This lesson feeds the exam's largest category, Algebraic Expressions, Equations and Inequalities (20%), and underpins the 30% Representations category. Section 2 of the CLEP exam (23 questions, 40 minutes) allows no calculator — every skill here has to run by hand, quickly.
| Rule | Statement | Example |
|---|---|---|
| Product | b^m · b^n = b^(m+n) | 5^x · 5^(2x) = 5^(3x) |
| Quotient | b^m / b^n = b^(m−n) | x^7/x^3 = x^4 |
| Power | (b^m)^n = b^(mn) | (2^x)^3 = 2^(3x) |
| Negative | b^(−n) = 1/b^n | x^(−2) = 1/x² |
| Rational | b^(m/n) = (ⁿ√b)^m | 16^(3/4) = (⁴√16)³ = 2³ = 8 |
Two habits pay off on exam day. First, with rational exponents, take the root first, then the power — 16^(3/4) as (16^(1/4))³ keeps the numbers small; (16³)^(1/4) forces you through 4096. Second, check what matches before applying a rule: exponents add when bases match (2^x · 2^y = 2^(x+y)); bases multiply when exponents match (2^x · 3^x = 6^x). Mixing these two produces classic wrong answers, and the exam stocks its distractors with them.
One identity worth owning because it makes different-looking expressions identical: b^(x+k) = b^k · b^x. So 2^(x+3) is the same expression as 8·2^x — a fact the exam tests directly as "which of the following is equivalent?"
To simplify √n, extract the largest perfect square: √48 = √(16·3) = 4√3. (Watch the misread √48 ≠ 16√3 — the 16 comes out as its square root.)
Rationalizing clears radicals from denominators:
6/(√5 − √2) · (√5 + √2)/(√5 + √2) = 6(√5 + √2)/(5 − 2) = 2(√5 + √2)
The denominator becomes 5 − 2 = 3 (subtraction, not addition — the most common slip).
Adding and subtracting polynomials is collecting like terms; multiplying is distributing every term. Three products to know on sight:
(a + b)² = a² + 2ab + b² (a − b)² = a² − 2ab + b²
(a + b)(a − b) = a² − b²
The persistent error: (a + b)² ≠ a² + b². Squaring does not distribute over addition — the middle term 2ab is always there. Degrees are predictable under multiplication: a degree-m polynomial times a degree-n polynomial has degree exactly m + n, because the leading terms multiply to a nonzero leading term.
Run these in order, every time:
| Step | Pattern | Example |
|---|---|---|
| 1. GCF first | pull out the greatest common factor | 2x³ − 8x = 2x(x² − 4) |
| 2. Difference of squares | a² − b² = (a − b)(a + b) | x² − 4 = (x − 2)(x + 2) |
| 3. Trinomial | find factors of ac summing to b | x² − x − 6 = (x − 3)(x + 2) |
| 4. Grouping (4 terms) | pair, factor each pair, extract common binomial | see below |
"Factor completely" means keep going until nothing factors further: 2x³ − 8x = 2x(x² − 4) = 2x(x − 2)(x + 2). Stopping at 2x(x² − 4) is the standard trap.
Grouping, worked: x³ + 3x² − 4x − 12 = x²(x + 3) − 4(x + 3) = (x + 3)(x² − 4) = (x + 3)(x − 2)(x + 2). Note the sign discipline: factoring −4 from −4x − 12 leaves (x + 3), matching the first group — that match is your signal grouping worked.
Simplify by factoring top and bottom, then canceling common factors (never terms):
(x² − 9)/(x² − x − 6) = (x − 3)(x + 3) / [(x − 3)(x + 2)] = (x + 3)/(x + 2), x ≠ 3, x ≠ −2
Canceling the x² and the constants separately — getting 9/6 or similar — is not algebra; only whole factors cancel. Excluded values come from the original denominator (here x ≠ 3 survives as a restriction even after the factor cancels).
Add and subtract over the least common denominator (LCD), and distribute subtraction across the entire numerator:
3/(x + 1) − 2/x = [3x − 2(x + 1)] / [x(x + 1)] = (x − 2) / [x(x + 1)]
The parentheses around 2(x + 1) are load-bearing: dropping them gives 3x − 2x + 2 = x + 2 instead of x − 2.
Dividing p(x) by lower-degree d(x) produces quotient q(x) and remainder r(x) with deg r < deg d:
p(x)/d(x) = q(x) + r(x)/d(x) equivalently p(x) = d(x)·q(x) + r(x)
Worked layout for (x³ − 5x + 6) ÷ (x − 2) — note the inserted 0x² placeholder for the missing term:
x³ + 0x² − 5x + 6 ÷ (x − 2):
x³ − 2x² ← x²(x − 2); bring down: 2x² − 5x
2x² − 4x ← 2x(x − 2); bring down: −x + 6
−x + 2 ← −1(x − 2); remainder: 4
Result: (x³ − 5x + 6)/(x − 2) = x² + 2x − 1 + 4/(x − 2).
Ten-second verification: evaluate both forms at x = 0. Left: 6/(−2) = −3. Right: −1 + 4/(−2) = −3. ✓ This check catches nearly every sign slip and costs almost nothing — run it whenever you divide.
Two consequences worth knowing:
Q1. E. By definition, x^(m/n) = (ⁿ√x)^m = ⁿ√(x^m); with m = 2, n = 3 this is ∛(x²). A) puts the 3 on the power and the 2 in the root — numerator and denominator swapped. B) x²/x³ = x^(−1), which subtracts exponents instead of forming a rational exponent. C) multiplies x by the fraction, confusing an exponent with a coefficient. D) nests the roots, giving x^(1/6) — the fraction 2/3 is a power over a root, not a root of a root. Fix: in x^(m/n), the denominator is the root and the numerator is the power — "denominator down in the root."
Q2. E. √48 = √(16·3) = √16·√3 = 4√3. A) extracts √4 = 2 from 48 = 4·12 but then miswrites the leftover √12 as √6 — and 2√12 itself would still not be simplest form. B) swaps the roles of the extracted and remaining factors. C) pulls 16 out without taking its square root. D) takes √16 = 4 but then also halves it. Fix: find the largest perfect square factor and bring out its square root — 16 comes out as 4.
Q3. C. 2^(x+3) = 2³·2^x = 8·2^x by the product rule read in reverse. A) converts multiplication of powers into addition of terms. B) multiplies the base by the added exponent's base, inventing 6^x. D) uses the exponent 3 itself as the coefficient instead of 2³. E) treats the sum x + 3 as the product 3x, misapplying the power rule. Fix: b^(x+k) = b^k·b^x — a shift in the exponent is a constant multiple out front.
Q4. A. Multiply by the conjugate: 6(√5 + √2)/[(√5)² − (√2)²] = 6(√5 + √2)/3 = 2(√5 + √2). B) keeps the minus sign in the numerator — but the conjugate you multiply by has the opposite sign, and it lands in the numerator. C) adds instead of subtracts in the denominator, getting 5 + 2 = 7. D) inverts the final arithmetic, dividing the conjugate by 3 instead of 6 by 3. E) both keeps the wrong sign and mishandles the constant. Fix: (√a − √b)(√a + √b) = a − b — the rationalized denominator is a subtraction.
Q5. A. GCF first: 2x(x² − 4); then difference of squares: 2x(x − 2)(x + 2). B) stops after the GCF — x² − 4 still factors, so the factorization is not complete. C) writes the difference of squares as a perfect square, but (x − 2)² = x² − 4x + 4 ≠ x² − 4. D) uses ±4 instead of ±2, forgetting to take the square root of 4. E) drops the GCF factor x entirely. Fix: after every factoring step, re-scan each factor — "completely" means no factor still factors.
Q6. D. Group: x²(x + 3) − 4(x + 3) = (x + 3)(x² − 4) = (x + 3)(x − 2)(x + 2). A) flips the sign of the binomial from the grouping step. B) writes x² + 4 for the leftover quadratic, but the grouping yields x² − 4 (and x² + 4 doesn't factor over the reals anyway). C) turns the difference of squares into a perfect square. E) combines both errors — wrong binomial sign and wrong quadratic factorization. Fix: after grouping, the two groups must expose the same binomial; then check the remaining quadratic for further factoring.
Q7. A. Factor both: (x − 3)(x + 3)/[(x − 3)(x + 2)]; cancel the common factor (x − 3), leaving (x + 3)/(x + 2). B) "cancels" the x² terms and reduces the constants −9/−6 — terms are not factors and cannot be canceled. C) cancels the wrong factors, keeping (x − 3) and inventing (x − 2) below. D) factors the denominator as (x + 3)(x − 2), a sign error in the trinomial. E) keeps the factor that should cancel and cancels the one that shouldn't. Fix: factor completely first; only identical whole factors cancel.
Q8. C. By the remainder theorem, the remainder of p(x) ÷ (x + 2) is p(−2) = 3. A) would require p(−2) = 0, not 3 — a factor means zero remainder. B) uses the wrong divisor: division by x − 2 leaves remainder p(2), which is unknown. D) same confusion as A: a zero at −2 means p(−2) = 0. E) the y-intercept is p(0), not p(−2). Fix: divisor (x − a) ↔ remainder p(a) — flip the sign of the constant in the divisor to find where to evaluate.
Q9. B. LCD is x(x + 1): [3x − 2(x + 1)]/[x(x + 1)] = (3x − 2x − 2)/[x(x + 1)] = (x − 2)/[x(x + 1)]. A) subtracts the numerators as bare constants (3 − 2 = 1) without adjusting for the LCD. C) fails to distribute the subtraction, computing 3x − 2x + 2. D) adds the fractions instead of subtracting. E) cancels an x across a sum after an incorrect combination. Fix: when subtracting fractions, put parentheses around the entire second numerator before distributing the minus sign.
Q10. D. Evaluating both forms at a convenient point (x = 0) is a valid equivalence check: 6/(−2) = −3 and −1 + 4/(−2) = −3 agree, confirming the division. A) degree of the quotient is less than the dividend's (2 vs. 3), so this "check" tests nothing true. B) x = 2 is excluded from both forms — both denominators vanish there. C) there is no such relationship between the remainder and the constant terms; the remainder comes from the full division. E) the quotient alone need not pass through points of the original function — y = x² + 2x − 1 gives −1 at x = 0, not 6. Fix: after any division, evaluate both forms at x = 0 (or another easy allowed value) and confirm they match.
Q11. B. The leading terms multiply: (ax³)(bx²) = ab·x⁵ with ab ≠ 0, so the degree is exactly 3 + 2 = 5. A) multiplies the degrees instead of adding them. C) squares-and-sums in some fashion — 3² = 9 confusions appear when the power rule is misapplied to degrees. D) adds then doubles, another power-rule misfire. E) no cancellation of leading terms is possible in a product — degrees add without exception, so it is determined. Fix: degrees add under multiplication (and the leading coefficients multiply to something nonzero, so no drop is possible).
Q12. E. (x + 3)² = (x + 3)(x + 3) = x² + 3x + 3x + 9 = x² + 6x + 9; the student's answer omits the middle term. A) squaring does not distribute over addition — that property holds for multiplication, not squaring a sum. B) includes a middle term but only counts 3x once; the term appears twice in the expansion. C) the identity fails for every x ≠ 0, not just negative x — try x = 1: 16 ≠ 10. D) uses 3·3 = 9 as the middle coefficient instead of 2·3 = 6. Fix: (a + b)² = a² + 2ab + b² — the middle term is twice the product, always.
Q1. E. By definition, x^(m/n) = (ⁿ√x)^m = ⁿ√(x^m); with m = 2, n = 3 this is ∛(x²). A) puts the 3 on the power and the 2 in the root — numerator and denominator swapped. B) x²/x³ = x^(−1), which subtracts exponents instead of forming a rational exponent. C) multiplies x by the fraction, confusing an exponent with a coefficient. D) nests the roots, giving x^(1/6) — the fraction 2/3 is a power over a root, not a root of a root. Fix: in x^(m/n), the denominator is the root and the numerator is the power — "denominator down in the root."
Q2. E. √48 = √(16·3) = √16·√3 = 4√3. A) extracts √4 = 2 from 48 = 4·12 but then miswrites the leftover √12 as √6 — and 2√12 itself would still not be simplest form. B) swaps the roles of the extracted and remaining factors. C) pulls 16 out without taking its square root. D) takes √16 = 4 but then also halves it. Fix: find the largest perfect square factor and bring out its square root — 16 comes out as 4.
Q3. C. 2^(x+3) = 2³·2^x = 8·2^x by the product rule read in reverse. A) converts multiplication of powers into addition of terms. B) multiplies the base by the added exponent's base, inventing 6^x. D) uses the exponent 3 itself as the coefficient instead of 2³. E) treats the sum x + 3 as the product 3x, misapplying the power rule. Fix: b^(x+k) = b^k·b^x — a shift in the exponent is a constant multiple out front.
Q4. A. Multiply by the conjugate: 6(√5 + √2)/[(√5)² − (√2)²] = 6(√5 + √2)/3 = 2(√5 + √2). B) keeps the minus sign in the numerator — but the conjugate you multiply by has the opposite sign, and it lands in the numerator. C) adds instead of subtracts in the denominator, getting 5 + 2 = 7. D) inverts the final arithmetic, dividing the conjugate by 3 instead of 6 by 3. E) both keeps the wrong sign and mishandles the constant. Fix: (√a − √b)(√a + √b) = a − b — the rationalized denominator is a subtraction.
Q5. A. GCF first: 2x(x² − 4); then difference of squares: 2x(x − 2)(x + 2). B) stops after the GCF — x² − 4 still factors, so the factorization is not complete. C) writes the difference of squares as a perfect square, but (x − 2)² = x² − 4x + 4 ≠ x² − 4. D) uses ±4 instead of ±2, forgetting to take the square root of 4. E) drops the GCF factor x entirely. Fix: after every factoring step, re-scan each factor — "completely" means no factor still factors.
Q6. D. Group: x²(x + 3) − 4(x + 3) = (x + 3)(x² − 4) = (x + 3)(x − 2)(x + 2). A) flips the sign of the binomial from the grouping step. B) writes x² + 4 for the leftover quadratic, but the grouping yields x² − 4 (and x² + 4 doesn't factor over the reals anyway). C) turns the difference of squares into a perfect square. E) combines both errors — wrong binomial sign and wrong quadratic factorization. Fix: after grouping, the two groups must expose the same binomial; then check the remaining quadratic for further factoring.
Q7. A. Factor both: (x − 3)(x + 3)/[(x − 3)(x + 2)]; cancel the common factor (x − 3), leaving (x + 3)/(x + 2). B) "cancels" the x² terms and reduces the constants −9/−6 — terms are not factors and cannot be canceled. C) cancels the wrong factors, keeping (x − 3) and inventing (x − 2) below. D) factors the denominator as (x + 3)(x − 2), a sign error in the trinomial. E) keeps the factor that should cancel and cancels the one that shouldn't. Fix: factor completely first; only identical whole factors cancel.
Q8. C. By the remainder theorem, the remainder of p(x) ÷ (x + 2) is p(−2) = 3. A) would require p(−2) = 0, not 3 — a factor means zero remainder. B) uses the wrong divisor: division by x − 2 leaves remainder p(2), which is unknown. D) same confusion as A: a zero at −2 means p(−2) = 0. E) the y-intercept is p(0), not p(−2). Fix: divisor (x − a) ↔ remainder p(a) — flip the sign of the constant in the divisor to find where to evaluate.
Q9. B. LCD is x(x + 1): [3x − 2(x + 1)]/[x(x + 1)] = (3x − 2x − 2)/[x(x + 1)] = (x − 2)/[x(x + 1)]. A) subtracts the numerators as bare constants (3 − 2 = 1) without adjusting for the LCD. C) fails to distribute the subtraction, computing 3x − 2x + 2. D) adds the fractions instead of subtracting. E) cancels an x across a sum after an incorrect combination. Fix: when subtracting fractions, put parentheses around the entire second numerator before distributing the minus sign.
Q10. D. Evaluating both forms at a convenient point (x = 0) is a valid equivalence check: 6/(−2) = −3 and −1 + 4/(−2) = −3 agree, confirming the division. A) degree of the quotient is less than the dividend's (2 vs. 3), so this "check" tests nothing true. B) x = 2 is excluded from both forms — both denominators vanish there. C) there is no such relationship between the remainder and the constant terms; the remainder comes from the full division. E) the quotient alone need not pass through points of the original function — y = x² + 2x − 1 gives −1 at x = 0, not 6. Fix: after any division, evaluate both forms at x = 0 (or another easy allowed value) and confirm they match.
Q11. B. The leading terms multiply: (ax³)(bx²) = ab·x⁵ with ab ≠ 0, so the degree is exactly 3 + 2 = 5. A) multiplies the degrees instead of adding them. C) squares-and-sums in some fashion — 3² = 9 confusions appear when the power rule is misapplied to degrees. D) adds then doubles, another power-rule misfire. E) no cancellation of leading terms is possible in a product — degrees add without exception, so it is determined. Fix: degrees add under multiplication (and the leading coefficients multiply to something nonzero, so no drop is possible).
Q12. E. (x + 3)² = (x + 3)(x + 3) = x² + 3x + 3x + 9 = x² + 6x + 9; the student's answer omits the middle term. A) squaring does not distribute over addition — that property holds for multiplication, not squaring a sum. B) includes a middle term but only counts 3x once; the term appears twice in the expansion. C) the identity fails for every x ≠ 0, not just negative x — try x = 1: 16 ≠ 10. D) uses 3·3 = 9 as the middle coefficient instead of 2·3 = 6. Fix: (a + b)² = a² + 2ab + b² — the middle term is twice the product, always.