Pre-Calculus Articles

Article / Updated 08-14-2023

As you work through pre-calculus, adopting certain tasks as habits can help prepare your brain to tackle your next challenge: calculus. In this article, you find ten habits that should be a part of your daily math arsenal. Perhaps you’ve been told to perform some of these tasks since elementary school — such as showing all your work — but other tricks may be new to you. Either way, if you remember these ten pieces of advice, you’ll be even more ready for whatever calculus throws your way. Figure Out What the Problem Is Asking Often, you’ll find that reading comprehension and the ability to work with multiple parts that comprise a whole is an underlying property of a math problem. That’s okay — that’s also what life is all about!! When faced with a math problem, start by reading the whole problem or all the directions to the problem. Look for the question inside the question. Keep your eyes peeled for words like solve, simplify, find, and prove, all of which are common buzz words in any math book. Don’t begin working on a problem until you’re certain of what it wants you to do. For example, take a look at this problem: The length of a rectangular garden is 24 feet longer than the garden’s width. If you add 2 feet to both the width and the length, the area of the garden is 432 square feet. How long is the new, bigger garden? If you miss any of the important information, you may start to solve the problem to figure out how wide the garden is. Or you may find the length but miss the fact that you’re supposed to find out how long it is with 2 feet added to it. Look before you leap! Underlining key words and information in the question is often helpful. This can’t be stressed enough. Highlighting important words and pieces of information solidifies them in your brain so that as you work, you can redirect your focus if it veers off-track. When presented with a word problem, for example, first turn the words into an algebraic equation. If you’re lucky and are given the algebraic equation from the get-go, you can move on to the next step, which is to create a visual image of the situation at hand. And, if you’re wondering what the answer to the example problem is, you’ll find out as you read further. Draw Pictures (the More the Better) Your brain is like a movie screen in your skull, and you’ll have an easier time working problems if you project what you see onto a piece of paper. When you visualize math problems, you’re more apt to comprehend them. Draw pictures that correspond to a problem and label all the parts so you have a visual image to follow that allows you to attach mathematical symbols to physical structures. This process works the conceptual part of your brain and helps you remember important concepts. As such, you’ll be less likely to miss steps or get disorganized. If the question is talking about a triangle, for instance, draw a triangle; if it mentions a rectangular garden filled with daffodils for 30 percent of its space, draw that. In fact, every time a problem changes and new information is presented, your picture should change, too. If you were asked to solve the rectangular garden problem from the previous section, you’d start by drawing two rectangles: one for the old, smaller garden and another for the bigger one. Putting pen or pencil to the paper starts you on the way to a solution. Plan Your Attack — Identify Your Targets When you know and can picture what you must find, you can plan your attack from there, interpreting the problem mathematically and coming up with the equations that you’ll be working with to find the answer: Start by writing a statement. In the garden problem from the last two sections, you’re looking for the length and width of a garden after it has been made bigger. With those in mind, define some variables: Let x = the garden’s width now. Let y = the garden’s length now. Now add those variables to the rectangle you drew of the old garden. But you know that the length is 24 feet greater than the width, so you can rewrite the variable y in terms of the variable x so that y = x + 24. You know that the new garden has had 2 feet added to both its width and length, so you can modify your equations: Let x + 2 = the garden’s new width. Let y + 2 = x + 24 + 2 = x + 26 the garden’s new length. Now add these labels to the picture of the new garden. By planning your attack, you’ve identified the pieces of the equation that you need to solve. Write Down Any Formulas If you start your attack by writing the formula needed to solve the problem, all you have to do from there is plug in what you know and then solve for the unknown. A problem always makes more sense if the formula is the first thing you write when solving. Before you can do that, though, you need to figure out which formula to use. You can usually find out by taking a close look at the words of the problem. In the case of the garden problem from the previous sections, the formula you need is that for the area of a rectangle. The area of a rectangle is A = lw. You’re told that the area of the new rectangle is 432 square feet, and you have expressions representing the length and width, so you can replace A = lw with 432 = (x + 26)(x + 2). As another example, if you need to solve a right triangle, you may start by writing down the Pythagorean Theorem if you know two sides and are looking for the third. For another right triangle, perhaps you’re given an angle and the hypotenuse and need to find the opposite side; in this situation, you’d start off by writing down the sine ratio. Show Each Step of Your Work Yes, you’ve been hearing it forever, but your third-grade teacher was right: Showing each step of your work is vital in math. Writing each step on paper minimizes silly mistakes that you can make when you calculate in your head. It’s also a great way to keep a problem organized and clear. And it helps to have your work written down when you get interrupted by a phone call or text message — you can pick up where you left off and not have to start all over again. It may take some time to write every single step down, but it’s well worth your investment. Know When to Quit Sometimes a problem has no solution. Yes, that can be an answer, too! If you’ve tried all the tricks in your bag and you haven’t found a way, consider that the problem may have no solution at all. Some common problems that may not have a solution include the following: Absolute-value equations This happens when the absolute value expression is set equal to a negative number. You may not realize the number is negative, at first, if it’s represented by a variable. Equations with the variable under a square-root sign If your answer has to be a real number, and complex numbers aren’t an option, then the expression under the radical may represent a negative number. Not allowed. Quadratic equations When a quadratic isn’t factorable and you have to resort to the quadratic formula, you may run into a negative under the radical; you can’t use that expression if you’re allowed only real answers. Rational equations Rational expressions have numerators and denominators. If there’s a variable in the denominator that ends up creating a zero, then that value isn’t allowed. Trig equations Trig functions have restrictions. Sines and cosines have to lie between –1 and 1. Secants and cosecants have to be greater than or equal to 1 or less than or equal to –1. A perfectly nice-looking equation may create an impossible answer. On the other hand, you may get a solution for some problem that just doesn’t make sense. Watch out for the following situations: If you’re solving an equation for a measurement (like length or area) and you get a negative answer, either you made a mistake or no solution exists. Measurement problems include distance, and distance can’t be negative. If you’re solving an equation to find the number of things (like how many books are on a bookshelf) and you get a fraction or decimal answer, then that just doesn’t make any sense. How could you have 13.4 books on a shelf? Check Your Answers Even the best mathematicians make mistakes. When you hurry through calculations or work in a stressful situation, you may make mistakes more frequently. So, check your work. Usually, this process is very easy: You take your answer and plug it back into the equation or problem description to see if it really works. Making the check takes a little time, but it guarantees you got the question right, so why not do it? For example, if you go back and solve the garden problem from earlier in this chapter by looking at the equation to be solved: 432 = (x + 26)(x + 2), you multiply the two binomials and move the 432 to the other side. Solving x2 + 28x – 380 = 0 you get x = 10 and x = –38. You disregard the x = –38, of course, and find that the original width was 10 feet. So, what was the question? It asks for the length of the new, bigger garden. The length of the new garden is found with y = x + 26. So, the length (and answer) is that the length is 36 feet. Does this check? If you use the new length or 36 and the new width of x + 2 = 12 and multiply 36 times 12 you get 432 square feet. It checks! Practice Plenty of Problems You’re not born with the knowledge of how to ride a bike, play baseball, or even speak. The way you get better at challenging tasks is to practice, practice, practice. And the best way to practice math is to work the problems. You can seek harder or more complicated examples of questions that will stretch your brain and make you better at a concept the next time you see it. Along with working along on the example problems in this book, you can take advantage of the For Dummies workbooks, which include loads of practice exercises. Check out Trigonometry Workbook For Dummies, by Mary Jane Sterling, Algebra Workbook and Algebra II Workbook For Dummies, both also by Mary Jane Sterling, and Geometry Workbook For Dummies, by Mark Ryan (all published by Wiley), to name a few. Even a math textbook is great for practice. Why not try some (gulp!) problems that weren’t assigned, or maybe go back to an old section to review and make sure you’ve still got it? Typically, textbooks show the answers to the odd problems, so if you stick with those you can always double-check your answers. And if you get a craving for some extra practice, just search the Internet for “practice math problems” to see what you can find! For example, if you search the Internet for “practice systems of equations problems” you’ll find more than a million hits. That’s a lot of practice! Keep Track of the Order of Operations Don’t fall for the trap that always is lying there by performing operations in the wrong order. For instance, 2 – 6 × 3 doesn’t become –4 × 3 = –12. You’ll reach those incorrect answers if you forget to do the multiplication first. Focus on following the order of PEMDAS every time, all the time: Parentheses (and other grouping devices) Exponents and roots Multiplication and Division from left to right Addition and Subtraction from left to right Don’t ever go out of order, and that’s an order! Use Caution When Dealing with Fractions Working with denominators can be tricky. It’s okay to write: But, on the other hand: Also, reducing or cancelling in fractions can be performed incorrectly. Every term in the numerator has to be divided by the same factor — the one that divides the denominator. So, it’s true that: because But , because the factor being divided out is just 4, not 4x. And, again, it has to be the same factor that’s dividing, throughout. Can you spot the error here? Some poor soul reduced each term and the term directly above it — separately. Nope, doesn’t work that way. The correct process is to factor the trinomials and then divide by the common factor:

