Every polynomial on the ACT can be conquered in under 45 seconds—if you know the patterns. Factoring appears in 4–6 questions per test, often disguised inside coordinate geometry, functions, or word problems. The good news: ACT polynomials follow predictable blueprints. There are exactly five factoring techniques you need, and the test recycles the same structures over and over. By the end of this chapter, you'll see polynomials not as algebra chaos but as systematic puzzles—spot the pattern, apply the technique, collect the points.
Every polynomial on the ACT can be conquered in under 45 seconds—if you know the patterns. Factoring appears in 4–6 questions per test, often disguised inside coordinate geometry, functions, or word problems. The good news: ACT polynomials follow predictable blueprints. There are exactly five factoring techniques you need, and the test recycles the same structures over and over. By the end of this chapter, you'll see polynomials not as algebra chaos but as systematic puzzles—spot the pattern, apply the technique, collect the points.
Before you try any factoring technique, pull out the greatest common factor. This is the single most important habit in polynomial factoring. It simplifies everything that follows and often reveals hidden patterns.
To find the GCF: take the largest number that divides all coefficients, and the lowest power of each variable that appears in every term.
6x³ + 12x² - 18x = 6x(x² + 2x - 3)
That scary cubic just became a simple trinomial. Factor it further: 6x(x + 3)(x - 1).
The ACT loves testing whether you'll spot the GCF before attempting something harder. About 20% of factoring problems are solved by this step alone.
A trinomial x² + bx + c (with leading coefficient 1) factors into (x + m)(x + n) where m and n are two numbers that:
Multiply to c (the constant term) Add to b (the middle coefficient)
This is the most common factoring pattern on the ACT. Train yourself to find these two numbers instantly.
Thirty seconds. That's how long you have per ACT math problem. But here's the secret about linear equations: once you recognize the patterns, most of them take less than twenty. Lines are one of the most important topics on the ACT math section—you'll see 2–3 questions directly testing linear graphing, with slope and intercept skills appearing throughout many other problems from systems to word problems to coordinate geometry. The tools are simple: slope, intercepts, and two forms of the equation. Master these, and you've just secured a significant chunk of your math score.
Slope measures steepness—how much a line rises (or falls) for each unit it moves right. The formula:
m = (y₂ - y₁)/(x₂ - x₁) = frac{rise}{run}
Positive slope: line goes up-left to down-right (like climbing a hill). Negative slope: line goes up-right to down-left (like going downhill). Zero slope: horizontal line (y = c). Undefined slope: vertical line (x = c).
The workhorse of linear equations:
y = mx + b
m = slope (steepness) b = y-intercept (where the line crosses the y-axis)
Given any linear equation, your first move should be to rearrange it into this form. It immediately tells you the slope and y-intercept, which is usually enough to answer the question.
Every polynomial graph tells a story—and the ACT expects you to read it in seconds. How many times does the curve cross the x-axis? Does it open up or down? Where's the vertex? What happens at the far left and far right? These aren't abstract questions: they're worth 4–6 points on your math section. The best part? You don't need to do much algebra. Most polynomial graph questions can be answered just by looking at the shape and counting key features. This chapter teaches you to decode any polynomial graph on sight.
The graph of y = ax² + bx + c is always a parabola—a U-shaped curve. Three things determine its shape:
The sign of a: positive to opens upward (happy face). Negative to opens downward (sad face). The size of |a|: larger means narrower, smaller means wider. The vertex: the turning point, located at x = -(b)/(2a).
The axis of symmetry is the vertical line through the vertex: x = -(b)/(2a). Everything on the parabola is symmetric about this line.
Vertex form makes the vertex obvious:
y = a(x - h)² + k
The vertex is at (h, k). No calculation needed—just read it off. Be careful with the sign: y = (x - 3)² + 1 has vertex (3, 1), not (-3, 1).
The ACT frequently asks you to identify the vertex from vertex form, or to convert from standard form to vertex form by completing the square.
The ACT weaves word problems throughout the test—roughly a third of all 45 questions require translating English into math. The equation-building and shortcut techniques in this chapter directly target 3–5 of those problems, and the skills carry into many more. Students who crack these patterns gain a massive advantage. Most test-takers freeze when they see a wall of text, but here's the secret: word problems are just algebra in costume. Every sentence translates into a mathematical expression. "Three more than twice a number" becomes 2x + 3. "Decreases by 15% each year" becomes P(0.85)^t. Once you learn the translation dictionary, the intimidation vanishes and you're left with straightforward equations you already know how to solve.
Word problems use the same phrases over and over. Once you learn the translations, setting up equations becomes automatic:
"is" or "equals" to = "more than" or "added to" to + "less than" or "decreased by" to - "times" or "of" or "product" to × "per" or "each" or "divided by" to ÷ "a number" or "some quantity" to x (your variable) "twice" to 2x "consecutive integers" to x, x+1, x+2, ... "consecutive odd/even" to x, x+2, x+4, ...
The most important step in any word problem is defining your variable clearly. Write it down: "Let x = the number of weeks" or "Let x = the original price." This keeps you organized and prevents the confusion that causes most errors.
Then translate the problem sentence by sentence into an equation. Don't try to write the whole equation at once—build it piece by piece.
Every ACT math test contains about 5 expression simplification problems, and students who master these score an average of 3 points higher overall. Why? Because simplification isn't just its own topic—it's the final step in nearly every algebra, function, and geometry problem. If you can't simplify cleanly, you'll get the right idea but the wrong answer. This chapter drills the core simplification skills until they're automatic: combining like terms, distributing, factoring, and handling rational expressions.
Like terms have the same variable(s) raised to the same power(s). You can add or subtract their coefficients, but you cannot combine unlike terms.
3x and -5x are like terms (both have x¹). Combine: 3x - 5x = -2x. 2y and 7y are like terms. Combine: 2y + 7y = 9y. 3x and 2y are NOT like terms. They stay separate. x² and x are NOT like terms (different powers). They stay separate.
The distributive property removes parentheses by multiplying the outside term by each term inside:
a(b + c) = ab + ac
This works with subtraction too: a(b - c) = ab - ac. And with negative signs: -(b + c) = -b - c. Watch that negative distribution carefully—it's the number one source of errors on simplification problems.
Every ACT features about 3 exponential or logarithmic questions, and here's the beautiful part: they almost all use the same trick. If you can rewrite both sides of an equation with the same base, the exponents must be equal. That one idea—match the bases, set the exponents equal—solves the majority of these problems in under 30 seconds. Logarithms are just the reverse: instead of asking "what is 2⁵?" they ask "2 raised to what power gives 32?" Same relationship, different direction.
If a^m = aⁿ, then m = n. This is the entire strategy: rewrite both sides with the same base, then set the exponents equal.
The key: know your powers of small numbers. 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64 3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81 5¹ = 5, 5² = 25, 5³ = 125
A logarithm answers the question: "what exponent do I need?"
log_b(a) = c means the same thing as b^c = a.
Examples: log₂(8) = 3 because 2³ = 8. log₁₀(1000) = 3 because 10³ = 1000. log₄(16) = 2 because 4² = 16.
To solve a log equation, convert to exponential form and solve the resulting equation.
Imagine scoring perfect on every linear equation the ACT throws at you. Sounds impossible? Here's the truth: linear equations and inequalities directly appear in 3–5 questions per test, with the solving techniques also powering many other problems across the section. The difference between solving these correctly and guessing comes down to three critical moves: isolating variables efficiently, handling negative coefficients without errors, and knowing when to flip an inequality sign. Most students stumble not on complex algebra but on simple sign errors and rushed arithmetic. Master the systematic approach in this chapter, and you'll solve any linear equation in under 30 seconds.
Every linear equation follows the same dance:
Step 1: Distribute and clear parentheses. 3(x - 2) = 15 becomes 3x - 6 = 15.
Step 2: Combine like terms. Move all variable terms to one side, all constants to the other.
Step 3: Isolate the variable by dividing by its coefficient.
That's it. Three steps, every time.
When both sides have variables, collect all variable terms on one side and all constants on the other. The key: perform the same operation on both sides to keep the equation balanced.
Every sixty seconds on the ACT, a student loses precious time wrestling with a rational or radical equation — not realizing there's a systematic trick for each type. Rational equations (fractions with x in the denominator) are solved by multiplying away the denominators. Radical equations (square roots of expressions with x) are solved by squaring both sides. Both types require checking for extraneous solutions — answers that satisfy the transformed equation but not the original. This chapter teaches you the method for each, plus the one extra step that separates correct answers from traps.
A rational equation has x in one or more denominators. The strategy: multiply both sides by the least common denominator (LCD) to eliminate all fractions, then solve the resulting polynomial equation.
Critical: before solving, note which x values make any denominator zero. These are excluded from the domain. If your solution is one of these excluded values, it's extraneous — discard it.
For simple proportions like (a)/(b) = (c)/(d), skip the LCD and cross-multiply directly: ad = bc. This shortcut saves valuable seconds on the ACT.
When the variable is under a square root, isolate the radical on one side, then square both sides to eliminate it.
√(expression) = something Square: expression = (something)²
Warning: squaring can introduce extraneous solutions. You MUST check every answer in the original equation.
Every ACT includes at least one systems question, but here's the kicker: mastering systems helps on far more than just that one problem. Systems thinking—the ability to work with two or more equations simultaneously—underlies word problems, geometry, and function analysis throughout the test. The two solving methods (substitution and elimination) are tools you'll reach for constantly. And the ACT loves making one method dramatically faster than the other, so learning to spot which one to use is half the battle.
Substitution works best when one equation already has a variable isolated (or nearly so). Solve one equation for one variable, then substitute that expression into the other equation.
Best when: one equation is already solved for y (like y = 3x + 1), or a variable has coefficient 1.
Elimination works by adding or subtracting the two equations so that one variable cancels. If the coefficients don't match, multiply one or both equations first.
Best when: coefficients of one variable are the same (or negatives of each other), or can be made equal with a simple multiplier.