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SAT Math: Tips! Tricks! Traps! Techniques!

SAT Math: Tips! Tricks! Traps! Techniques!

Introduction

This guide is a collection of tricks, techniques, and facts intended to help strengthen your mathematical background for standardized exams like the SAT. While it is not comprehensive, it aims to supplement your existing high school math knowledge and provide you with additional strategies to succeed. This guide is continuously updated, so be sure to check for the latest information.

Test-Taking Techniques

0.1 Skip Questions You Don’t Know (For Now)

If you encounter a question that you have no idea how to approach, skip it temporarily. This strategy allows you to focus on questions you are confident about, maximizing the points you can earn. Interestingly, working on different questions later in the test may remind you of a technique you can use to solve a previously skipped question. If you still can’t solve it, make an educated guess, but remember to eliminate as many answer choices as possible first. Even eliminating one option increases your probability of guessing correctly!

0.2 Read the Answer Choices First

After reading the question, quickly scan the available answer choices. Doing so can sometimes help you eliminate obviously wrong answers, narrowing down your options. Additionally, the choices might give you clues about how to approach the problem or alert you if your calculations are going in the wrong direction.

0.3 Skip Over Wordy Questions Initially

If a math problem looks more like a reading passage, skip it on your first pass through the section. This tactic helps you maximize your score by tackling simpler, more straightforward problems first. Visually scan for questions with less text and focus on those initially.

0.4 Be Careful with Tricky Wording

SAT questions often use tricky wording to mislead students. For example, you may correctly solve for a variable like xx only to find out that the question actually asks for x2x^2, half of xx, or a sum involving xx. Carefully read the question to ensure you understand exactly what is being asked before selecting an answer. The correct value of xx may appear as an answer choice to tempt you, but don’t be fooled!

Key Points Recap

  • Skip Hard Questions Initially: Focus on questions you can solve confidently and come back to difficult ones later.
  • Read the Choices: This can help you eliminate incorrect options and guide your approach to the problem.
  • Watch Out for Tricky Questions: Pay attention to exactly what the question is asking.
  • Pace Yourself: Time management is crucial. Don’t spend too long on one problem.

By following these strategies, you can approach the SAT math section with more confidence and maximize your score potential. Remember that practice makes perfect—apply these techniques consistently to build your test-taking skills.

0.5 Saved Seconds Add Up!

Even small savings in time during the test can make a big difference. If you save 5 seconds on a calculation multiple times, those seconds quickly accumulate. Saving 5 seconds 5 times gives you an extra 25 seconds, and saving 5 seconds 12 times gives you a full extra minute. Many of the strategies and shortcuts in this guide can help you save time. Don’t underestimate how valuable these time savings can be!

0.6 Be Smart with Calculator Use

Use your calculator wisely, even on the calculator section. Here are a few guidelines:

0.6.1 Trust Yourself with Basic Arithmetic

At this point, you should have confidence in your ability to perform basic arithmetic without using a calculator. Simple operations like 2+3=52 + 3 = 5, 4×25=1004 \times 25 = 100, or dividing 1000 by 2 to get 500 can and should be done mentally. Excessive use of your calculator can actually slow you down. Save calculator use for complex calculations. For example, if you have to convert 22\frac{\sqrt{2}}{2} into a decimal (0.7071) for a free-response question, that’s a good use of your calculator.

0.6.2 Beware of a False Sense of Security

Calculators can sometimes lead to mistakes, such as incorrect inputs, missed parentheses, or errors in settings. Over-reliance on a calculator can create a false sense of security. To minimize errors, rely on your calculator only for the final steps of your calculations or for more complex operations that you’re less confident about handling manually.

0.7 Practice Under Real Test Conditions

This is a crucial part of preparing for the SAT. Simulate actual test conditions whenever possible by:

  • Taking timed practice tests on paper. This helps you become comfortable with managing the test booklet, answer sheet, and calculator while keeping track of time.
  • Practicing gridding answers. This skill is important for free-response questions, as the answer sheet scanner is unforgiving of gridding errors.
  • Managing skipped questions. Learn to skip and return to questions effectively without losing your place or misaligning your answers.

Key Considerations for Online Practice

While online practice tests can help with refreshing and testing various skills, they don’t fully replicate the SAT experience. Here are some challenges with online practice:

  • Clicking and scrolling. Navigating questions and diagrams online takes extra time.
  • Switching between questions. In some cases, you may not be able to skip around easily.
  • Writing down complex problems. Geometry and algebra problems often need to be written down for easier solving.
  • Reading and grammar sections. On paper, you can underline phrases and annotate passages. This isn’t as easily done online.

When practicing, be mindful of these differences. Proper gridding and managing skipped questions on paper are critical skills. Misaligning your answers due to incorrect gridding can lead to disastrous results. Ensuring you have solid practice managing these aspects will give you the confidence and skillset needed to perform your best on test day.

0.8 Practice New Techniques

Learning any new technique takes practice, and this holds true for SAT math strategies as well as life skills. If you come across a new method or trick in this guide that makes you say, “Aha!” but you’ve never used it before, take the time to practice it. Go back to practice test problems and apply the new method. Practice it again and again until it feels natural. To truly master a new skill, you may need to practice it on 25 to 50 different problems. There are many practice books, websites, and resources available to help. If you find new calculator tricks in this guide, use them in your homework or any math work you do to build proficiency. Keep practicing!

1. The Nature of Numbers and Numerical Operations

Understanding how numbers work and interact is important for the SAT. Comfort with numbers and basic operations is part of the test.

1.1 Combinations of Odd and Even Integers

When working with odd and even numbers, here’s what you need to know:

1.1.1 Addition and Subtraction Rules

  • Odd + Odd = Even (Example: 3 + 5 = 8)
  • Even + Even = Even (Example: 4 + 6 = 10)
  • Odd + Even = Odd (Example: 3 + 4 = 7)

1.1.2 Multiplication Rules

  • Odd × Odd = Odd (Example: 3 × 5 = 15)
  • Even × Even = Even (Example: 4 × 6 = 24)
  • Odd × Even = Even (Example: 3 × 4 = 12)

If you forget these rules during the test, you can always create quick examples in the margins of your exam booklet to remind yourself. While these basic rules may seem simple, they can be crucial for quickly eliminating wrong answers and understanding how numbers interact in SAT problems.

1.2 Understanding Percentages and How They Work Together

Percentages are simply fractions expressed with a denominator of 100. While they may seem tricky at times, especially when dealing with increases, decreases, markups, and markdowns, there are useful shortcuts and principles to keep in mind:

  • Percentage Increase and Markups: To find the value after a percentage increase, multiply the original value by 1+percentage1001 + \frac{\text{percentage}}{100}

    • Example: A 13% markup on $10 is calculated as 10×1.13=$11.3010 \times 1.13 = \$11.30

  • Percentage Decrease and Markdowns: When you decrease an amount by a percentage, what remains is a fraction of the original amount. Think about what’s left.

    • Example: A 30% markdown on $10 leaves 70% of the price. The new price is 10×0.7=$710 \times 0.7 = \$7 .

  • Stacking Percentage Changes: When multiple percentage changes are applied sequentially (like increases or decreases), you multiply each percentage change.

    • Example: Start with $13, increase it by 11%, then decrease by 25%, increase by 1%, and finally decrease by 66%: $13×1.11×0.75×1.01×0.34=$3.72\$13 \times 1.11 \times 0.75 \times 1.01 \times 0.34 = \$3.72 

  • Order of Operations with Percentages: The order in which you apply percentage changes does not matter, as the operations involve multiplication, which is commutative (meaning the order can be changed).

    • Example:
      • Applying a 2% increase after a 10% increase results in: New salary=1.02×(1.10×s)=1.122×s\text{New salary} = 1.02 \times (1.10 \times s) = 1.122 \times s 
      • Applying a 10% increase after a 2% increase also results in: New salary=1.10×(1.02×s)=1.122×s\text{New salary} = 1.10 \times (1.02 \times s) = 1.122 \times s 
      Both give the same result!

1.3 Exponents, Roots, and Values Between -1 and 1

Exponents typically increase numbers when applied to values greater than 1. However, for numbers between 0 and 1 (fractions), exponents actually make them smaller. Here’s how it works:

  • Examples:
    • (12)2=14\left(\frac{1}{2}\right)^2 = \frac{1}{4} and 14<12\frac{1}{4} < \frac{1}{2}
    • (25)4=16625\left(\frac{2}{5}\right)^4 = \frac{16}{625} and 16625<25\frac{16}{625} < \frac{2}{5}
    • (23)3=827\left(\frac{2}{3}\right)^3 = \frac{8}{27}  and 827<23\frac{8}{27} < \frac{2}{3}

This happens because raising fractions (numbers between 0 and 1) to positive powers results in smaller values. It’s helpful to visualize this effect on a graph, where lines become more pronounced with increasing odd powers of xx .

SAT Math: Tips! Tricks! Traps! Techniques!

Graphs and Behavior of Powers and Roots

When examining powers of values, especially within different intervals, their behavior relative to the line y=xy = x reveals interesting characteristics:

  • For values of xx  between 0 and 1: Raising a number in this interval to a power (such as squaring it or raising it to a higher exponent) results in a value that is less than the original xx -value. Graphically, curves representing these powers lie below the line y=xy = x .

  • For values of xx  between −1-1  and 0: Interestingly, raising these values to a power results in values that are larger (more positive) than the original xx -values themselves. This is because multiplying negative fractions by themselves results in positive values.

  • At x=−1x = -1  and x=1x = 1 : The powers of these values stabilize, as raising −1-1  or 11  to any power remains −1-1  or 11 , respectively. This behavior is evident at the outermost parts of the graph.

  • For values of x>1x > 1 : Raising these values to positive powers results in values that are greater than the original xx -values. The curves representing these powers lie above the line y=xy = x .

  • For values of x<−1x < -1 : The powers of these values are smaller than the xx -values themselves due to the exponential growth of negative values (with increasing even powers making the values more negative).

Behavior of Roots

  • For positive values greater than 1: Taking the square root or higher roots of these values yields results that are smaller than the original values. For example:

    • 25=±5\sqrt{25} = \pm5 
    • −273=−3\sqrt[3]{-27} = -3 

  • For fractions between 0 and 1: The roots of these values yield results that are larger than the original values. This is because taking roots reduces the denominator faster than the numerator in fractional values.

    • Examples:
      • 110=1010≈0.3162\sqrt{\frac{1}{10}} = \frac{\sqrt{10}}{10} \approx 0.3162  and 0.3162>1100.3162 > \frac{1}{10}
      • 143≈0.62996\sqrt[3]{\frac{1}{4}} \approx 0.62996  and 0.62996>140.62996 > \frac{1}{4}

Graphical Representation of Roots

The graph of roots, similar to that of powers, showcases curves that behave in the opposite manner. For values between 0 and 1, the curves for roots are above the line y=xy = x , indicating that the roots are greater than the original values. As the order of the root increases, the curves become more pronounced, emphasizing this effect further.

SAT Math: Tips! Tricks! Traps! Techniques!

Complex and Imaginary Numbers: The Basics

1.4.1 The Basics of Complex Numbers and ii 

The fundamental concept of complex arithmetic starts with the definition of the imaginary unit:

i=−1i = \sqrt{-1}

A complex number is any number that combines a real part and an imaginary part. It is generally expressed in the form:

z=a+biz = a + bi 

Here:

  • aa  is the real part of the complex number
  • bb  is the imaginary part, with bibi representing the imaginary term

Quick Facts About Complex Numbers:

  • Addition and Subtraction: When adding or subtracting complex numbers, combine their respective real and imaginary parts.
    • Example: (3+4i)+(2+5i)=5+9i(3 + 4i) + (2 + 5i) = 5 + 9i 
  • Multiplication: To multiply complex numbers, treat the terms like binomials and apply the rule i2=−1i^2 = -1  whenever you encounter i2i^2.
    • Example: (2+3i)(1+4i)=2+8i+3i+12i2=2+11i−12=−10+11i(2 + 3i)(1 + 4i) = 2 + 8i + 3i + 12i^2 = 2 + 11i – 12 = -10 + 11i 
  • Complex Conjugate: The conjugate of a complex number a+bia + bi is a−bia – bi . Multiplying a complex number by its conjugate results in a real number.
    • Example: (3+4i)(3−4i)=9−16i2=9+16=25(3 + 4i)(3 – 4i) = 9 – 16i^2 = 9 + 16 = 25 

Why Complex Numbers Are Important

Complex numbers extend our understanding of real numbers and are crucial in fields such as engineering, physics, and mathematics. They allow solutions to equations that have no real solutions (e.g., x2+1=0x^2 + 1 = 0 ) and are fundamental in representing waveforms, electrical currents, and much more.

1.4.2 The Trick for High Powers of ii

Evaluating high powers of ii can seem challenging at first, but there’s a helpful pattern to make it easy. Here’s a quick overview of powers of ii:

  • i=−1i = \sqrt{-1}
  • i2=−1i^2 = -1 
  • i3=−ii^3 = -i
  • i4=1i^4 = 1 

The pattern repeats every four powers: i,−1,−i,1i, -1, -i, 1 . To find the value of higher powers of ii , divide the exponent by 4 and observe the remainder:

  • If the remainder is 0, the result is i4=1i^4 = 1.
  • If the remainder is 1, the result is ii .
  • If the remainder is 2, the result is i2=−1i^2 = -1 .
  • If the remainder is 3, the result is i3=−ii^3 = -i .

Examples:

  • i3459i^{3459} : Divide 3459 by 4. The remainder is 3, so i3459=i3=−ii^{3459} = i^3 = -i .
  • i37i^{37} : Divide 37 by 4. The remainder is 1, so i37=ii^{37} = i.
  • i101i^{101} : Divide 101 by 4. The remainder is 1, so i101=ii^{101} = i.

1.4.3 Rationalizing the Denominator (Complex Numbers)

In complex arithmetic, it’s conventional not to have complex numbers in the denominator of fractions. The technique used to “realize” the denominator is similar to rationalizing a denominator with roots.

To rationalize a complex denominator, multiply the fraction (numerator and denominator) by the complex conjugate of the denominator. The complex conjugate of a+bia + bi  is a−bia – bi .

Example:

Given 1−7+3i\frac{1}{-7 + 3i}, rationalize the denominator:

  1. Multiply the numerator and denominator by the conjugate −7−3i-7 – 3i : 1−7+3i×−7−3i−7−3i=−7−3i(−7)2+(3)2\frac{1}{-7 + 3i} \times \frac{-7 – 3i}{-7 – 3i} = \frac{-7 – 3i}{(-7)^2 + (3)^2}
  2. Calculate the denominator using the difference of squares formula: (−7)2+(3)2=49+9=58(-7)^2 + (3)^2 = 49 + 9 = 58 
  3. Simplify the expression: −7−3i58=−758−358i\frac{-7 – 3i}{58} = -\frac{7}{58} – \frac{3}{58}i 

Key Points

  • Complex Conjugate: The complex conjugate of a+bia + bi is a−bia – bi .
  • Difference of Squares: Multiplying a complex number by its conjugate results in a real number: (a+bi)(a−bi)=a2+b2(a + bi)(a – bi) = a^2 + b^2 
  • Rationalizing complex denominators ensures no complex terms remain in the denominator, making the expression simpler to work with.

2. Important Algebra Concepts: Forms, Rules, and Techniques

2.1 Splitting Apart Fractions

It’s essential to understand how fractions can be split apart correctly because you may encounter answers in different forms than you expect. Fractions can only be split across addition or subtraction in the numerator. Here’s the correct way to split a fraction:

a±bc=ac±bc\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

Important Reminder: You cannot split fractions across addition or subtraction in the denominator. The following is not valid:

ab±c≠ab±ac\frac{a}{b \pm c} \neq \frac{a}{b} \pm \frac{a}{c}

In general terms:

a±bf(x)=af(x)±bf(x)\frac{a \pm b}{f(x)} = \frac{a}{f(x)} \pm \frac{b}{f(x)}

This rule emphasizes that the expression f(x)f(x)  in the denominator must remain intact for each term.

2.2 Cancelling in Fractions

Building on the splitting of fractions, it’s crucial to know how to properly factor, reduce, or cancel terms in fractions. Here’s what you need to know:

  • When you have a sum or difference in the numerator, you must consider the entire term before attempting to cancel anything with the denominator. You cannot individually cancel terms unless the entire expression is factored appropriately.

Incorrect Method:

5x+48≠5x+12(Incorrectly cancelling the 4 and 8)\frac{5x + 4}{8} \neq \frac{5x + 1}{2} \quad (\text{Incorrectly cancelling the 4 and 8}) 

Correct Method:

  1. Split the fraction: 5x+48=5x8+48\frac{5x + 4}{8} = \frac{5x}{8} + \frac{4}{8}
  2. Reduce each term where possible: 5x8+12\frac{5x}{8} + \frac{1}{2}
  3. (Optional) Rearrange terms: 58x+12\frac{5}{8}x + \frac{1}{2}

Example with Proper Reduction:

Given:

16x−48\frac{16x – 4}{8}

Method 1 (Splitting and Reducing):

  1. Split the fraction: 16x8−48\frac{16x}{8} – \frac{4}{8}
  2. Reduce each term: 2x−122x – \frac{1}{2}

2.2 Cancelling in Fractions (Continued)

Method 2 for Reducing Fractions:

Given:

16x−48\frac{16x – 4}{8}

  1. Factor the numerator: 4(4x−1)8\frac{4(4x – 1)}{8}
  2. Reduce by cancelling common factors (4 in this case): 4x−12\frac{4x – 1}{2}
  3. Split the fraction (optional): 4x2−12\frac{4x}{2} – \frac{1}{2}
  4. Reduce each term: 2x−122x – \frac{1}{2}

2.3 Quadratic Equations

2.3.1 Forms of Quadratic Equations

  • Standard Form: ax2+bx+c=0ax^2 + bx + c = 0 
  • Factored Form: k(x−a)(x−b)=0k(x – a)(x – b) = 0 
    • Where the roots are x=ax = a  and x=bx = b , and kk  is a constant.
  • Vertex Form: y−k=4p(x−h)2y – k = 4p(x – h)^2 
    • Where the vertex is at (h,k)(h, k).

2.3.2 The Difference of Squares Pattern for Quadratics

This pattern is useful for quickly factoring and multiplying binomials:

(x−a)(x+a)=x2−a2(x – a)(x + a) = x^2 – a^2 

Examples:

  • x2−16=(x+4)(x−4)x^2 – 16 = (x + 4)(x – 4) 
  • x2−100=(x+10)(x−10)x^2 – 100 = (x + 10)(x – 10) 
  • For more complex cases: x2−2=(x+2)(x−2)x^2 – 2 = (x + \sqrt{2})(x – \sqrt{2}) 
    16x2−4=(4x+2)(4x−2)16x^2 – 4 = (4x + 2)(4x – 2) 
  • General Form: ax2−b=(a⋅x+b)(a⋅x−b)ax^2 – b = (\sqrt{a} \cdot x + \sqrt{b})(\sqrt{a} \cdot x – \sqrt{b}) 

2.3.3 The Quadratic Formula

This formula is a must-know for solving quadratic equations of the form ax2+bx+c=0ax^2 + bx + c = 0 :

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}

  • Where aa, bb, and cc are the coefficients of the equation.

2.3.4 The Nature of the Roots of Quadratics Using the Quadratic Formula

The discriminant (b2−4acb^2 – 4ac ) determines the nature of the roots of a quadratic:

  1. If b2−4ac>0b^2 – 4ac > 0 (positive):
    • Two distinct real roots.
  2. If b2−4ac=0b^2 – 4ac = 0 :
    • One real, repeated root (a double root).
  3. If b2−4ac<0b^2 – 4ac < 0  (negative):
    • Two complex (imaginary) roots, as this involves taking the square root of a negative number.

 

2.3.5 Complex/Imaginary Roots

  • Imaginary roots come in conjugate pairs: If a polynomial has an imaginary root such as x=a+bix = a + bi , then it must also have x=a−bix = a – bi  as a root.
  • Roots from the sum of squares: A quadratic of the form ax2+b2=0ax^2 + b^2 = 0  produces imaginary roots of the form x=±a⋅ix = \pm \sqrt{a} \cdot i . This contrasts with the difference of squares, which has real roots.

2.4 Solving for xx  in terms of yy  and nn 

  • This type of problem simply asks you to isolate xx  on one side of the equation or inequality, while moving all other terms (like yy , nn , and constants) to the other side. It tests your algebraic manipulation skills.

2.5 Solving for Multiple Variables

  • While you cannot generally find unique values for more than one variable without enough equations, you can often simplify or rearrange expressions with multiple variables. Keep working through the algebraic manipulations; the expressions may simplify as you proceed.

2.6 Direct and Inverse Proportionality

  • Direct Proportionality: When yy is directly proportional to xx, the relationship is expressed as: y=k⋅xy = k \cdot x  where kk  is a constant.
  • Inverse Proportionality: When yy  is inversely proportional to xx , the relationship is expressed as: x⋅y=kx \cdot y = k  where kk  is a constant.

Example Scenario: If distance (dd ) varies directly with time (tt ) and inversely with a force (ff ), we might start with:

d=k1⋅tandd⋅f=k2d = k_1 \cdot t \quad \text{and} \quad d \cdot f = k_2

This simplifies to:

d=k2fd = \frac{k_2}{f}

Combining these gives:

d=k1⋅k2⋅tf=Ktfd = \frac{k_1 \cdot k_2 \cdot t}{f} = \frac{Kt}{f}

where K=k1⋅k2K = k_1 \cdot k_2 .

2.7 Function Notation and Custom Symbols

  • Function Notation Examples:
    • Notation like f(x)f(x) or f(a,b)f(a, b)  is common. If you see a custom symbol with a definition, treat it like function notation.
    • Example: If a custom operation is defined as a !!! b=ab+2ba \, !!! \, b = ab + 2b , and you need to find −6 !!! 7-6 \, !!! \, 7 , this is equivalent to finding: f(−6,7)=(−6)×7+2×7=−42+14=−28f(-6, 7) = (-6) \times 7 + 2 \times 7 = -42 + 14 = -28 

This section helps you understand how to handle complex numbers, proportionality, and custom function notations effectively by using standard algebraic techniques and principles.

3. Basic Statistics & Data Reasoning: Mean, Median, Mode, Central Tendency

Definitions:

  • Mean (Average):
    • Calculated by summing all the data points and then dividing by the total number of data points.
    • Example: If the data points are 2, 3, and 8, the mean is 2+3+83=133=4.33\frac{2 + 3 + 8}{3} = \frac{13}{3} = 4.33 .
  • Median:
    • This is the middle value of a data set that has been arranged in numerical order.
    • If there is an odd number of data points, the median is the middle value.
    • If there is an even number of data points, the median is the average of the two middle values.
    • Example: For the data set 1, 3, 5, 7, the median is 3+52=4\frac{3 + 5}{2} = 4 .
  • Mode:
    • The value that appears most frequently in a data set.
    • A data set can have one mode, more than one mode (bimodal or multimodal), or no mode at all if all values are unique.
    • Example: For the data set 2, 4, 4, 5, 7, the mode is 4.

Central Tendency Measures:

  • Mean, median, and mode are measures of central tendency used to describe what a “typical” value in a data set looks like.
  • These measures often vary from each other and are used to understand the overall distribution and trends in a data set.

Important Points:

  • Symmetrical Distribution:
    • In a perfectly symmetrical data distribution, the mean, median, and mode are all the same.
  • Skewed Distributions:
    • Left-Skewed (Negative Skew):
      • The “tail” is on the left side.
      • The order of mean, median, and mode (left to right) follows alphabetical order: mean<median<mode\text{mean} < \text{median} < \text{mode}.
      • The mean is pulled toward the tail (smaller values), while the mode is at the peak (hump).
    • Right-Skewed (Positive Skew):
      • The “tail” is on the right side.
      • The order of mean, median, and mode (left to right) is reverse alphabetical order: mode<median<mean\text{mode} < \text{median} < \text{mean}.
      • The mean is pulled toward the tail (larger values), while the mode is at the peak (hump).

Resistance to Change or Outliers:

  • Median:
    • The median is resistant to outliers, meaning it remains relatively stable even when extreme values are added or removed.
    • Example: Adding a very large or very small value to a data set has a limited impact on the median.
  • Mean:
    • The mean is sensitive to outliers, meaning it can be significantly affected by very large or small values.
    • Example: Adding an extremely high or low value to a data set can drastically shift the mean.

4. Counting Problems

Counting problems are divided into two major methods: Permutations and Combinations. These approaches relate to arranging and selecting objects or elements in different orders and groupings.

4.1 Permutations

  • Permutations are arrangements where the order of selection matters.
    • Keywords often used in permutation problems include “assigning ranks,” “finishing order,” etc.
    • Remember: Permutation and Place both begin with “P.” If order didn’t matter, switching positions (e.g., 1st and 3rd place in a race) wouldn’t cause a reaction. But in reality, it does.

4.2 Calculating Permutations

  • Permutations are calculated using the formula:
    nPr = first r factors of n!
4.2.1 P-notation (Shortcut Method)
  • This shortcut is faster and more intuitive than using factorials for larger numbers:
    • Examples:

      • 10P2=10×9=9010P2 = 10 \times 9 = 90

         

         

      • 10P7=10×9×8×7×6×5×4=604,80010P7 = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 604,800

         

         

      • 100P5=100×99×98×97×96=9,034,502,400100P5 = 100 \times 99 \times 98 \times 97 \times 96 = 9,034,502,400

         

         

4.3 Calculating Permutations “Positionally”

  • Position also starts with “P.” This method works for problems involving ordered arrangements like license plates, social security numbers, etc.
    • Examples:
      1. License Plates: How many plates can be made with 3 letters followed by 3 digits?
        • Solution:
          26×26×26×10×10×10=17,576,00026 \times 26 \times 26 \times 10 \times 10 \times 10 = 17,576,000

           

           
        • Explanation: 26 letters in the alphabet and 10 digits [0–9].
      2. Custom License Plates: How many plates with 2 letters, an “A,” two digits, ending in “3”?
        • Solution:
          26×26×1×10×10×1=67,60026 \times 26 \times 1 \times 10 \times 10 \times 1 = 67,600

           

           
      3. Phone Numbers: How many phone numbers in one area code?
        • Solution: Calculate the total possible phone numbers, then subtract those not allowed (e.g., those starting with 911 or 0).

4.4 Combinations

  • Combinations are groupings where the order of selection doesn’t matter.
    • Keywords include “teams,” “delegations,” “handshakes,” “diagonals,” etc.
    • Combinations are useful for avoiding double-counting (e.g., shaking hands—only counted once).

4.5 Calculating Combinations

  • Combinations are calculated using the formula:
4.5.1 C-notation (Shortcut Method)
  • This is a useful and efficient way to calculate combinations:
    • Example:

      • (52)=5P22!=5×42×1=10 \binom{5}{2} = \frac{5P2}{2!} = \frac{5 \times 4}{2 \times 1} = 10

         

         

Summary:

  • Permutations are used when the order matters.
  • Combinations are used when the order does not matter.
  • Shortcut methods (P-notation and C-notation) help simplify and speed up calculations, especially on standardized tests.

4.6 Combinations Examples

Using the results from the earlier examples, here are some calculations with combinations:

Examples:

  1. 10C2=10×92!=902=4510C2 = \frac{10 \times 9}{2!} = \frac{90}{2} = 45 
  2. 10C7=10×9×8×7×6×5×47!=6048005040=12010C7 = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{7!} = \frac{604800}{5040} = 120
  3. 100C5=100×99×98×97×965!=9034502400120=75287520100C5 = \frac{100 \times 99 \times 98 \times 97 \times 96}{5!} = \frac{9034502400}{120} = 75287520 

5. Probability Problems

Probability is the likelihood of a particular outcome or set of outcomes compared to the total number of possible outcomes. The general formula for probability is given by:

Probability of event X=P(X)=Number of ways event X can occurTotal number of possible outcomes\text{Probability of event X} = P(X) = \frac{\text{Number of ways event X can occur}}{\text{Total number of possible outcomes}}

5.1 “AND” Means Multiply

  • Example: What is the probability of rolling a 1 and then a 6 on two consecutive rolls of a fair 6-sided die?
    • Solution: 16×16=136\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}

5.2 “OR” Means Add

  • Example: What is the probability of rolling a 1 or a 6 on a single roll of a fair 6-sided die?
    • Solution: 16+16=26=13\frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}

5.3 “AND” Combined with “OR” Means a Combination of Multiplying and Adding

  • Example: What is the probability of rolling a 1 and then a 6 on two consecutive rolls, or a 2 and a 5 on two consecutive rolls?
    • Solution: (16×16)+(16×16)=136+136=236=118\left(\frac{1}{6} \times \frac{1}{6}\right) + \left(\frac{1}{6} \times \frac{1}{6}\right) = \frac{1}{36} + \frac{1}{36} = \frac{2}{36} = \frac{1}{18}

5.4 Ratios and Odds

  • Odds are different from probability calculations. While probability compares a specific outcome to the total possible outcomes, odds compare one set of outcomes against another set (e.g., wins
     
    , red
     
    , males
     
    ).
    • Example: An urn contains 3 blue marbles and 2 red marbles. Here are some things we can calculate about the urn:
      • Total number of blue marbles: 3
      • Total number of red marbles: 2
      • Total number of marbles: 5
      • Ratio of blue to red marbles (blue
         
        ) = 3:23 : 2

Probability and Ratio Example Summarization

  • Ratio of red marbles to blue marbles (red) = 2:32:3 
  • Probability of pulling a blue marble on one draw: P(blue)=35P(\text{blue}) = \frac{3}{5}
  • Probability of pulling a red marble on one draw: P(red)=25P(\text{red}) = \frac{2}{5}
  • Percentage of blue marbles in the urn: 35=0.6=60%\frac{3}{5} = 0.6 = 60\% 
  • Percentage of red marbles in the urn: 25=0.4=40%\frac{2}{5} = 0.4 = 40\% 
  • Odds of picking a blue marble: 3:23:2  or 1.51.5  in favor of. Note that this cannot be a probability since it is greater than 1.
  • Odds of picking a red marble: 2:32:3 or approximately 0.660.66  in favor of. This differs from the probability calculation above.
  • Comparison: The odds favor picking a blue marble over a red marble when all 5 marbles are in the urn.

Example Problem

Problem: A room has a male-to-female ratio of 23:723:7 . What percentage of the group is female?

  • Solution:
    • Total number of people in the room: 23+7=3023 + 7 = 30 
    • Fraction of the group that is female: 730\frac{7}{30}
    • Percentage of the group that is female: 730≈0.233=23.3%\frac{7}{30} \approx 0.233 = 23.3\% 

Systems of Equations Tips

6.1 One Equation is a Multiple of the Other (All Terms)

  • Quick Tip: If one equation is a multiple of another in a system, it means there is an infinite number of solutions since both equations represent the same line.

6.2 Graphing Both Equations

  • Quick Tip: If both equations can be easily graphed, the solution is the point where both graphs intersect. This method is viable if you can quickly transform the equations into the slope-intercept form y=mx+by = mx + b .

 

Systems of Equations and Geometry Strategies

6.3 Substitution Method

When one variable has a coefficient of ±1\pm 1 , substitution is usually efficient. Examples:

  • For an equation like x+7y=−8x + 7y = -8 , solve for xx .
  • For an equation like −9x−y=37-9x – y = 37 , solve for yy .
  • Equations already in a simplified form, such as y=−23x−1y = -\frac{2}{3}x – 1  or x=y2+11x = \frac{y}{2} + 11  are ready for substitution.
  • If x=40x = 40  or y=−2y = -2 , substitution becomes even easier.

6.4 Elimination Method

If one variable in each equation can be related by a constant, elimination is a great approach. This technique often involves multiplying one or both equations to align terms for easy addition or subtraction to eliminate a variable. While effective, this method can be time-consuming.

7. Important Geometry Concepts

7.1 Common Pythagorean Triples

Memorizing these triples can save valuable time on tests:

  • (3, 4, 5) and (5, 12, 13) are frequently used.
  • These triples can be scaled up or down by multiplying or dividing each length by the same number, due to triangle similarity rules:
    • Examples: (5,12,13)(5, 12, 13) , (10,24,26)(10, 24, 26) , (52,122,132) 
    • Other examples: (6,8,10)(6, 8, 10) , (1,43,53)(1, \frac{4}{3}, \frac{5}{3}) 
  • Note: In a triple, the last number is always the hypotenuse length. Example: For (5,13,x)(5, 13, x) , x=194x = \sqrt{194}.

7.2 Special Right Triangles

  • 45°-45°-90° Triangles:
    • These triangles have legs of equal length, and the hypotenuse is 2\sqrt{2}  times the length of one leg.
    • For legs of length ss, the hypotenuse is 2⋅s\sqrt{2} \cdot s .
    • This relationship holds true for all isosceles right triangles.
SAT Math: Tips! Tricks! Traps! Techniques!

30°-60°-90° Triangles

In a 30°-60°-90° triangle, the sides follow a consistent ratio:
  • The shortest leg is always opposite the 30° angle.
  • If the shortest leg has a length of 1 unit, the other leg (opposite the 60° angle) has a length of 3\sqrt{3}  .
  • The hypotenuse (the longest side) has a length of 2 units.

General Rule:

If the shortest leg of the triangle has length ss :
  • The length of the other leg (opposite the 60° angle) is 3⋅s\sqrt{3} \cdot s  .
  • The length of the hypotenuse is 2s2s  .
This consistent ratio helps quickly solve problems involving 30°-60°-90° triangles by scaling the side lengths up or down based on the value of the shortest leg ss .
SAT Math: Tips! Tricks! Traps! Techniques!

7.3 Finding Right Triangles in Geometric Figures (2D & 3D)

Identifying right triangles within more complex geometric shapes is a common challenge on the SAT. Here are some scenarios where right triangles are often found:

  • Cube Inscribed in a Sphere:

    • The diameter of the sphere is equal to the diagonal of the cube.
    • The sphere touches each vertex of the cube (8 points of contact).
    • You can find the diagonal of the cube using the formula for the space diagonal: Diagonal=a2+a2+a2=a3\text{Diagonal} = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3} , where aa is the side length of the cube.

  • Sphere Inscribed in a Cube:

    • The diameter of the sphere is equal to the edge length of the cube.
    • The sphere touches the cube at the center of each of its six faces.

  • Pyramids with Rectangular Bases:

    • A right triangle can always be found by considering:
      • The height of the pyramid (one leg).
      • Half of the diagonal of the base (another leg).
      • The slant height (hypotenuse).

 

  • The measure of an external angle of a triangle is equal to the sum of the measures of the two non-adjacent (or remote) interior angles.
    • This property can save significant time in problems involving triangle angles, as it allows you to find unknown angles quickly without needing to calculate each interior angle individually.

Example:

  • If you know two interior angles of a triangle are 40° and 60°, then the external angle adjacent to the third (unknown) interior angle is 40°+60°=100°40° + 60° = 100° .

In addition to the concept that the measure of an external angle is equal to the sum of the two non-adjacent interior angles, there is a useful shortcut derived from the relationship between supplementary angles and triangle angle sums:

  • Given:

    • The angles within a triangle △ABC\triangle ABC  sum to 180∘180^\circ : m∠A+m∠B+m∠α=180∘m\angle A + m\angle B + m\angle \alpha = 180^\circ .
    • Angles ∠α\angle \alpha  and ∠β\angle \beta  are supplementary: m∠α+m∠β=180∘m\angle \alpha + m\angle \beta = 180^\circ .

  • Derivation:

    • By subtracting m∠αm\angle \alphafrom 180∘180^\circ , we find that 180∘−m∠α=m∠β180^\circ – m\angle \alpha = m\angle \beta .
    • Inside the triangle, 180∘−m∠α=m∠A+m∠B180^\circ – m\angle \alpha = m\angle A + m\angle B .

  • Conclusion:

    • By the transitive property of equality, m∠β=m∠A+m∠Bm\angle \beta = m\angle A + m\angle B .
    • This result follows directly from the “overlap” between the triangle’s angle sum and the supplementary angle relationship, meaning you can find the external angle m∠βm\angle \beta  without needing additional steps.
SAT Math: Tips! Tricks! Traps! Techniques!

7.5 Useful Properties of n-gons

  • Definition: An n-gon is a convex polygon with nn  sides and nn  interior angles. A regular n-gon is a special type where all nn  sides and angles are congruent.
  • Sum of Interior Angles: The sum of all the interior angles of an n-gon is given by: 180×(n−2) degrees180 \times (n – 2) \text{ degrees} 
    • Easy Way to Remember: Consider a triangle (3-gon) as an example. Since all angles in a triangle sum to 180∘180^\circ , the formula holds true when n=3n = 3 .
  • Sum of Exterior Angles: The sum of all exterior angles of any n-gon is always 360∘360^\circ . Therefore, each exterior angle of a regular n-gon is given by: 360n degrees\frac{360}{n} \text{ degrees} 

7.6 Doing “Shape Math”

  • General Concept: Problems involving differences in areas and volumes, such as inner/outer rings of circles or differences in geometric figures, often require setting up expressions for each part of the shape.
    • Example: The difference between the area of an outer circle and an inner circle for concentric circles gives the area of the ring (or a “donut” shape).
    • Approach: Create expressions or values for each component of the problem (e.g., areas, perimeters) and use subtraction or other operations as needed to find desired quantities.

7.7 Non-Obvious Parallel Lines Cut by Transversals

  • Reminder: If the problem mentions shapes such as parallelograms, squares, rhombuses, or trapezoids, there are likely parallel lines involved. Similarly, terms like “line AB∥CDAB \parallel CD ” or “line ℓ∥m\ell \parallel m ” indicate parallelism.
  • Diagram Tips:
    • Draw a Diagram: If the problem is described verbally, sketch it out. Extend lines as necessary to uncover hidden relationships between angles and lines.
    • Look for Relationships: Identifying parallel lines and transversals can help reveal congruent angles, supplementary angles, and other properties that simplify the problem.

 

SAT Math: Tips! Tricks! Traps! Techniques!

7.8 Circles

7.8.1 Degrees and Radians Conversion

  • Converting between degrees and radians can be easily done using the following formula: π180=rd\frac{\pi}{180} = \frac{r}{d} where dd  represents degrees and rr  represents radians. Simply solve for the desired variable. Remember to set your calculator to the correct mode (degrees or radians) when performing trigonometric calculations.

7.8.2 Quick Conversion from Radians to Degrees

  • Since π\pi  radians is equivalent to 180∘180^\circ , you can quickly convert radians to degrees by treating π\pi  as 180∘180^\circ .
    • Example 1: Convert 7π8\frac{7\pi}{8} radians to degrees. d=7⋅1808=157.5∘d = \frac{7 \cdot 180}{8} = 157.5^\circ 
    • Example 2: Convert 11π16\frac{11\pi}{16} radians to degrees. d=11⋅18016=123.75∘d = \frac{11 \cdot 180}{16} = 123.75^\circ 

7.8.3 Area of a Sector

  • The area of a sector of a circle is found by taking the proportion of the circle’s total area based on the central angle of the sector.
    • For an angle measured in degrees: m∠sector360∘=Area of Sectorπr2\frac{\text{m}\angle \text{sector}}{360^\circ} = \frac{\text{Area of Sector}}{\pi r^2}
    • When using radians, the formula simplifies to: Area of Sector=θ⋅r22\text{Area of Sector} = \theta \cdot \frac{r^2}{2}
    where θ\theta  is the angle in radians.

8 Useful Calculator Moves

8.1 Math / Frac (TI-84 Family)

  • [Section to be completed soon]

8.2 Finding Remainders When Dividing Integers (General Technique)

  • This technique helps find remainders manually, even when calculators may not excel at displaying remainders directly:
    1. Divide the dividend by the divisor as usual.
    2. Take the integer part (whole number part) of the result.
    3. Multiply the integer part by the divisor.
    4. Subtract this product from the original number to get the remainder.
    • Note: The remainder will always be within the range 00 to divisor−1\text{divisor} – 1 . If it falls outside this range, there is likely a mistake.

Example: Finding the Quotient and Remainder of Division Problems

Example Problem

Find the quotient and remainder when dividing 12091209  by 77 . Solution:
  • First, divide 12091209  by 77  using a calculator. The result is approximately 172.7142857172.7142857 .
  • The integer part of the quotient is 172172 .
  • Calculate 172×7=1204172 \times 7 = 1204 .
  • Subtract this product from the original number: 1209−1204=51209 – 1204 = 5 .
Answer: 172172  remainder 55  (or 172r5172r5). Check: Multiply and add to verify: 172×7+5=1209172 \times 7 + 5 = 1209

8.3 Finding Remainders Using the TI-84 Family of Calculators

Steps for Finding Remainders

  1. Divide the numbers as usual.
  2. Convert the result to a fraction using the calculator:
    • Press Math and select Frac (or press 1) to convert to a fraction.
  3. Change the result to a mixed number:
    • Go to Math, then Frac, and select Un/d to convert it into a mixed number.
    • The integer part represents the quotient, and the numerator of the fraction is the remainder.
Important Note:
  • The TI-84 simplifies fractions automatically. If the denominator in the fraction differs from the original divisor, multiply the numerator and denominator by a factor to return it to the original form. This step ensures the correct remainder.

Example

Find the quotient and remainder of 249249  divided by 66.
  • The calculator shows the result as 41.541.5.
  • Converting this to a mixed number yields 411241\frac{1}{2}. Since the original divisor is 66, convert 12\frac{1}{2} to sixths: 12=36\frac{1}{2} = \frac{3}{6}
  • The remainder is 33.
Answer: 41r341r3. Check: Verify the result: 41×6+3=24941 \times 6 + 3 = 249

9. Optional Topics

9.1 Working with Absolute Value

9.1.1 Absolute Value Equations
  • Absolute value equations represent two possible equations: ∣x∣=a  ⟹  x=aorx=−a|x| = a \implies x = a \quad \text{or} \quad x = -a
  • The absolute value bars act like parentheses when isolating a variable. Rearrange to get the equation into the form ∣f(x)∣=a|f(x)| = a and split accordingly.
9.1.2 Absolute Value Inequalities
  • For inequalities, do not split the terms on the opposite side of the equation/inequality. Instead, negate the argument within the absolute value bars.

9.2 Rational Roots Theorem

  • Useful for factoring polynomials of degree greater than 22. The theorem relates the constant term of a polynomial and its leading coefficient.
  • For a polynomial of the form: axn+bxn−1+…+c=0ax^n + bx^{n-1} + \ldots + c = 0where aa is the coefficient of the highest degree term, the possible rational roots are given by: Possible roots=±(factors of the constant termfactors of a)\text{Possible roots} = \pm \left(\frac{\text{factors of the constant term}}{\text{factors of } a}\right)
  • Example: If a=1a = 1  and the constant term is 22 , possible roots are ±1,±2\pm1, \pm2 .
 

9.2 The Rational Roots Theorem

The Rational Roots Theorem is a useful concept from the Fundamental Theorem of Algebra, often taught in Algebra II. It is particularly helpful for factoring polynomials of degree greater than two. The theorem establishes a relationship between the constant term and the leading coefficient aa of a polynomial written in standard form, where aa is the coefficient of the term with the highest degree.

Standard Forms of Polynomials:

  • Quadratic: ax2+bx+c=0ax^2 + bx + c = 0  (where cc is the constant term)
  • Cubic: ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0  (where dd  is the constant term)
  • Fourth-degree: ax4+bx3+cx2+dx+f=0ax^4 + bx^3 + cx^2 + dx + f = 0  (where ff  is the constant term)
  • And so on…

The Rational Roots Theorem Explained:

The theorem states that possible values of xx  (rational roots) can be found by forming fractions where:

  • Numerator: Factors of the constant term.
  • Denominator: Factors of the leading coefficient aa.

To construct a list of possible roots, we form combinations of one element from the list of factors of the constant term divided by an element from the list of factors of the leading coefficient.

Example 1:

  • Consider a polynomial where a=1a = 1  and the constant term is 22 .
    • Factors of the constant term 22 : ±1,±2\pm1, \pm2 
    • Factors of the leading coefficient 11 : ±1\pm1 
    • Possible roots: x=±1,±2x = \pm1, \pm2 

Example 2:

  • Consider a polynomial where a=12a = 12  and the constant term is 4949 .
    • Factors of the constant term 4949 : ±1,±7,±49\pm1, \pm7, \pm49 
    • Factors of the leading coefficient 1212 : ±1,±2,±3,±4,±6,±12\pm1, \pm2, \pm3, \pm4, \pm6, \pm12 
    • Possible roots: x=±1,±12,±13,±14,±16,±112,±7,±72,±73,±74,…x = \pm1, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{1}{4}, \pm\frac{1}{6}, \pm\frac{1}{12}, \pm7, \pm\frac{7}{2}, \pm\frac{7}{3}, \pm\frac{7}{4}, \ldots 

Finding Roots:

To find the actual roots of a polynomial, substitute possible values of xx  into the polynomial expression and check if it results in zero (this defines a root). Once a root is found, a corresponding binomial factor (x−root)(x – \text{root}) can be constructed. From there, polynomial division or synthetic division can be used to simplify the polynomial further.

Constructing Binomials:

For possible roots from the list (e.g., x=±1,±12,±13,…x = \pm1, \pm\frac{1}{2}, \pm\frac{1}{3}, \ldots ), the binomial factors would be:

(x±1),(x±12),(x±13),…(x \pm 1), \left(x \pm \frac{1}{2}\right), \left(x \pm \frac{1}{3}\right), \ldots 

These can be used to divide the original polynomial.

9.3 The Polynomial Remainder Theorem

The Polynomial Remainder Theorem offers insight into dividing polynomials by a binomial of the form (x−r)(x – r) . According to the theorem:

  • For any polynomial P(x)P(x) , evaluating P(r)P(r)  gives the remainder when P(x)P(x)  is divided by (x−r)(x – r) .
  • If P(r)=0P(r) = 0 , it means that (x−r)(x – r)  is a factor of P(x)P(x)  and divides the polynomial evenly, leaving no remainder.

This theorem can speed up the process of factoring polynomials or verifying if a particular value is a root. On standardized tests like the SAT, questions may directly ask for the remainder when dividing a polynomial by a binomial, making this theorem highly useful for a quick solution.

Example:

Given P(x)=2x3−3x2+4x−5P(x) = 2x^3 – 3x^2 + 4x – 5 , find the remainder when dividing by (x−2)(x – 2) :

  • Substitute r=2r = 2  into the polynomial: P(2)=2(2)3−3(2)2+4(2)−5=16−12+8−5=7P(2) = 2(2)^3 – 3(2)^2 + 4(2) – 5 = 16 – 12 + 8 – 5 = 7 
  • The remainder is 7.

9.4 The Binomial Theorem

The Binomial Theorem is useful for expanding expressions raised to a power, such as (x+y)n(x + y)^n . It generalizes binomial expansions using the formula:

(x+y)n=∑k=0n(nk)xn−kyk(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k 

Where:

  • (nk)\binom{n}{k} is the binomial coefficient, representing “n choose k” combinations.
  • xn−kx^{n-k}  and yky^k  are terms in the expansion.

This theorem is particularly helpful for finding specific terms in large expansions without fully expanding the expression.

Example:

Find the 3rd term in the expansion of (x+1)5(x + 1)^5 .

  • The general term in the expansion is given by: Tk=(5k−1)x5−(k−1)⋅1k−1T_k = \binom{5}{k-1} x^{5-(k-1)} \cdot 1^{k-1} 
  • For the 3rd term (k=3k = 3 ): T3=(52)x5−2⋅12=10x3T_3 = \binom{5}{2} x^{5-2} \cdot 1^2 = 10x^3 

The 3rd term is 10x310x^3 .

9.4.1 Using Pascal’s Triangle

Pascal’s Triangle is a useful tool for finding the coefficients of terms in a binomial expansion without having to manually calculate binomial coefficients. Here’s how to use it:

Basics of Pascal’s Triangle:

  • Each row corresponds to the power of the binomial expansion.
  • The rows are indexed starting with n=0n = 0 .
  • For example:
    • n=0n = 0 : 11 
    • n=1n = 1 : 111 \quad 1 
    • n=2n = 2 : 1211 \quad 2 \quad 1 
    • n=3n = 3 : 13311 \quad 3 \quad 3 \quad 1 
    • n=4n = 4 : 146411 \quad 4 \quad 6 \quad 4 \quad 1 
    • n=5n = 5 : 151010511 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1 

Example: Expanding (x+1)2(x + 1)^2 

  • Go to the n=2n = 2 row of Pascal’s Triangle: 1,2,11, 2, 1 .
  • This gives the expansion: 1x2+2x+11x^2 + 2x + 1 .

More Complex Example: Expanding (3x−7)3(3x – 7)^3 

  1. Use the n=3n = 3 row: 1,3,3,11, 3, 3, 1 .
  2. Apply these coefficients to the terms of the binomial (3x−7)(3x – 7) : 1(3x)3+3(3x)2(−7)+3(3x)(−7)2+1(−7)31(3x)^3 + 3(3x)^2(-7) + 3(3x)(-7)^2 + 1(-7)^3 
  3. Calculate each term:
    • 1×(3x)3=27x31 \times (3x)^3 = 27x^3 
    • 3×(3x)2×(−7)=−189x23 \times (3x)^2 \times (-7) = -189x^2 
    • 3×(3x)×(−7)2=441x3 \times (3x) \times (-7)^2 = 441x 
    • 1×(−7)3=−3431 \times (-7)^3 = -343 
  4. The expanded form is: 27x3−189x2+441x−34327x^3 – 189x^2 + 441x – 343 

Pascal’s Triangle is quick to set up and can be used to find coefficients easily during a test.

10 Great Places for Reviewing Concepts

Here are some recommended resources for SAT Math preparation and general concept review:

  • The College Board: College Board
    • The official SAT test creator offers free practice materials, sample questions, and full-length tests.
  • Khan Academy: Khan Academy
    • Provides free SAT practice in partnership with The College Board, with personalized practice recommendations.
  • Paul’s Online Math Notes: Paul’s Online Math Notes
    • A comprehensive set of math notes and tutorials, great for revising calculus, algebra, and other math concepts.
  • McGraw-Hill “Top 50 Skills for a Top Score: SAT Math”: Amazon Link
    • A book focused on building key skills for the SAT Math section.
  • Desmos Online Graphing Calculator: Desmos
    • A powerful graphing tool that works directly in your web browser and is excellent for visualizing math problems.

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