Article / Updated 05-03-2023

Once you have used the rational root theorem to list all of the possible rational roots of any polynomial, the next step is to test the roots. One way is to use long division of polynomials and hope that when you divide you get a remainder of 0. Once you have a list of possible rational roots, you then pick one and assume that it’s a root. For example, consider the equation f(x) = 2x4 – 9x3 – 21x2 + 88x + 48, which has the following possible rational roots: If x = c is a root, then x – c is a factor. So if you pick x = 2 as your guess for the root, x – 2 should be a factor. You can use long division to test if x – 2 is actually a factor and, therefore, x = 2 is a root. Dividing polynomials to get a specific answer isn’t something you do every day, but the idea of a function or expression that’s written as the quotient of two polynomials is important for pre-calculus. If you divide a polynomial by another and get a remainder of 0, the divisor is a factor, which in turn gives a root. In math lingo, the division algorithm states the following: If f(x) and d(x) are polynomials such that d(x) isn’t equal to 0, and the degree of d(x) isn’t larger than the degree of f(x), there are unique polynomials q(x) and r(x) such that In plain English, the dividend equals the divisor times the quotient plus the remainder. You can always check your results by remembering this information. Remember the mnemonic device Dirty Monkeys Smell Bad when doing long division to check your roots. Make sure all terms in the polynomial are listed in descending order and that every degree is represented. In other words, if x2 is missing, put in a placeholder of 0x2 and then do the division. (This step is just to make the division process easier.) To divide two polynomials, follow these steps: Divide. Divide the leading term of the dividend by the leading term of the divisor. Write this quotient directly above the term you just divided into. Multiply. Multiply the quotient term from Step 1 by the entire divisor. Write this polynomial under the dividend so that like terms are lined up. Subtract. Subtract the whole line you just wrote from the dividend. You can change all the signs and add if it makes you feel more comfortable. This way, you won’t forget signs. Bring down the next term. Do exactly what this says; bring down the next term in the dividend. Repeat Steps 1–4 over and over until the remainder polynomial has a degree that’s less than the dividend’s. The following list explains how to divide 2x4 – 9x3 – 21x2 + 88x + 48 by x – 2. Each step corresponds with the numbered step in the illustration in this figure. The process of long division of polynomials. (Note that using Descartes’s rule of signs, you find that this particular example may have positive roots, so it’s efficient to try a positive number here. If Descartes’s rule of signs had said that no positive roots existed, you wouldn’t test any positives!) Divide. What do you have to multiply x in the divisor by to make it become 2x4 in the dividend? The quotient, 2x3, goes above the 2x4 term. Multiply. Multiply this quotient by the divisor and write it under the dividend. Subtract. Subtract this line from the dividend: (2x4 – 9x3) – (2x4 – 4x3) = –5x3. If you’ve done the job right, the subtraction of the first terms always produces 0. Bring down. Bring down the other terms of the dividend. Divide. What do you have to multiply x by to make it –5x3? Put the answer, –5x2, above the –21x2. Multiply. Multiply the –5x2 times the x – 2 to get –5x3 + 10x2. Write it under the remainder with the degrees lined up. Subtract. You now have (–5x3 – 21x2) – (–5x3 + 10x2) = –31x2. Bring down. The +88x takes its place. Divide. What to multiply by to make x become –31x2? The quotient –31x goes above –21x2. Multiply. The value –31x times (x – 2) is –31x2 + 62x; write it under the remainder. Subtract. You now have (–31x2 + 88x) – (–31x2 + 62x), which is 26x. Bring down. The +48 comes down. Divide. The term 26x divided by x is 26. This answer goes on top. Multiply. The constant 26 multiplied by (x – 2) is 26x – 52. Subtract. You subtract (26x + 48) – (26x – 52) to get 100. Stop. The remainder 100 has a degree that’s less than the divisor of x – 2. Wow . . . now you know why they call it long division. You went through all that to find out that x – 2 isn’t a factor of the polynomial, which means that x = 2 isn’t a root. If you divide by c and the remainder is 0, then the linear expression (x – c) is a factor and that c is a root. A remainder other than 0 implies that (x – c) isn’t a factor and that c isn’t a root.

Article / Updated 10-06-2022

At some point, your pre-calculus teacher will ask you to find the general formula for the nth term of an arithmetic sequence without knowing the first term or the common difference. In this case, you will be given two terms (not necessarily consecutive), and you will use this information to find a1 and d. The steps are: Find the common difference d, write the specific formula for the given sequence, and then find the term you're looking for. For instance, to find the general formula of an arithmetic sequence where a4 = –23 and a22 = 40, follow these steps: Find the common difference. You have to be creative in finding the common difference for these types of problems. a.Use the formula an = a1 + (n – 1)d to set up two equations that use the given information. For the first equation, you know that when n = 4, an = –23: –23 = a1 + (4 – 1)d –23 = a1 + 3d For the second equation, you know that when n = 22, an = 40: 40 = a1 + (22 – 1)d 40 = a1 + 21d b.Set up a system of equations and solve for d. The system looks like this: You can use elimination or substitution to solve the system. Elimination works nicely because you can multiply either equation by –1 and add the two together to get 63 = 18d. Therefore, d = 3.5. Write the formula for the specific sequence. This step involves a little work. a.Plug d into one of the equations to solve for a1. You can plug 3.5 back into either equation: –23 = a1 + 3(3.5), or a1 = –33.5. b.Use a1 and d to find the general formula for an. This step becomes a simple three-step simplification: an = –33.5 + (n – 1)3.5 an = –33.5 + 3.5n – 3.5 an = 3.5n – 37 Find the term you were looking for. In this example, you weren't asked to find any specific term (always read the directions!), but if you were, you could plug that number in for n and then find the term you were looking for.

Article / Updated 09-22-2022

You can use the sum and difference formulas for cosine to calculate the cosine of the sums and differences of angles similarly to the way you can use the sum and difference formulas for sine, because the formulas look very similar to each other. When working with sines and cosines of sums and differences of angles, you're simply plugging in given values for the variables (angles). Just make sure you use the correct formula based on the information you're given in the question. Here are the sum and difference formulas for cosines: The sum and difference formulas for cosine (and sine) can do more than calculate a trig value for an angle not marked on the unit circle (at least for angles that are multiples of 15 degrees). They can also be used to find the cosine (and sine) of the sum or difference of two angles based on information given about the two angles. For such problems, you'll be given two angles (call them A and B), the sine or cosine of A and B, and the quadrant(s) in which the two angles are located. Use the following steps to find the exact value of cos(A + B), given that cos A = –3/5, with A in quadrant II of the coordinate plane, and sin B = –7/25, with B in quadrant III: Choose the appropriate formula and substitute the information you know to determine the missing information. then substitutions result in this equation: To proceed any further, you need to find cos B and sin A. Draw pictures representing right triangles in the quadrant(s). Drawing pictures helps you visualize the missing pieces of info. You need to draw one triangle for angle A in quadrant II and one for angle B in quadrant III. Using the definition of sine as opp/hyp and cosine as adj/hyp, this figure shows these triangles. Notice that the value of a leg is missing in each triangle. To find the missing values, use the Pythagorean theorem. The length of the missing leg in Figure a is 4, and the length of the missing leg in Figure b is –24. Determine the missing trig ratios to use in the sum or difference formula. You use the definition of cosine to find that cos B = –24/25 and the definition of sine to find that sin A = 4/5. Substitute the missing trig ratios into the sum or difference formula and simplify. You now have this equation: Follow the order of operations to get this answer: This equation simplifies to cos(A + B) = 4/5.

Article / Updated 08-08-2022

The 30-60-90 triangle is shaped like half of an equilateral triangle, cut straight down the middle along its altitude. It has angles of 30 degrees, 60 degrees, and 90 degrees, thus, its name! In any 30-60-90 triangle, you see the following: The shortest leg is across from the 30-degree angle, the length of the hypotenuse is always double the length of the shortest leg, and you can find the length of the long leg by multiplying the short leg by the square root of 3. The hypotenuse is the longest side in a right triangle, which is different from the long leg. The long leg is the leg opposite the 60-degree angle. Two of the most common right triangles are 30-60-90 and the 45-45-90-degree triangles. All 30-60-90 triangles have sides with the same basic ratio. If you look at the 30–60–90-degree triangle in radians, it translates to the following: The figure illustrates the ratio of the sides for the 30-60-90-degree triangle. If you know one side of a 30-60-90 triangle, you can find the other two by using shortcuts. Here are the three situations you come across when doing these calculations: Type 1: You know the short leg (the side across from the 30-degree angle). Double its length to find the hypotenuse. You can multiply the short side by the square root of 3 to find the long leg. Type 2: You know the hypotenuse. Divide the hypotenuse by 2 to find the short side. Multiply this answer by the square root of 3 to find the long leg. Type 3: You know the long leg (the side across from the 60-degree angle). Divide this side by the square root of 3 to find the short side. Double that figure to find the hypotenuse. In the triangle TRI in this figure, the hypotenuse is 14 inches long; how long are the other sides? Because you have the hypotenuse TR = 14, you can divide by 2 to get the short side: RI = 7. Now you multiply this length by the square root of 3 to get the long side:

Cheat Sheet / Updated 04-06-2022

Pre-calculus uses the information you know from Algebra I and II and ratchets up the difficulty level to prepare you for calculus. This cheat sheet is designed to help you review key formulas and functions on the fly as you study. It includes formulas, the laws of logarithmic functions, trigonometric values of basic angles, conic section equations, and interval notation.

Cheat Sheet / Updated 02-24-2022

Pre-calculus draws from algebra, geometry, and trigonometry and combines these topics to prepare you for the techniques you need to succeed in calculus. This cheat sheet provides the most frequently used formulas, with brief descriptions of what the letters and symbols represent. Counting techniques are also here, letting you count numbers of events without actually having to list all the ways to do them. Also, you find a step-by-step description of how to complete the square — most useful when you’re working with conic sections and other equations with specific formats.

Article / Updated 12-21-2021

The fundamental theorem of algebra can help you find imaginary roots. Imaginary roots appear in a quadratic equation when the discriminant of the quadratic equation — the part under the square root sign (b2 – 4ac) — is negative. If this value is negative, you can’t actually take the square root, and the answers are not real. In other words, there is no real solution; therefore, the graph won’t cross the x-axis. Using the quadratic formula always gives you two solutions, because the plus/minus sign means you’re both adding and subtracting and getting two completely different answers. When the number underneath the square-root sign in the quadratic formula is negative, the answers are called complex conjugates. One is r + si and the other is r – si. These numbers have both real (the r) and imaginary (the si) parts. The complex number system consists of all numbers r + si where r and s are real numbers. Observe that when s = 0, you simply have the real numbers. Therefore the real numbers are a subset of the complex number system. The fundamental theorem of algebra says that every polynomial function has at least one root in the complex number system. The highest degree of a polynomial gives you the highest possible number of distinct complex roots for the polynomial. Between this fact and Descartes’s rule of signs, you can get an idea of how many imaginary roots a polynomial has. Here’s how Descartes’s rule of signs can give you the numbers of possible real roots, both positive and negative: Positive real roots. For the number of positive real roots, look at the polynomial, written in descending order, and count how many times the sign changes from term to term. This value represents the maximum number of positive roots in the polynomial. For example, in the polynomial f(x) = 2x4 – 9x3 – 21x2 + 88x + 48, you see two changes in sign (don’t forget to include the sign of the first term!) — from the first term (+2x4) to the second (-9x3) and from the third term (-21x2) to the fourth term (88x). That means this equation can have up to two positive solutions. Descartes’s rule of signs says the number of positive roots is equal to changes in sign of f(x), or is less than that by an even number (so you keep subtracting 2 until you get either 1 or 0). Therefore, the previous f(x) may have 2 or 0 positive roots. Negative real roots. For the number of negative real roots, find f(–x) and count again. Because negative numbers raised to even powers are positive and negative numbers raised to odd powers are negative, this change affects only terms with odd powers. This step is the same as changing each term with an odd degree to its opposite sign and counting the sign changes again, which gives you the maximum number of negative roots. The example equation becomes f(–x) = 2x4 + 9x3 – 21x2 – 88x + 48, which changes signs twice. There can be, at most, two negative roots. However, similar to the rule for positive roots, the number of negative roots is equal to the changes in sign for f(–x), or must be less than that by an even number. Therefore, this example can have either 2 or 0 negative roots. Pair up every possible number of positive real roots with every possible number of negative real roots; the remaining number of roots for each situation represents the number of imaginary roots. For example, the polynomial f(x) = 2x4 – 9x3 – 21x2 + 88x + 48 has a degree of 4, with two or zero positive real roots, and two or zero negative real roots. With this information, you can pair up the possible situations: Two positive and two negative real roots, with zero imaginary roots Two positive and zero negative real roots, with two imaginary roots Zero positive and two negative real roots, with two imaginary roots Zero positive and zero negative real roots, with four imaginary roots The following chart makes the information easier to picture: Positive real roots Negative real roots Imaginary roots 2 2 0 2 0 2 0 2 2 0 0 4 Complex numbers are written in the form r + si and have both a real and an imaginary part, which is why every polynomial has at least one root in the complex number system. Real and imaginary numbers are both included in the complex number system. Real numbers have no imaginary part, and pure imaginary numbers have no real part. For example, if x = 7 is one root of the polynomial, this root is considered both real and complex because it can be rewritten as x = 7 + 0i (the imaginary part is 0). The fundamental theorem of algebra gives the total number of complex roots (say there are seven); Descartes’s rule of signs tells you how many possible real roots exist and how many of them are positive and negative (say there are, at most, two positive roots but only one negative root). Now, assume you’ve found them all: x = 1, x = 7, and x = –2. These roots are real, but they’re also complex because they can all be rewritten. The first two columns in the chart find the real roots and classify them as positive or negative. The third column is actually finding, specifically, the non-real numbers: complex numbers with non-zero imaginary parts.

Article / Updated 12-21-2021

Sometimes your geometry teacher may spice things up a bit with complicated polar coordinates — points with negative angles and/or radii. The following list shows you how to plot in three situations — when the angle is negative, when the radius is negative, and when both are negative. When the angle is negative: Negative angles move in a clockwise direction. This figure shows an example point, D. To locate the polar coordinate point D at first locate the angle and then find the location of the radius, 1, on that line. When the radius is negative: When graphing a polar coordinate with a negative radius, you move from the pole in the direction opposite the given positive angle (on the same line as the given angle but in the direction opposite to the angle from the pole). For example, check out point F at in the figure. Some teachers prefer to teach their students to move right along the x- (polar) axis for positive numbers (radii) and left for negative. Then you do the rotation for the angle in a positive direction. You’ll get to the same spot with that method. For example, take a look at point F in the figure. Because the radius is negative, move along the left x-axis 1/2 of a unit. Then rotate the angle in the positive direction (counterclockwise) pi/3 radians. You should arrive at your destination, point F. When both the angle and radius are negative: To express a polar coordinate with a negative radius and a negative angle, locate the terminal side of the negative angle first and then move in the opposite direction to locate the radius. For example, point G in the figure has these characteristics at Indeed, except the origin, each given point can have the following four types of representations: Positive radius, positive angle Positive radius, negative angle Negative radius, positive angle Negative radius, negative angle For example, point E in the figure can have three other polar coordinate representations with different combinations of signs for the radius and angle: When polar graphing, you can change the coordinate of any point you’re given into polar coordinates that are easy to deal with (such as positive radius, positive angle).

Article / Updated 12-21-2021

Depending on how many times you must multiply the same binomial — a value also known as an exponent — the binomial coefficients for that particular exponent are always the same. The binomial coefficients are found by using the combinations formula. If the exponent is relatively small, you can use a shortcut called Pascal's triangle to find these coefficients. If not, you can always rely on algebra! Pascal's triangle, named after the famous mathematician Blaise Pascal, names the binomial coefficients for the binomial expansion. It is especially useful when raising a binomial to lower degrees. For example, if a sadistic teacher asked you to find (3x + 4)10, you probably wouldn't want to use Pascal's triangle; instead, you'd just use the algebraic formula described shortly. The figure illustrates this concept. The top number of the triangle is 1, as well as all the numbers on the outer sides. To get any term in the triangle, you find the sum of the two numbers above it. Each row gives the coefficients to (a + b)n, starting with n = 0. To find the binomial coefficients for (a + b)n, use the nth row and always start with the beginning. For instance, the binomial coefficients for (a + b)5 are 1, 5, 10, 10, 5, and 1 — in that order. If you need to find the coefficients of binomials algebraically, there is a formula for that as well. The rth coefficient for the nth binomial expansion is written in the following form: You may recall the term factorial from your earlier math classes. If not, here is a reminder: n!, which reads as "n factorial," is defined as You read the expression for the binomial coefficient as "n choose r." You usually can find a button for combinations on a calculator. If not, you can use the factorial button and do each part separately. To make things a little easier, 0! is defined as 1. Therefore, you have these equalities: For example, to find the binomial coefficient given by substitute the values into the formula: