Call Now to Set Up Tutoring:

(888) 888-0446

High School Math : High School Math

Study concepts, example questions & explanations for High School Math

Create An Account

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Understanding Zeros Of A Polynomial

A polyomial with leading term $x^{3}$ has 6 as a triple root. What is this polynomial?

Possible Answers:

$x ^{3} + 18 x^{2}- 108x -216$

$x ^{3} + 18 x^{2}+ 108x +216$

$x ^{3} - 18 x^{2}+ 108x -216$

$x ^{3} +216$

$x ^{3} -216$

Correct answer:

$x ^{3} - 18 x^{2}+ 108x -216$

Explanation:

Since 6 is a triple root, and the degree of the polynomial is 3, the polynomial is $(x-6)^{3}$ , which we can expland using the cube of a binomial pattern.

$(x-6)^{3} =x ^{3} - 3\cdot x^{2} \cdot 6 + 3\cdot x\cdot 6 ^{2} -6^{3}$

$x ^{3} - 18 x^{2}+ 108x -216$

Report an Error

Example Question #1 : Express A Polynomial As A Product Of Linear Factors

A polyomial with leading term $x^ {3}$ has 5 and 7 as roots; 7 is a double root. What is this polynomial?

Possible Answers:

$x^{3}+ 19x^{2} +119x +245$

$x^{3}- 17x^{2} +95x -175$

$x^{3} - 19x^{2} + 119x -245$

$x^{3} -12x^{2} + 35x$

$x^{3}+ 17x^{2} +95x +175$

Correct answer:

$x^{3} - 19x^{2} + 119x -245$

Explanation:

Since 5 is a single root and 7 is a double root, and the degree of the polynomial is 3, the polynomial is $(x-5)(x-7)^{2}$ . To put this in expanded form:

$(x-7)^{2} = x^{2} - 2 \cdotx\cdot 7 + 7^{2} = x^{2} - 14x + 49$

$(x-5)(x-7)^{2} =(x-5) \left ( x^{2} - 14x + 49 \right )$

$=x \left ( x^{2} - 14x + 49 \right ) -5\left( x^{2} - 14x + 4 9 \right )$

$= x^{3} - 14x^{2} + 49x - \left( 5x^{2} - 70x + 245 \right )$

$= x^{3} - 14x^{2} + 49x - 5x^{2} + 70x -245$

$= x^{3} - 19x^{2} + 119x -245$

Report an Error

Example Question #1 : Understanding Zeros Of A Polynomial

What are the solutions to $y = x^{2} + x -6$ ?

Possible Answers:

x = -6, x = 1

x = 6, x = -1

x = -3, x = 2

x = 3, x = -2

x = 5, x = 1

Correct answer:

x = -3, x = 2

Explanation:

When we are looking for the solutions of a quadratic, or the zeroes, we are looking for the values of such that the output will be zero. Thus, we first factor the equation.

$0 = x^{2} + x - 6 = (x+3)(x-2)$

Then, we are looking for the values where each of these factors are equal to zero.

x + 3 = 0 implies x = -3

and x - 2 = 0 implies x = 2

Thus, these are our solutions.

Report an Error

Example Question #11 : Pre Calculus

Find the zeros of the following polynomial:

$f(x) = x^{4} - 4x^{3} - 7x^{2} + 22x + 24$

Possible Answers:

x = -2, -1, 3, 4

x = -3, -2, 1, 4

x = -4, -3, 1, 2

x = -4, -1, 2, 3

x = -4, -2, 1, 3

Correct answer:

x = -2, -1, 3, 4

Explanation:

First, we need to find all the possible rational roots of the polynomial using the Rational Roots Theorem:

$\frac{\textup{factors of the constant}}{\textup{factors of the leading coefficient}}$

Since the leading coefficient is just 1, we have the following possible (rational) roots to try:

±1, ±2, ±3, ±4, ±6, ±12, ±24

When we substitute one of these numbers for , we're hoping that the equation ends up equaling zero. Let's see if is a zero:

$f(-1)=(-1)^{4}-4(-1)^{3}-7(-1)^{2}+22(-1)+24$

f(-1)=1+4-7-22+24

f(-1)=0

Since the function equals zero when is , one of the factors of the polynomial is (x+1) . This doesn't help us find the other factors, however. We can use synthetic substitution as a shorter way than long division to factor the equation.

$\textup{-1 } | \textup{ 1 -4 -7 22 24}$

$\small \textup{ -1 }\textup{ 5 }\textup{ 2 }\textup{ -24}$

$\frac{ }{\textup{ 1 -5 -2 24 }\textup{ 0}}$

Now we can factor the function this way:

$f(x)=(x+1)(x^{3}-5x^{2}-2x+24)$

We repeat this process, using the Rational Roots Theorem with the second term to find a possible zero. Let's try :

$f(-2)=[(-2)+1][(-2)^{3}-5(-2)^{2}-2(-2)+24]$

f(-2)=(-1)(-8-20+4+24)=0

When we factor using synthetic substitution for x=-2 , we get the following result:

$f(x)=(x+1)(x+2)(x^{2}-7x+12)$

Using our quadratic factoring rules, we can factor completely:

f(x)=(x+1)(x+2)(x-4)(x-3)

Thus, the zeroes of f(x) are x=-2, -1, 3, 4.

Report an Error

Example Question #12 : Pre Calculus

Simplify the following polynomial:

$(\frac{-a^{-2}b^{3}c^{-1}}{3a^{-4}b^{-1}c^{3}})^{-2}$

Possible Answers:

$\frac{9a^{4}}{c^{8}b^{4}}$

$\frac{9b^{8}}{c^{8}a^{4}}$

$\frac{c^{8}}{9a^{4}b^{8}}$

$\frac{9a^{4}}{c^{8}b^{8}}$

$\frac{9c^{8}}{a^{4}b^{8}}$

Correct answer:

$\frac{9c^{8}}{a^{4}b^{8}}$

Explanation:

To simplify the polynomial, begin by combining like terms:

$(\frac{-a^{-2}b^{3}c^{-1}}{3a^{-4}b^{-1}c^{3}})^{-2}$

$(\frac{-a^{2}b^{4}}{3c^{4}})^{-2}$

Report an Error

Example Question #21 : Pre Calculus

Simplify the following polynomial function:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

Possible Answers:

$\frac{a^{3}}{b}+\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}+b^{2}-\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}-\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}-b+\frac{a^{4}}{b^{3}}$

Correct answer:

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

Explanation:

First, multiply the outside term with each term within the parentheses:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

$a^{3}b^{-1} + b-a^{4}b^{-3}$

Rearranging the polynomial into fractional form, we get:

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

Report an Error

Example Question #2031 : High School Math

You are given that $\log_{10} x = A$ and $\log_{10} y = B$ .

Which of the following is equal to $100^{2A-B}$ ?

Possible Answers:

$\frac{2x}{y}$

$x^{4}y^{2}$

$\frac{x}{\sqrt{y}}$

$\frac{x^{4}}{y^{2}}$

$\frac{y^{2}}{x^{4}}$

Correct answer:

$\frac{x^{4}}{y^{2}}$

Explanation:

Since $\log_{10} x = A$ and $\log_{10} y = B$ , it follows that $10^{A} = x$ and $10^{B} = y$

$100^{2A-B} = \left( 10^{2} \right ) ^{2A-B} =10^{2 (2A-B)} =10^{4A-2B}=\frac{10^{4A}}{10^{2B}} =\frac{\left ( 10^{A}\right )^4}{\left (10^{B}\right )^2}=\frac{x^4}{y^2}$

Report an Error

Example Question #2032 : High School Math

$\textup{Solve for }x\textup{: }\,\log_{2}32=x$

Possible Answers:

Correct answer:

Explanation:

$\textup{This can be re-written as an exponent problem: } 2^{x}=32$

$2\times2\times2\times2\times2=32\:\textup{ so }32=2^{5}\:\:\:x=5$

Report an Error

Example Question #2 : Solving Logarithms

What is $log_{2}(8)$ ?

Possible Answers:

$\frac{1}{8}$

Correct answer:

Explanation:

Recall that by definition a logarithm is the inverse of the exponential function. Thus, our logarithm corresponds to the value of in the equation:

$2^{x} = 8$ .

We know that $2^{3} = 8$ and thus our answer is .

Report an Error

Example Question #2033 : High School Math

Solve for : $\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

Possible Answers:

The correct solution set is not included among the other choices.

$x = 4 \textrm{ or }x = 128$

$x = 1 \textrm{ or }x = \log_{2}7$

$x = -7 \textrm{ or }x = -2$

$x = 2 \textrm{ or }x = 7$

Correct answer:

The correct solution set is not included among the other choices.

Explanation:

$\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\log_{2} \left[ \left ( x+4 \right )\left ( x+5 \right ) \right ] = \log_{2} 6$

$2^{ \log_{2} \left[ \left ( x+4 \right )\left ( x+5 \right ) \right ] }=2^{ \log_{2} 6}$

(x + 4)(x + 5) = 6

FOIL: $x^{2} + 4x + 5x + 4 \cdot 5 = 6$

$x^{2} + 9x + 20 = 6$

$x^{2} + 9x + 20 - 6 = 6- 6$

$x^{2} + 9x + 14 = 0$

(x + 2) (x + 7) = 0

$x + 2 = 0 \textrm{ or }x + 7 = 0$

$x = -2 \textrm{ or } x = -7$

These are our possible solutions. However, we need to test them.

$\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\log_{2}\left ( -2+4 \right ) + \log_{2}\left ( -2+5 \right ) = \log_{2} 6$

$\log_{2}2 + \log_{2}3 = \log_{2} 6$

$\log_{2}2 = 1$

$\log_{2}3 = \frac{\ln 3}{\ln 2} \approx \frac{1.10}{0.69} \approx 1.59$

$\log_{2}6 = \frac{\ln 6}{\ln 2} \approx \frac{1.79}{0.69} \approx 2.59$

The equation becomes 1 + 1.59 = 2.59 . This is true, so x = -2 is a solution.

x = -7 :

$\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\log_{2}\left ( -7+4 \right ) + \log_{2}\left ( -7+5 \right ) = \log_{2} 6$

$\log_{2}\left (-3 \right ) + \log_{2}\left (-2 \right ) = \log_{2} 6$

However, negative numbers do not have logarithms, so this equation is meaningless. x = -7 is not a solution, and x = -2 is the one and only solution. Since this is not one of our choices, the correct response is "The correct solution set is not included among the other choices."

Report an Error

View Tutors

Richie
Certified Tutor

Full Sail University, Bachelor of Science, Computer Science.

View Tutors

Bobby
Certified Tutor

University of Houston - Downtown, Bachelor of Science, Applied Mathematics.

View Tutors

Arushi
Certified Tutor

Emory University, Bachelor of Science, Chemistry.

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ISEE Tutors in New York City, SAT Tutors in Washington DC, Math Tutors in Denver, Computer Science Tutors in San Francisco-Bay Area, Spanish Tutors in Miami, Calculus Tutors in Boston, English Tutors in Phoenix, Chemistry Tutors in Chicago, Biology Tutors in Miami, Reading Tutors in Chicago

Popular Courses & Classes

ACT Courses & Classes in Miami, Spanish Courses & Classes in Seattle, MCAT Courses & Classes in Los Angeles, ACT Courses & Classes in Washington DC, MCAT Courses & Classes in New York City, GRE Courses & Classes in Phoenix, SSAT Courses & Classes in San Diego, ACT Courses & Classes in Boston, SAT Courses & Classes in Philadelphia, ISEE Courses & Classes in San Francisco-Bay Area

Popular Test Prep

MCAT Test Prep in Houston, ISEE Test Prep in Chicago, MCAT Test Prep in New York City, SSAT Test Prep in Houston, MCAT Test Prep in Chicago, SSAT Test Prep in Denver, ACT Test Prep in Miami, ACT Test Prep in Phoenix, ISEE Test Prep in Atlanta, LSAT 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

High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #2 : Understanding Zeros Of A Polynomial

Example Question #1 : Express A Polynomial As A Product Of Linear Factors

Example Question #1 : Understanding Zeros Of A Polynomial

Example Question #11 : Pre Calculus

Example Question #12 : Pre Calculus

Example Question #21 : Pre Calculus

Example Question #2031 : High School Math

Example Question #2032 : High School Math

Example Question #2 : Solving Logarithms

Example Question #2033 : High School Math

All High School Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: