engineering.

View College Algebra Tutors

Samantha
Certified Tutor

Middlesex County College, Associate in Science, Sustainability Studies.

All College Algebra Resources

5 Diagnostic Tests 84 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Computer Science Tutors in New York City, Algebra Tutors in New York City, GRE Tutors in Denver, GRE Tutors in Seattle, GRE Tutors in Atlanta, Biology Tutors in New York City, Algebra Tutors in San Diego, MCAT Tutors in New York City, MCAT Tutors in Atlanta, MCAT Tutors in San Diego

Popular Courses & Classes

GMAT Courses & Classes in Boston, Spanish Courses & Classes in Seattle, ACT Courses & Classes in Chicago, ACT Courses & Classes in Atlanta, SAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Chicago, ACT Courses & Classes in Philadelphia, ACT Courses & Classes in Miami, GMAT Courses & Classes in Miami, SSAT Courses & Classes in Seattle

Popular Test Prep

GMAT Test Prep in Boston, LSAT Test Prep in Miami, ACT Test Prep in San Francisco-Bay Area, ACT Test Prep in Dallas Fort Worth, LSAT Test Prep in Washington DC, ISEE Test Prep in Atlanta, SAT Test Prep in San Diego, ACT Test Prep in Philadelphia, GMAT Test Prep in Miami, SAT Test Prep in Chicago

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

College Algebra : Zeros/Roots of a Polynomial

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #31 : Zeros/Roots Of A Polynomial

Example Question #225 : College Algebra

Example Question #32 : Zeros/Roots Of A Polynomial

Example Question #33 : Zeros/Roots Of A Polynomial

Example Question #231 : College Algebra

Example Question #232 : College Algebra

Example Question #51 : Polynomial Functions

Example Question #52 : Polynomial Functions

Example Question #53 : Polynomial Functions

Example Question #233 : College Algebra

All College Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: