The Remainder Theorem says that if you divide some polynomial *p*(*x*) by the linear factor *x* – *a*, the remainder that you get is equal to *p*(*a*). For example, consider the problem below where the polynomial *p*(*x*) = *x**² – *2*x* – 11 is divided by *x* – 5 using either Long Division or Synthetic Division.

The remainder in both cases is 4. The Remainder Theorem says that if the remainder when you divide by *x* – 5 is 4, then it is also true that *p*(5) = 4. It is easy to check that this is indeed true.

*p*(*x*) = *x**² – *2*x* – 11

*p*(5) = 5² – 2(5) – 11

*p*(5) = 25 – 10 – 11

*p*(5) = 4

There is also a relationship between the polynomial *p*(*x*), the divisor (*x *– *a*), the quotient *q*(*x*) and the remainder *r*. Specifically

*p*(*x*) = *q*(*x*)⋅(*x* – *a*) + r

Using the example above, you can write

*p*(*x*) = (*x* + 3)(*x* – 5) + 4

*p*(*x*) = *x**² – *5*x* + 3*x* – 15 + 4

*p*(*x*) = *x**² – *2*x* – 11

OK, let’s try a few problems. When you’re done, scroll down for the answers.

1. A student used division to divide a polynomial *p*(*x*) by *x* – 3 and got a remainder of 5. Which of the following statements about *p*(*x*) must be true?

A) *x* – 3 is a factor of *p*(*x*).

B) *x* – 5 is a factor of *p*(x)

C) The value of *p*(-3) is 5.

D) The value of *p*(3) is 5.

2. When a polynomial *p*(*x*) is divided by* x* – 4, the quotient is another polynomial *q*(*x*), and the remainder is -6. Which of the following must be true of *p*(*x*)?

A) *p*(*x*) = (*x* – 4)(*x* – 6)

B) *p*(x) = (*x* – 4)(*x* + 6)

C) *p*(*x*) = *q*(*x*)(*x* – 4) – 6

D) *p*(*x*) = *q*(*x*)·(*x* – 4) + 6

3. When a polynomial *p*(*x*) is divided by *x* + 3, the quotient *q*(*x*) is *x* – 8 and the remainder *r* is 5. Which of the following is the polynomial *p*(*x*)?

A) *p*(*x*) = *x*^{2} – 5*x* – 29

B) *p*(*x*) = *x*^{2} – 5*x* – 19

C) *p*(*x*) = *x*^{2} + 5*x* + 19

D) *p*(*x*) = *x*^{2} + 5*x* + 29

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(Scroll down for answers)

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1. D

2. C

3. B

*p*(*x*) = (*x* + 3)(*x* – 8) + 5

*p*(*x*) = *x*² – 8*x* + 3*x* – 24 + 5

*p*(*x*) = *x**² –* 5*x* – 19

If you have questions about these problems or anything else related to the SAT, send us an email at info@cardinalec. com.

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*f*(*x*) = a·b^*x*

In this equation, *a* and *b* are constants and the variable, *x*, appears as an exponent.

The number *a* is the initial value (when *x* = 0). For a population growth problem, it would represent the initial population of whatever group you are looking at. For a compound interest problem, the *a* represents the amount of the deposit or the loan.

The *b* number is the *growth factor* or *decay factor*. Let’s take a minute to look at how this number *b* is computed. You should add the percent of growth (or subtract the percent of decay) to 100 percent and then convert that percent to a decimal.

Example: A population increase of 4.5 percent per year.

*b* = 100% + 4.5% = 104.5% = 1.045

Example: A population decrease of 3 percent per year.

*b* = 100% – 3% = 97% = 0.97

Here are some practice problems involving exponential growth that you can try on your own. When you’re done, scroll down for the answers.

1) You deposit $500 into a bank account that pays 3% interest compounded annually. Which of the following equations could you use to compute the amount in the account (in dollars), *A*, at the end of two years?

A) *A* = 500(.02)³

B) *A *= 500(.03)²

C) *A* = 500(1.02)³

D) *A* = 500(1.03)²

2) Grid-In

Professor Roberts has been studying ways to encourage the growth of a population of endangered California seals. The current population of the seals is 600, and Professor Roberts believes that the population will grow at 3 percent if a conservation plan is put in place. She models the population, *P*, of the seals after *t* years using the function *P* = *a·b^t. *What would be the value of *b* in this function?

3) Grid-In

Professor Jenkins believes that the population of seals in the problem above will grow at a rate of 3.5 percent rather than at 3 percent. Professor Roberts and Professor Jenkins both use their models to predict the population of seals after ten years. How many *more* seals will Professor Jenkins’ model predict compared to the amount predicted by Professor Roberts’ model?

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Scroll down for answers

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1) D

2) 1.03

*b* = 100% + 3% = 103% = 1.03

3) 40

Jenkins: 600(1.035)^10 ≈ 846 seals

Roberts: 600(1.03)^10 ≈ 806 seals

846 – 806 = 40

If you have questions about these problems or anything else to do with the SAT, send us an email at info@cardinalec.com.

]]>The circle below has its center at *O* and triangle *AOB* is equilateral with sides that are each 6 inches.

1. What is the area, in square inches, of the shaded region?

A. 6π – 3√3

B. 9π – 3√3

C. 6π – 6√3

D. 6π – 9√3

E. 9π – 9√3

2. What is the perimeter, in inches, of the shaded region?

F. 2π + 6

G. 2π + 18

H. 4π + 6

J. 4π + 18

K. 6π + 6

When you’re ready, scroll down for the solutions.

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Solutions

1. The area of the shaded region is equal to the area of the triangle subtracted from the area of the sector. Begin by finding each of these areas. To find the area of the sector you can either use the formula for sector area or view the sector as some part of the total area of the circle.

By formula:

*A* =(1/2)(*r*^{2})(*θ*)

*A* = (1/2)(6^{2})(π/3) = 6π

(Note: The triangle is equilateral so each of its angles has a degree measure of 60°. A degree measure of 60° is equivalent to π/3 radians.)

As part of a circle:

Full circle:

*A* = πr^{2
}*A* = π(6^{2}) = 36π

Sector:

*A* = (60/360)(36π)

*A* = (1/6)(36π)

*A* = 6π

(Note: The angle in the triangle, which is also the central angle of the sector is 60°. Therefore, the sector’s area is 60/360 or 1/6 the area of the full circle.)

To find the triangle’s area, use the formula for the area of an equilateral triangle:

A = (6²√3)/4

A = 36√3/4

A = 9√3

Therefore, the area of the shaded region is equal to 6π – 9√3 and the correct answer is Choice D.

2. The perimeter of the shaded region consists of one side of the triangle and the arc length for the central angle of 60°. The side of the triangle is 6. The arc length can be found using the arc length formula or by thinking of it as some part of the full circumference of the circle.

By formula:

s = r·θ

s = 6(π/3)

s = 2π

As part of the circumference:

Full circumference:

C = 2πr

C = 2π(6)

C = 12π

Arc Length:

s = (60/360)(12π)

s = (1/6)(12π)

s = 2π

So the perimeter of the shaded region is 2π + 6 and the correct answer is Choice F.

If you have questions about this problem or anything else to do with the SAT, send us an email at info@cardinalec.com .

]]>Let’s look at three problems based on data in a histogram.

The next three problems are based on the histogram below.

The library at the local college is trying to determine how frequently the library is used by students. They surveyed 25 students asking them “How many times did you use the library this week?” The results are summarized in the histogram below.

1. What is the mean number of times the students used the library?

(A) 2.12

(B) 2.50

(C) 2.52

(D) 2.61

(E) 2.74

2. What is the median number of times the students used the library?

(A) 1.5

(B) 2

(C) 2.5

(D) 3

(E) 3.5

3. What is the mode of the data presented in the table?

(A) 1 only

(B) 2 only

(C) 3 only

(D) Both 4 and 5

(E) There is no mode for the data in the table.

Solutions:

To find the mean, we need to know the total number of times that the library was used. We can find that by multiplying the number of uses by the number of students for each bar of the histogram. For instance, there were 6 students who used the library 3 times this week, so that’s a total of 6(3) = 18 uses for that bar. The calculation would look like this:

0(2) + 1(4) + 2(7) + 3(6) + 4(3) + 5(3)

= 0 + 4 + 14 + 18 + 12 + 15

= 63 total uses of the library

Now divide the number of uses by the number of students: 63/25 = 2.52. The mean number of uses is 2.52 and the correct answer to the first question is Choice C.

To find the median, you could list all of the numbers in order and find the middle number taking into consideration that many of the numbers occur more than once. While it’s possible to do that for this problem, with 25 numbers it might take a while. Is there an easier way? If you divide 25 (the number of students) by 2, you get 12.5. Rounding that up to 13 tells you that the 13th number is the middle number or median.

(Note: If there are an even number of data points, you still divide by 2. Take that term and the next one and average them to get the median. For instance, suppose there are 40 data points. Dividing by 2 gets you 20. The median is the average of the 20th and 21st terms.)

So which term is the 13th in our list? Begin adding from the left-most bar of the histogram. There are 2 students who never visited the library and another 4 who visited once. That’s 6 students. Next, we see that 7 students visited the library twice. That brings us up to 13 students, so the 13th number in the list would be a 2. The median is 2 and the correct answer is Choice B.

The mode is the number that appears most frequently. What answer did the students give more than any other? From the histogram we can see that more students went to the library twice than any other number of times. The mode is 2 and the correct answer to the third question is Choice B.

If you have questions about this problem or anything else to do with the SAT, send us an email at info@cardinalec.com .

]]>**Mean** – When people say they are finding the average, what they are usually finding is the mean, which is also called the *arithmetic mean*. To find the mean, you add up all of the data points and divide by the number of numbers.

**Median** – The median is the middle number when the numbers are ordered from least to greatest. If you have an even number of numbers, there is no “middle” number. In this case, the median is the mean of the two middle numbers.

**Mode** – The mode is the number that occurs the most frequently. It is possible to have more than one mode if there is a tie for what occurs most frequently. If each of the numbers appears only once, then we say there is *no mode*.

**Range** – The range is the difference between the largest number and the smallest number.

Now, let’s look a few problems that involve data that is presented to you in a frequency table.

The next three questions refer to the data presented in the table below.

A survey of 15 elementary students asked them how many pets they have. Their responses are summarized in the table below.

1. What is the mean number of pets that the students have?

(A) 1.4

(B) 1.6

(C) 1.8

(D) 2.0

(E) 2.5

2. What is the median number of pets that the students have?

(A) 1

(B) 1.5

(C) 2

(D) 2.5

(E) 3

3. What is the mode of the data presented in the table?

(A) 1

(B) 2

(C) 2.5

(D) 3

(E) There is no mode for the data in the table.

Solutions:

To find the mean, we need to know how many total pets the students had. We can find that by multiplying the number of students by the number of pets for each row of the table. For instance, there were 2 students who had 3 pets, so that’s a total of 2(3) = 6 pets for that row. The calculation would look like this:

0(2) + 1(7) + 2(3) + 3(1) + 4(2)

= 0 + 7 + 6 + 3 + 8

= 24 total pets

Now divide the number of pets by the number of students: 24/15 = 1.6. The mean number of pets is 1.6 and the correct answer to the first question is Choice B.

To find the median, you need to list all of the numbers in order and find the middle number. Make sure you take into consideration that many of the numbers occur more than once. The median is *not* simply 2 because that number is in the middle of the table. Here are all the students’ answers listed in order:

0 0 1 1 1 1 1 1 1 2 2 2 3 4 4

Counting in from each side, you will see that the middle number (the 8th term in the list) is 1, so the correct answer to the second question is Choice A.

The mode is the number that appears most frequently. What answer did the students give more than any other? From the table we can see that more students had one pet than any other number of pets. The mode is 1 and the correct answer to the third question is Choice A.

If you have questions about this problem or anything else to do with the SAT, send us an email at info@cardinalec.com .

]]>Here’s a challenging problem involving a quadratic function. Give it a try and then scroll down for the solution.

*f*(*x*) = *a*(*x* + *b*)(*x* – 8)

The equation above represents a quadratic function where *a* and *b* are constants. If the function is graphed in the *xy*-plane, its vertex is at the point (2, -54). What is the value of *a* + *b*?

(A) 4

(B) 5.5

(C) 6

(D) 7.5

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(Scroll down for the solution)

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Solution:

In order to get this problem correct, you need to know about the symmetry properties of the graphs of quadratic functions. The graph of a quadratic function is always a parabola centered on a line called the *axis of symmetry* which passes through the vertex.

The *x*-intercepts (or “roots” or “zeroes”) of the function must be the same distance away from the axis of symmetry. In this case, the axis of symmetry is the vertical line *x* = 2. We know this because the vertex of the function is (2, -54).

For this function, because one of the factors is (*x* – 8), we know that one of the *x*-intercepts of the graph must be (8, 0). This point lies 6 horizontal units away from the axis of symmetry. The other *x*-intercept must also be 6 horizontal units away from the axis of symmetry but in the other direction, so it has to be (-4, 0). That makes the first factor (*x* + 4).

Here’s a graph of the function:

Now that we know the value of *b*, let’s see if we can figure out the value of *a*. We can do this using the fact that one of the points on the graph is (2, -54). Let’s plug that point into the function and solve for *a*.

*f*(*x*) = *a*(*x* + 4)(*x* – 8)

-54 = *a*(2 + 4)(2 – 8)

-54 = *a*(6)(-6)

-54 = *a*(-36)

-54/-36 = *a*

1.5* = a*

So, the value of *a* is 1.5 and the value of *b* is 4. That makes *a* + *b* = 1.5 + 4 = 5.5. The correct answer is Choice B.

If you want to get some more practice working with quadratic functions and their graphs, you might want to check out our e-book, SAT Math: Focus on Quadratics & Parabolas. It’s available on Amazon for only $0.99.

If you have questions about this problem or anything else to do with the SAT, send us an email at info@cardinalec.com

]]>A regular polygon is covered by a piece of paper so that only one of its interior angles is visible. You are able to determine that the interior angle is four times as large as the exterior angle. How many sides does the polygon have?

A. 8

B. 9

C. 10

D. 11

E. 12

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(Scroll down for the solution)

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Solution:

The interior angle and the exterior angle form a linear pair and the sum of their angle measures is 180 degrees. You can write and solve an equation to find the value of *x*:

4*x* + *x* = 180

5*x* = 180

*x* = 36

You now know that the interior angle is 4 x 36 = 144 degrees and the exterior angle is 36 degrees. You could work with the interior angles formula, but it’s much easier if you remember that the exterior angles of any polygon *always* sum to 360 degrees. Dividing 36 into 360 tells us that the polygon has 10 exterior angles and, therefore, 10 sides.

If you want to try some challenging problem similar to those you’ll see on ACT test day, you might want to check out our book, *200 Challenging ACT Math Problems*, available on Amazon.

If you have questions about this problem or anything else on the ACT, send us an email at info@cardinalec.com .

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This book is the second in a series that addresses the topics that appear most frequently in the Math section of the New SAT. You’ll review the properties of lines, including their graphs, and look at the different methods for solving systems of linear equations. Then you’ll apply what you know as you solve 25 problems — some done without your calculator and some with, just as it is on test day. Each problem comes with a full explanation of how to arrive at the correct answer.

We’d like to help as many students as possible achieve success on the SAT, so we’ve made the price of each of the volumes in this series just $0.99.

The book can be purchased on Amazon by clicking on the image to the left or going here.

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This book will be the first in a series that addresses the topics that appear most frequently in the Math section of the New SAT. You’ll get a chance to review the most important concepts of quadratic equations and the graphs of parabolas. Then you’ll apply what you know to 25 problems — some done without your calculator and some with, just as it is on test day. Each problem comes with a full explanation of how to arrive at the correct answer.

We’d like to help as many students as possible achieve success on the SAT, so we’ve made the price of this volume just $0.99.

The book can be purchased on Amazon by clicking on the image to the left or going here.

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Complete the Square Problems

Complete the Square Solutions

If you have any questions about these problems, completing the square in general, or anything else to do with the SAT, send us an email at info@cardinalec.com .

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