CLEP College Algebra Quick Review

Every exponent rule is just repeated multiplication or division written compactly, so you can always fall back on expanding a small case to check yourself.

$a^m · a^n = a^m+n$; $/a^ma^n = a^m-n$; $(a^m)^n = a^mn$; $(ab)^n = a^n b^n$; $a^0 = 1$ ($a ≠ 0$); $a^-n = /1a^n$; $a^1/n = sqrt[n]a$; $a^m/n = ≤ft(sqrt[n]a)^m$.

$(a^m)^n$ (multiply exponents) versus $a^m · a^n$ (add exponents) --- these are opposite operations that students routinely swap.

Writing $(x^3)^2 = x^5$ instead of $x^6$, or distributing an exponent over a sum, e.g. treating $(x+y)^2$ as $x^2+y^2$ instead of expanding it.

Same base, multiply $arrow$ add exponents; power of a power $arrow$ multiply exponents.

Factoring reverses multiplication, and CLEP rewards recognizing a handful of patterns instantly rather than guessing.

GCF first, always. Difference of squares: $a^2-b^2=(a-b)(a+b)$. Perfect square trinomial: $a^2± 2ab+b^2=(a± b)^2$. Sum/difference of cubes: $a^3± b^3=(a± b)(a^2 ab+b^2)$. Trinomial $x^2+bx+c=(x+p)(x+q)$ where $pq=c,\ p+q=b$.

Difference of squares $a^2-b^2$ (factors) versus sum of squares $a^2+b^2$ (does NOT factor over the reals).

Forgetting to pull out the GCF before applying a pattern, leaving a factorable expression only partially factored.

GCF first $arrow$ then pattern-match: squares, cubes, or $p· q$/$p+q$.

Expanding is factoring in reverse: multiply every term of one factor by every term of the other and combine like terms.

FOIL for binomials: $(a+b)(c+d)=ac+ad+bc+bd$. Square of a binomial: $(a+b)^2=a^2+2ab+b^2$ (NOT $a^2+b^2$). Cube of a binomial: $(a+b)^3=a^3+3a^2b+3ab^2+b^3$.

$(a+b)^2$ versus $a^2+b^2$ --- the middle term $2ab$ is the single most commonly dropped term on the exam.

Sign errors when expanding $(a-b)^2$ or $(a-b)^3$: the middle terms alternate sign, e.g. $(a-b)^2 = a^2-2ab+b^2$.

Square a binomial $arrow$ first, twice-the-product, last; never just square each term.

Rational (fractional) algebraic expressions follow the same rules as numeric fractions, but every operation requires fully factored expressions first.

Simplify by factoring numerator/denominator and canceling common factors. Multiply: $/ab·/cd=/acbd$. Divide: multiply by the reciprocal. Add/subtract: find a common denominator first.

Canceling common factors (allowed) versus canceling common terms added or subtracted inside an expression (not allowed), e.g. $/x+2x≠ 2$.

Canceling an $x$ that is part of a sum rather than a factor, or forgetting that a canceled factor cannot equal zero (excluded value).

Factor everything first --- only factors cancel, never terms.

Absolute value measures distance from zero, so it is always nonnegative and splits every equation or inequality into two cases.

$|x| = x$ if $x≥q 0$; $|x|=-x$ if $x<0$. $|x|=a$ (with $a≥q 0$) $⇒ x=a$ or $x=-a$. $|a-b|$ = distance between $a$ and $b$ on the number line.

$|x|=a$ having two solutions when $a>0$ versus $|x|=a$ having NO solution when $a<0$ (absolute value can never equal a negative number).

Solving $|x-3|=-5$ and reporting solutions instead of recognizing the equation has no solution because the right side is negative.

Absolute value $arrow$ split into two cases, check $a≥q 0$ first.

Logarithms are exponents in disguise, so every log property is really an exponent rule translated into log language.

$_b(xy)=_b x+_b y$; $_b≤ft(/xy)=_b x-_b y$; $_b(x^n)=n_b x$; $_b b=1$; $_b 1=0$; $b^_b x=x$; change of base $_b x=/ x b$.

$_b(x+y)$ (no simplification exists) versus $_b(xy)=_b x+_b y$ --- logs distribute over products, never over sums.

