ACT Math Quiz: Quadratics And Polynomials

What this quiz covers

This quiz focuses on Quadratics And Polynomials, giving you a quick way to practice the rules, question types, and explanations that matter most for ACT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $x^2+3x+10$
2. $x^2-3x-10$
3. $x^2+3x-10$ (correct answer)
4. $x^2+7x-10$

Explanation: Expanding a product of two binomials means multiplying every term of the first by every term of the second, which FOIL organizes as First, Outer, Inner, Last. For $(x+5)(x-2)$ the First product is $x^2$, the Outer is $-2x$, the Inner is $5x$, and the Last is $(5)(-2)=-10$, so combining the two middle terms gives $x^2+3x-10$. The result $x^2+3x+10$ keeps the correct middle term but multiplies $5$ and $-2$ as if both were positive, and $x^2-3x-10$ reverses which middle term is negative, giving $-5x+2x$ instead of $5x-2x$. The result $x^2+7x-10$ comes from adding $5$ and $2$ for the middle coefficient instead of combining $+5x$ with $-2x$. When you expand binomials, compute the constant product with its signs first, then add the two cross terms rather than adding the raw numbers, since the sign on each cross term decides the middle coefficient.

Question 2

1. $-5$ and $2$
2. $5$ and $-2$ (correct answer)
3. $5$ and $2$
4. $-5$ and $-2$

Explanation: This is a polynomial factors and zeros question testing synthetic division and factoring. Choice B (5 and −2) is correct — since (x − 1) is a factor, divide p(x) by (x − 1) using synthetic division: coefficients 1, −4, −7, 10 divided by root 1: bring down 1; 1×1 − 4 = −3; −3×1 − 7 = −10; −10×1 + 10 = 0 ✓. Quotient: x² − 3x − 10 = (x − 5)(x + 2). Zeros: x = 5 and x = −2. Choice A (−5 and 2) flips both signs of the correct zeros. Choice C (5 and 2) gets x = 5 correct but uses x = 2 instead of x = −2, likely from factoring x² − 3x − 10 as (x − 5)(x − 2) (wrong sign on the second factor). Choice D (−5 and −2) negates both correct zeros. Pro tip: After dividing out the known factor, you're left with a simpler polynomial to factor. Here: (x − 1)(x² − 3x − 10) = 0. Factor x² − 3x − 10: find two numbers that multiply to −10 and sum to −3: −5 and +2. So (x − 5)(x + 2) = 0 → x = 5 or x = −2.

Question 3

1. $3x^2 - 7x + 8$ (correct answer)
2. $3x^2 + x + 6$
3. $7x^2 + x + 6$
4. $7x^2 - 7x + 8$

Explanation: This is a polynomial operations question testing distribution of a negative sign. Choice A (3x² − 7x + 8) is correct — distribute the negative sign: (5x² − 3x + 7) − (2x² + 4x − 1) = 5x² − 3x + 7 − 2x² − 4x + 1. Key step: −(−1) = +1. Combine: (5−2)x² + (−3−4)x + (7+1) = 3x² − 7x + 8. Choice B (3x² + x + 6) correctly subtracts x² terms but fails to distribute the negative on the x term: −3x − (4x) treated as −3x + 4x = +x, and −(−1) = −1 instead of +1. Choice C (7x² + x + 6) adds x² terms instead of subtracting (5 + 2 = 7) and makes both sign errors. Choice D (7x² − 7x + 8) adds x² terms (5 + 2 = 7) but correctly handles x and constant terms. Pro tip: When subtracting a polynomial, distribute the negative sign to EVERY term inside the parentheses. Rewrite −(2x² + 4x − 1) as (−2x² − 4x + 1) before combining. The term that trips most students is −(−1) = +1.

Question 4

1. 1 (correct answer)
2. 0
3. 4
4. 3

Explanation: To find $f(3)$ for $f(x) = x^2 - 4x + 4$, we substitute $x = 3$ into the function. We get $f(3) = (3)^2 - 4(3) + 4 = 9 - 12 + 4 = 1$. Notice that this function can also be written as $(x - 2)^2$, so $f(3) = (3 - 2)^2 = 1^2 = 1$.

Question 5

1. 7 (correct answer)
2. 3
3. 0
4. 4

Explanation: To find $f(0)$ for $f(x) = 3x^2 - 4x + 7$, we substitute $x = 0$ into the function. We get $f(0) = 3(0)^2 - 4(0) + 7 = 0 - 0 + 7 = 7$. When evaluating polynomial functions at zero, all terms with $x$ become zero, leaving only the constant term.

Question 6

1. $(x + 4)(x + 6)$
2. $(x + 3)(x + 7)$
3. $(x - 4)(x - 6)$
4. $(x - 3)(x - 7)$ (correct answer)

Explanation: To factor x² - 10x + 21, we need two numbers that multiply to 21 and add to -10. The numbers are -3 and -7 since (-3)(-7) = 21 and -3 + (-7) = -10. Therefore, x² - 10x + 21 = (x - 3)(x - 7). Choice C has factors that multiply to 24, not 21.

Question 7

1. $-4$ and $\frac{3}{2}$
2. $-\frac{3}{2}$ and $4$ (correct answer)
3. $-3$ and $2$
4. $3$ and $-2$

Explanation: The correct answer is B (−3/2 and 4). Factor 2x² − 5x − 12 = 0. Look for factors of (2)(−12) = −24 that add to −5: those are −8 and +3. Rewrite: 2x² − 8x + 3x − 12 = 0 → 2x(x − 4) + 3(x − 4) = 0 → (2x + 3)(x − 4) = 0. Solutions: x = −3/2 and x = 4. A (−4 and 3/2) flips the signs of both solutions. C (−3 and 2) ignores the leading coefficient of 2, incorrectly factoring as (x + 3)(x − 2) = 0. D (3 and −2) also ignores the leading 2, factoring as (x − 3)(x + 2) = 0. When the leading coefficient isn't 1, always account for it in the factoring process.

Question 8

1. $x = 2, x = 3$
2. $x = -2, x = 3$ (correct answer)
3. $x = 2, x = -3$
4. $x = -2, x = -3$

Explanation: When a polynomial has a given factor, you can find the other factors by polynomial division or by testing the roots. Since (x - 1) is a factor, x = 1 is a root. To find the other roots, divide P(x) by (x - 1) using synthetic division with 1: you get x² - x - 6, which factors as (x - 3)(x + 2). Therefore, the complete factorization is P(x) = (x - 1)(x - 3)(x + 2), giving roots x = 1, 3, -2. The question asks for the "other two roots" (besides x = 1), which are -2 and 3.

Question 9

1. 180 (correct answer)
2. 182
3. 200
4. 729

Explanation: This is a perimeter and area question requiring setting up and solving an equation. Choice A (180) is correct — let w = width, then length = w + 3. Perimeter: 2(w) + 2(w + 3) = 54 → 4w + 6 = 54 → 4w = 48 → w = 12. Length = 15. Area = 12 × 15 = 180 sq ft. Choice B (182) results from an off-by-one error in the dimensions, using 13 × 14 = 182 — likely from a perimeter setup error that produces w = 13. Choice C (200) may come from a perimeter equation error that yields dimensions like 10 × 20. Choice D (729) results from treating the half-perimeter (27) as both dimensions of a square and squaring it: 27² = 729. Pro tip: Set up the perimeter equation explicitly — don't try to guess dimensions. Write 2w + 2(w + 3) = 54, solve for w, then compute the area from the actual dimensions.

Question 10

1. $-21$ (correct answer)
2. $-13$
3. $-5$
4. $11$

Explanation: This is a polynomial substitution question testing careful evaluation with negative inputs. Choice A (−21) is correct — substitute x = −2 term by term: 2(−2)³ = 2(−8) = −16. −(−2)² = −(4) = −4. 3(−2) = −6. +5 = 5. Sum: −16 − 4 − 6 + 5 = −21. Choice B (−13) results from computing 2(−2)³ as −8 rather than −16 — forgetting the coefficient of 2: (−2)³ = −8, but 2 × (−8) = −16. Choice C (−5) comes from multiple sign errors throughout, often including treating (−2)³ as +8 for some terms. Choice D (11) likely results from treating (−2)³ as +8 (a sign error on the cube of a negative): 2(8) − 4 − 6 + 5 = 16 − 5 = 11. Pro tip: Evaluate each term separately and write out each result before adding. Compute 2(−2)³ as: step 1: (−2)³ = −8, step 2: 2 × (−8) = −16. Rushing through in one mental step is where sign errors sneak in.

Question 11

1. $4x^2 - 9$
2. $4x^2 + 9$
3. $4x^2 - 6x + 9$
4. $4x^2 - 12x + 9$ (correct answer)

Explanation: This is an expanding binomials question testing the FOIL method on a perfect square. Choice D (4x² − 12x + 9) is correct — (2x − 3)² = (2x − 3)(2x − 3). Multiply: First: 2x × 2x = 4x². Outer: 2x × (−3) = −6x. Inner: (−3) × 2x = −6x. Last: (−3) × (−3) = 9. Sum: 4x² − 6x − 6x + 9 = 4x² − 12x + 9. Choice A (4x² − 9) is the classic "difference of squares" error — treating (a − b)² as a² − b², which only works for (a − b)(a + b), not (a − b)². Choice B (4x² + 9) squares each term and adds, ignoring the cross term and getting the sign of 9 wrong. Choice C (4x² − 6x + 9) halves the middle term — computing only one of the two −6x cross terms instead of combining both. Pro tip: A perfect square binomial (a − b)² ALWAYS produces three terms: a² − 2ab + b². The middle term is negative 2 times the product of the two terms. Never skip to a² − b² — that's a different product entirely.

Question 12

1. $(x + 6)(x - 4)$ (correct answer)
2. $(x - 6)(x + 4)$
3. $(x + 8)(x - 3)$
4. $(x - 12)(x + 2)$

Explanation: This is a factoring quadratics question testing factor-pair identification. Choice A ((x + 6)(x − 4)) is correct — find two numbers that multiply to −24 (the constant) and add to +2 (the x-coefficient): +6 and −4 satisfy both: 6 × (−4) = −24 ✓ and 6 + (−4) = +2 ✓. Check: (x + 6)(x − 4) = x² − 4x + 6x − 24 = x² + 2x − 24 ✓. Choice B ((x − 6)(x + 4)) has factors that multiply to −24 but add to −2, not +2 — this factors x² − 2x − 24. Choice C ((x + 8)(x − 3)) has factors that multiply to −24 but add to +5, not +2. Choice D ((x − 12)(x + 2)) has factors that multiply to −24 but add to −10. Pro tip: For ax² + bx + c with a = 1, list factor pairs of c (with signs) and find the pair that sums to b. The sign of c tells you the signs of your factors: negative c means one positive and one negative factor; positive c means both same sign.

Question 13

1. $-2$ (correct answer)
2. $1$
3. $3$
4. $5$

Explanation: The correct answer is A (−2). Since both sides share the same denominator (x − 3), multiply both sides by (x − 3): x² − x = 6 → x² − x − 6 = 0 → (x − 3)(x + 2) = 0. Solutions: x = 3 or x = −2. Check for extraneous solutions: x = 3 makes the denominator zero, so it is extraneous. The only valid solution is x = −2. B (1) is the sum of 3 + (−2) = 1 — the student finds both roots but reports their sum without excluding the extraneous one. C (3) accepts the extraneous solution x = 3, ignoring that it makes the denominator zero. D (5) may result from incorrect factoring. Always check solutions in rational equations — any value that zeros the denominator must be rejected.

Question 14

1. $-5$
2. $-3$ (correct answer)
3. $3$
4. $5$

Explanation: The correct answer is B (−3). By the Polynomial Remainder Theorem, if (x − 2) is a factor of P(x), then P(2) = 0. Substitute: P(2) = (2)³ − 3(2)² + k(2) + 10 = 8 − 12 + 2k + 10 = 6 + 2k. Set equal to zero: 6 + 2k = 0 → 2k = −6 → k = −3. A (−5) results from arithmetic errors in evaluating P(2), possibly computing 8 − 12 + 10 incorrectly. C (3) correctly reaches 2k = −6 but reports the positive value, dropping the negative sign. D (5) likely comes from substituting x = −2 (confusing the factor x − 2 with root x = −2) or from a sign error. Remember: the root of (x − 2) is x = +2, not x = −2.

Question 15

1. $(x + 3)(x + 3)$
2. $(x + 4)(x + 2)$
3. $(x - 3)(x - 3)$
4. $(x - 4)(x - 2)$ (correct answer)

Explanation: To factor x² - 6x + 8, we need two numbers that multiply to 8 and add to -6. The numbers are -4 and -2 since (-4)(-2) = 8 and -4 + (-2) = -6. Therefore, x² - 6x + 8 = (x - 4)(x - 2). Choice C represents a perfect square that would give x² - 6x + 9.

Question 16

1. $(-3, 7)$
2. $(7, -3)$
3. $(-7, 3)$ (correct answer)
4. $(3, -7)$

Explanation: To solve $x^2 + 4x - 21 = 0$ by factoring, we need two numbers that multiply to -21 and add to 4. The numbers 7 and -3 work: (7)(-3) = -21 and 7 + (-3) = 4. Factoring gives $(x + 7)(x - 3) = 0$, so $x = -7$ or $x = 3$, making the solutions $(-7, 3)$.

Question 17

1. $x^2 + 3x + 10$
2. $x^2 + 5x - 2$
3. $x^2 - 3x + 10$
4. $x^2 + 3x - 10$ (correct answer)

Explanation: To multiply $(x + 5)(x - 2)$ using FOIL: First terms: $x \cdot x = x^2$; Outer terms: $x \cdot (-2) = -2x$; Inner terms: $5 \cdot x = 5x$; Last terms: $5 \cdot (-2) = -10$. Combining these gives $x^2 - 2x + 5x - 10 = x^2 + 3x - 10$.

Question 18

1. $x=1$ only
2. $x=-1$ only
3. $x=1$ and $x=-1$ (correct answer)
4. $x=0$ and $x=1$

Explanation: The equation x² - 1 = 0 is a difference of squares since 1 = 1². Using the pattern a² - b² = (a - b)(a + b), we get x² - 1 = (x - 1)(x + 1) = 0. Setting each factor equal to zero gives x = 1 and x = -1. Choices A and B give only one solution each.

Question 19

1. $(x - 5)(x + 3)$
2. $(x + 5)(x - 3)$ (correct answer)
3. $(x - 15)(x + 1)$
4. $(x + 15)(x - 1)$

Explanation: Factoring quadratics is a fundamental algebra skill tested frequently on the ACT. To factor x² + 2x - 15, you need two numbers that multiply to -15 and add to +2. Those numbers are 5 and -3, giving you (x + 5)(x - 3). You can verify by using FOIL: (x + 5)(x - 3) = x² - 3x + 5x - 15 = x² + 2x - 15 ✓. Strategy tip: If you're unsure, you can always expand each answer choice to see which one matches the original expression.

Question 20

1. $3x^2 - x + 3$
2. $7x^2 - x + 3$
3. $3x^2 + 5x - 5$ (correct answer)
4. $-3x^2 - 5x + 5$

Explanation: Polynomial subtraction requires distributing the negative sign to every term in the subtracted polynomial—this is where most errors occur. (5x² + 2x - 1) - (2x² - 3x + 4) becomes 5x² + 2x - 1 - 2x² + 3x - 4. Combining like terms: (5x² - 2x²) + (2x + 3x) + (-1 - 4) = 3x² + 5x - 5. The most common mistake is forgetting to distribute the negative to the +4, which would give -5 instead of the correct answer. Strategy: Rewrite subtraction as addition of the opposite to avoid sign errors.