Writing $(x+y)= x+ y$, or splitting $_b(x^n)$ incorrectly as $(_b x)^n$ instead of $n_b x$.

Log of a product $arrow$ sum of logs; log of a power $arrow$ exponent moves out front.

Linear equations and inequalities are solved with the same steps --- isolate the variable --- except an inequality flips direction when multiplied or divided by a negative.

Solve by combining like terms, then isolating $x$. For inequalities: $a<b$ and $c<0$ $⇒$ $ac>bc$ (flip the sign). Compound inequalities: $-a<x+b<a$ solved by operating on all three parts simultaneously.

Solving an equation (single value, use $=$) versus solving an inequality (range of values, use interval or inequality notation).

Forgetting to flip the inequality sign when multiplying or dividing both sides by a negative number.

Multiply/divide by negative $arrow$ flip the inequality sign.

Every quadratic can be solved by factoring, completing the square, or the quadratic formula --- pick the fastest method for the given form.

Quadratic formula: for $ax^2+bx+c=0$ ($a≠ 0$), $x=/-b±sqrtb^2-4ac2a$. Discriminant $D=b^2-4ac$: $D>0$ two real roots, $D=0$ one repeated real root, $D<0$ two complex conjugate roots.

The discriminant $b^2-4ac$ (tells you the number/type of roots) versus the quadratic formula itself (gives the value of the roots).

Forgetting the $±$ in the quadratic formula, or computing $-b$ incorrectly when $b$ is already negative (sign error on the outer negative).

$x=/-b±sqrtb^2-4ac2a$ --- always check the discriminant sign first.

A quadratic inequality is solved by finding the zeros of the related equation, then testing the sign of the expression in each interval those zeros create.

Steps: (1) set equal to 0 and solve for the boundary points; (2) plot boundary points on a number line; (3) test a point in each interval in the original inequality; (4) include boundary points only if the inequality is $≤q$ or $≥q$.

Strict inequalities ($<,>$, open circles/parentheses) versus non-strict inequalities ($≤q,≥q$, closed circles/brackets) in the final interval notation.

Assuming the solution set is always ``between the roots'' --- for an upward parabola, $ax^2+bx+c>0$ is true OUTSIDE the roots, not between them.

Find zeros $arrow$ test each interval $arrow$ never assume which side is the solution.

Absolute value inequalities split into two cases just like absolute value equations, but ``less than'' gives an AND compound inequality while ``greater than'' gives an OR.

$|x|<a$ ($a>0$) $ -a<x<a$. $|x|>a$ ($a>0$) $ x<-a$ or $x>a$. $|x|≤q a$ and $|x|≥q a$ follow the same pattern with closed endpoints.

$|x|<a$ (bounded interval, ``AND'') versus $|x|>a$ (two rays going to infinity, ``OR'') --- these produce structurally different solution sets.

Writing the ``greater than'' case as a single bounded interval instead of two disjoint rays, or forgetting to reverse one inequality when splitting into cases.

Less than $arrow$ between (AND); greater than $arrow$ split apart (OR).

A system of linear equations is solved by combining equations (substitution or elimination) to eliminate one variable at a time.

Substitution: solve one equation for one variable, plug into the other. Elimination: multiply equations so a variable's coefficients are opposites, then add. A system has one solution (lines intersect), no solution (parallel, $0=$ nonzero), or infinitely many solutions (same line, $0=0$).

No solution (equations reduce to a false statement like $0=5$) versus infinitely many solutions (equations reduce to a true statement like $0=0$).

Multiplying only one side of an equation by a constant when setting up elimination, which destroys the equation's balance.

Line up coefficients, eliminate one variable, back-substitute.

When a system mixes a linear and a nonlinear (quadratic) equation, substitution is almost always the fastest path, since it converts the system into a single-variable quadratic.

Solve the linear equation for one variable, substitute into the quadratic equation, then solve the resulting quadratic (up to 2 solutions, so up to 2 intersection points).

A line intersecting a parabola in 0, 1, or 2 points versus a system of two lines, which intersects in 0, 1, or infinitely many points --- the maximum number of solutions differs by system type.

Finding only one solution to the resulting quadratic and forgetting the second root, which corresponds to a second valid intersection point.

Substitute linear into quadratic $arrow$ solve $arrow$ report both intersection points.

Exponential equations are solved by matching bases (then equating exponents) or by taking a logarithm of both sides when bases cannot be matched.

If $b^m=b^n$ then $m=n$ (same base $b>0,\ b≠ 1$). If bases differ, take $$ or $$ of both sides: $b^x=k ⇒ x=/ k b$.

Same-base equations (equate exponents directly) versus different-base equations (must apply logarithms) --- do not take a log when matching bases would solve it faster.

Rewriting $9^x$ as $3^2x$ incorrectly, or forgetting to apply the log to the ENTIRE side of the equation, not just part of it.

Same base $arrow$ drop it, equate exponents; different base $arrow$ take logs.

A logarithmic equation converts to an exponential equation using the definition of a logarithm, and every solution must be checked against the domain of the original logs.

Definition: $_b x = y b^y = x$ ($x>0,\ b>0,\ b≠ 1$). Combine multiple logs into one using log properties before converting to exponential form.

An algebraically valid solution versus an EXTRANEOUS solution --- a solution that makes the argument of any original log negative or zero must be rejected.

Solving the equation correctly but forgetting to check the result against the domain restriction $x>0$, keeping a solution that is actually extraneous.

Convert log to exponential form $arrow$ solve $arrow$ always check the domain.

A function is a rule that assigns exactly one output to each input, and $f(x)$ notation is just a label meaning ``the output of $f$ when the input is $x$.''

Vertical Line Test: a graph represents a function if and only if no vertical line crosses it more than once. $f(x)$ is read ``$f$ of $x$,'' not ``$f$ times $x$.''

$f(x)$ (function notation, output value) versus $f· x$ (multiplication) --- these look similar but mean completely different things.

Treating $f(x+2)$ as $f(x)+2$; function notation requires substituting the ENTIRE input expression wherever the variable appears.

$f(x+2)$ means plug in $(x+2)$ everywhere $x$ appears --- never distribute $f$.

The domain is every legal input value and the range is every output value the function actually produces; both must be checked against the function's specific form.

Exclude from the domain: denominators equal to 0, negative numbers under an even root, and non-positive arguments of a logarithm. Domain of a polynomial is all real numbers.

Domain restrictions from a DENOMINATOR ($≠ 0$) versus domain restrictions from a RADICAL ($≥q 0$) versus domain restrictions from a LOG ($>0$) --- each type uses a different inequality.

Finding the domain of $sqrtx-3$ and writing $x≥q 3$ but then applying the same $≥q$ logic to a rational function's denominator restriction (which needs $≠$, not $≥q$).

Denominator $≠ 0$; even radicand $≥q 0$; log argument $>0$.

Evaluating a function means substituting a specific input; composing functions means feeding the output of one function into another as its input.

$(f g)(x) = f(g(x))$: evaluate $g$ first, then plug that result into $f$. In general $(f g)(x) ≠ (g f)(x)$.

$(f g)(x)=f(g(x))$ (apply $g$ first) versus $(g f)(x)=g(f(x))$ (apply $f$ first) --- composition order matters and is almost never commutative.

Computing $f(g(x))$ by evaluating $f$ first instead of $g$, effectively reversing the composition order.

$(f g)(x)=f(g(x))$: work from the inside out.

Functions can be combined with the four basic operations, and each combination's domain is restricted by the domains of both original functions.

$(f+g)(x)=f(x)+g(x)$; $(f-g)(x)=f(x)-g(x)$; $(fg)(x)=f(x)· g(x)$; $≤ft(/fg)(x)=/f(x)g(x)$, requiring $g(x)≠ 0$.

$(f-g)(x)$ (subtract the whole function $g$) versus $f(x-g)$ (nonsensical without clarification) --- keep function combination notation separate from composition notation.

Finding the domain of $/fg$ but forgetting to also exclude values already outside the domains of $f$ or $g$ individually.

Combine functions $arrow$ domain is the overlap of both domains, minus any new zero denominators.

A linear function has a constant rate of change, so its graph is always a straight line fully described by slope and one point (or the $y$-intercept).

Slope-intercept form: $y=mx+b$, where $m$ = slope, $b$ = $y$-intercept. Slope formula: $m=/y_2-y_1x_2-x_1$. Point-slope form: $y-y_1=m(x-x_1)$.

Slope-intercept form $y=mx+b$ (best for graphing quickly) versus point-slope form $y-y_1=m(x-x_1)$ (best when given a point and slope, not the $y$-intercept).

Computing slope as $/x_2-x_1y_2-y_1$ (inverted) instead of $/y_2-y_1x_2-x_1$.

Slope $=/riserun=/y_2-y_1x_2-x_1$.

Every quadratic function graphs as a parabola whose vertex is the maximum or minimum point, easiest to see in vertex form.

Vertex form: $f(x)=a(x-h)^2+k$, vertex $(h,k)$. From standard form $f(x)=ax^2+bx+c$, the vertex $x$-coordinate is $x=-/b2a$. Opens upward if $a>0$ (minimum), downward if $a<0$ (maximum).

Vertex form $(x-h)^2$ (vertex at $x=+h$, note the sign flip) versus reading $h$ directly off $(x+3)^2$ as $h=+3$ instead of the correct $h=-3$.

Given $f(x)=(x+3)^2-5$, reporting the vertex as $(3,-5)$ instead of the correct $(-3,-5)$ --- the sign inside the parentheses flips.

$f(x)=a(x-h)^2+k arrow$ vertex $(h,k)$, watch the sign flip on $h$.

A polynomial's degree and leading coefficient control its end behavior, while its factored form reveals its zeros (roots).

Degree $n$ with positive leading coefficient: as $x→∞$, $f(x)→∞$; behavior as $x→-∞$ depends on whether $n$ is even (same direction) or odd (opposite direction). A polynomial of degree $n$ has at most $n$ real zeros.

Even-degree end behavior (both ends point the same direction) versus odd-degree end behavior (ends point opposite directions).

Assuming every zero found by factoring is a distinct $x$-intercept, without checking for repeated (multiplicity $>1$) roots, which change whether the graph crosses or touches the axis.

Even degree $arrow$ ends match; odd degree $arrow$ ends opposite. Leading coefficient sign sets the right-hand direction.

A rational function's graph is governed by where its denominator is zero (asymptotes/holes) and by comparing the degrees of numerator and denominator (end behavior).

Vertical asymptote at any zero of the denominator not canceled by the numerator. Horizontal asymptote: if $(num)<(den)$, $y=0$; if equal degrees, $y=/leading coeff. of numleading coeff. of den$; if $(num)>(den)$, no horizontal asymptote.

A vertical asymptote (denominator zero NOT canceled) versus a hole/removable discontinuity (a common factor DOES cancel from numerator and denominator).

Setting only the denominator to zero for vertical asymptotes without first checking whether that factor also cancels with the numerator, which would make it a hole instead.

Compare degrees for horizontal asymptote; factor and cancel first to tell asymptotes from holes.

An exponential function $f(x)=ab^x$ changes by a constant multiplicative factor for every unit increase in $x$, producing rapid growth or decay.

$f(x)=ab^x$: $a$ = initial value ($f(0)=a$), $b>0,\ b≠ 1$ = growth factor. $b>1$ models growth; $0<b<1$ models decay. Domain: all reals; range: $y>0$ (if $a>0$); horizontal asymptote $y=0$.

Exponential growth/decay (variable is the EXPONENT, e.g. $2^x$) versus polynomial growth (variable is the BASE, e.g. $x^2$) --- exponential functions eventually outgrow any polynomial.

Confusing the growth factor $b$ with the growth rate; e.g., $f(x)=100(1.05)^x$ represents 5% growth per period, not a factor of 1.05 being ``105% growth.''

$f(x)=ab^x$: $a$ is the start, $b$ is the multiplier each step; $y=0$ is always the asymptote.

A logarithmic function is the inverse of an exponential function, so its graph and domain/range are the exponential's graph reflected over the line $y=x$.

$f(x)=_b x$ is defined only for $x>0$ (domain), range is all real numbers, vertical asymptote at $x=0$. $f(x)=_b x b^f(x)=x$.

The domain/range of $y=_b x$ (domain $x>0$, range all reals) versus the domain/range of its inverse $y=b^x$ (domain all reals, range $y>0$) --- inverses swap domain and range.

Trying to evaluate $_b x$ for a negative $x$ or $x=0$ instead of recognizing the expression is undefined there.

Log functions: domain $x>0$, vertical asymptote at $x=0$ --- exponential's mirror image.

A piecewise function is defined by different formulas on different intervals of the domain, so you must first identify which piece applies to a given input.

Evaluate by checking which interval condition the input satisfies, then apply only that piece's formula. Graphs may have open or closed endpoints at the boundary between pieces, matching the inequality symbols used.

An open circle (boundary value NOT included, strict inequality) versus a closed circle (boundary value included, $≤q$ or $≥q$) at the transition point between pieces.

Plugging an input into the wrong piece of the function because the boundary condition was misread (e.g., using $x<2$ rule when $x=2$ exactly, which belongs to the $x≥q 2$ piece).

Find the correct interval first, then evaluate --- never guess which rule applies.

An inverse function reverses the input/output pairs of the original function, undoing whatever the original function did.

To find $f^-1(x)$: swap $x$ and $y$ in $y=f(x)$, then solve for $y$. Verify with $f(f^-1(x))=x$ and $f^-1(f(x))=x$. Graphically, $f^-1$ is the reflection of $f$ over the line $y=x$. Only one-to-one functions have inverses that are also functions.

$f^-1(x)$ (the inverse function) versus $/1f(x)$ (the reciprocal) --- the $-1$ exponent notation on a function name never means ``take the reciprocal.''

Writing $f^-1(x)=/1f(x)$ instead of correctly swapping variables and solving, especially under time pressure.

Swap $x$ and $y$, solve for $y$ $arrow$ that is $f^-1(x)$, NOT $/1f(x)$.

Every transformation of a parent function's graph can be read directly from how constants are attached inside or outside the function notation.

For $y=af(b(x-h))+k$: $k$ shifts up/down; $h$ shifts right/left (opposite of the sign shown); $a$ stretches/compresses vertically (and reflects over the $x$-axis if $a<0$); $b$ stretches/compresses horizontally (and reflects over the $y$-axis if $b<0$).

Transformations INSIDE the function argument, e.g. $f(x-h)$ (affects $x$-direction, sign is opposite of what it looks like) versus transformations OUTSIDE the function, e.g. $f(x)+k$ (affects $y$-direction, sign is exactly as it looks).

Reading $f(x+3)$ as a shift RIGHT 3 instead of the correct shift LEFT 3 --- inside-the-parentheses shifts always move opposite to the sign shown.

Inside the parentheses $arrow$ shifts $x$, opposite sign; outside $arrow$ shifts $y$, same sign.

Every real number belongs to a nested hierarchy of number sets, and knowing that hierarchy tells you which properties and operations are guaranteed to apply.

Natural $$ Whole $$ Integers $$ Rational $$ Real. Rational numbers can be written as $/pq$ ($q≠ 0$, terminating or repeating decimals); irrational numbers cannot (e.g., $$, $sqrt2$). Key properties: commutative, associative, distributive, identity, and inverse.

Rational numbers (terminating or REPEATING decimals) versus irrational numbers (non-terminating, NON-repeating decimals) --- a repeating decimal like $0.3$ is still rational.

Assuming any decimal that ``goes on forever'' is irrational, without checking whether the digits actually repeat in a pattern.

Repeats or terminates $arrow$ rational; never repeats, never terminates $arrow$ irrational.

Complex numbers extend the reals by defining $i=sqrt-1$, letting every quadratic (even with a negative discriminant) have a solution.

$i=sqrt-1$, so $i^2=-1$. Standard form: $a+bi$. Add/subtract by combining real and imaginary parts separately. Multiply using FOIL and $i^2=-1$. Conjugate of $a+bi$ is $a-bi$; multiplying a complex number by its conjugate gives a real number: $(a+bi)(a-bi)=a^2+b^2$.

$i^2=-1$ versus $i=-1$ --- $i$ itself is $sqrt-1$, not $-1$; only its SQUARE equals $-1$.

Simplifying $sqrt-9$ as $-3$ instead of the correct $3i$, or forgetting to multiply by the conjugate when dividing complex numbers.

$i^2=-1$; to divide complex numbers, multiply top and bottom by the conjugate.

A factorial counts the number of ways to arrange a set of distinct objects in order, growing extremely fast as $n$ increases.

$n! = n·(n-1)·(n-2)·s 2· 1$ for a positive integer $n$. Special case: $0!=1$. $/n!(n-k)! = n(n-1)·s(n-k+1)$ (used in permutations).

$0!=1$ (a defined special case, NOT zero) versus $0· n=0$ --- students often assume $0!$ should equal 0 by analogy with multiplication by zero.

Computing $/6!3!$ as $2!$ (dividing the numbers 6 and 3) instead of correctly canceling: $/6!3!=6·5·4=120$.

$n!=n(n-1)(n-2)·s 1$; and by definition $0!=1$.

The Binomial Theorem gives a direct formula for expanding $(x+y)^n$ without multiplying it out term by term, using combinations as coefficients.

$(x+y)^n=_k=0^nnkx^n-ky^k$, where $nk=/n!k!(n-k)!$. The expansion has $n+1$ terms; exponents on $x$ decrease from $n$ to 0 while exponents on $y$ increase from 0 to $n$.

$nk$ (the binomial coefficient/combination, order doesn't matter) versus a straightforward exponent --- $nk$ must be computed with the factorial formula, not read off directly.

Losing track of alternating signs when expanding $(x-y)^n$ --- every odd power of $(-y)$ contributes a negative term.

$(x+y)^n$ has $n+1$ terms with coefficients $nk$; watch signs when it's $(x-y)^n$.

No formula sheet is provided during the actual CLEP exam (only the built-in TI-30XS MultiView calculator), so this short list must be memorized cold.

Quadratic formula: $x=/-b±sqrtb^2-4ac2a$. Discriminant: $D=b^2-4ac$. Vertex of a parabola: $x=-/b2a$, or read $(h,k)$ from $a(x-h)^2+k$. Slope: $m=/y_2-y_1x_2-x_1$. Exponent rules: $a^m a^n=a^m+n$, $(a^m)^n=a^mn$, $a^-n=/1a^n$. Log rules: $_b(xy)=_b x+_b y$, $_b(x^n)=n_b x$. Difference of squares: $a^2-b^2=(a-b)(a+b)$. Binomial theorem: $(x+y)^n=_k=0^nnkx^n-ky^k$.

No reference sheet on test day $arrow$ these formulas must live in memory, not on paper.

Most CLEP College Algebra errors trace back to one of a small set of look-alike pairs --- drilling these pairs directly prevents the most common point losses.

When two ideas look alike on the exam, that similarity is usually the trap itself.

Every function on the exam belongs to one of a handful of families, each with a signature graph shape, domain/range, and asymptote pattern.

Identify the family first --- the family instantly tells you the expected domain, range, and shape.

Knowing exactly what tools the CLEP exam gives you (and what it assumes you already know) prevents wasted time hunting for support that isn't there.

Calculator: yes. Formula sheet: no. Prerequisite: one semester of college algebra (or equivalent).

A consistent five-step method prevents the most common source of errors on a timed, no-reference-sheet exam: skipping a step under pressure.

Identify $arrow$ simplify $arrow$ choose method $arrow$ solve $arrow$ check against the original problem.

Knowing the exact structure of the test lets you pace yourself precisely instead of guessing how much time each question deserves.

60 questions, 90 minutes, half routine/half conceptual --- pace at about 90 seconds per question.

Different equation types (linear, quadratic, absolute value, exponential, logarithmic) all reduce to the same underlying pathway: isolate, then apply the inverse operation for that function family.

Every equation type: isolate the special structure first, then undo it with its inverse operation.

When time is short, domain questions can be answered by running through the same three-question checklist every time, regardless of how complicated the function looks.

Denominator $≠ 0$, even root $≥q 0$, log $>0$ --- check all three, every time.

These statements sound true because they resemble a correct rule, but each one is a common misapplication that costs points on the actual exam.

If a rule ``feels obvious'' but involves a sign, a sum, or a square root of a negative --- double-check it.