ACT Math study guide · Chapter 9
Other Types of Equations
A polynomial equation of degree three or higher is usually solved by factoring. Factoring by grouping pairs terms so that a common factor can be removed from each pair, revealing a shared binomial factor. The factor theorem connects a polynomial's zeros to its factors directly: if x = a is a zero of a polynomial, then (x − a) is a factor, and substituting the known zero into the polynomial and setting the result equal to zero solves for an unknown coefficient.
An exponential equation is solved by rewriting both sides with a common base, then setting the exponents equal to each other. This method applies whenever both sides can be expressed as powers of the same base, and it requires no logarithms at this level.
A radical equation is solved by isolating the radical and squaring both sides. Squaring both sides of an equation can introduce a solution that satisfies the squared equation but not the original equation, so every candidate solution must be checked in the original equation before being accepted.
Where this appears as a model: a quantity decaying by a fixed proportion over time. A medication amount modeled by A(t)=200(0.5)^(t/6) drops to 25 milligrams after 18 hours, found by writing 25 as 200 times a power of one-half and matching exponents.
▸ Factoring higher-degree polynomials
A cubic equation with no constant term factors immediately by removing the common variable factor. A cubic equation with four terms often factors by grouping the first two terms and the last two terms separately.
Worked Example Solve by factoring: x³ − 4x = 0.
Factor out the common variable factor, then factor the remaining difference of squares.
x(x² − 4) = 0
x(x − 2)(x + 2) = 0
x = 0, 2, or −2
▸ Exponential equations and the factor theorem
Rewriting both sides of an exponential equation with the same base converts the equation into a simpler statement about exponents alone, since two powers of the same base are equal only when their exponents are equal.
Worked Example Solve 9^x = 27^(x−1).
Rewrite both sides with base 3.
(3²)^x = (3³)^(x−1)
3^(2x) = 3^(3x−3)
2x = 3x − 3
x = 3
A radical equation that appears alongside an exponential or polynomial equation on this test still follows the same rule: isolate the radical, square both sides, then check every candidate solution against the original equation before reporting a final answer.
▸ Practice
Question 1
- AA. x = 0, 3, −2
- BB. x = 0, −3, 2
- CC. x = 3, −2
- DD. x = 0, 3, 2
Question 2
- FF. x = 4
- GG. x = −1
- HH. x = 7
- JJ. x = 1
Question 3
- AA. x = 2, −3, 3
- BB. x = −2, 3, −3
- CC. x = 2, 3
- DD. x = −2, −3
Question 4
- FF. k = 5
- GG. k = −5
- HH. k = 3
- JJ. k = −3
Question 5
- AA. x = 2
- BB. x = −2
- CC. x = 1
- DD. x = 6
Question 6
- FF. 6
- GG. 32
- HH. 2
- JJ. 68
Question 7
- AA. x = 5, −2, 2
- BB. x = −5, 2, −2
- CC. x = −5, −2, −2
- DD. x = 5, 2, 2
Question 8
- FF. 3
- GG. 5
- HH. 4
- JJ. 2
Question 9
- AA. −2
- BB. 4
- CC. 2
- DD. −4
Question 10
- FF. 1
- GG. 1.5
- HH. 2
- JJ. 0.5
Question 11
- AA. x = 3, −3
- BB. x = −3, 3, 0
- CC. x = 0, 3
- DD. x = 0, −3
Question 12
- FF. x = 0 only
- GG. x = 5 or x = 0
- HH. x = 5 only
- JJ. no solution
Question 13
- AA. −1
- BB. 1
- CC. 7
- DD. −7
Question 14
- FF. 25
- GG. 26
- HH. 24
- JJ. 27
Question 15
- AA. x = −3, 5, 2
- BB. x = 3, −5, 2
- CC. x = −3, −5, 2
- DD. x = 3, 5, 2
Question 16
- FF. 1
- GG. 3
- HH. 2
- JJ. 0
Question 17
- AA. −2
- BB. 4
- CC. −4
- DD. 2
Question 18
- FF. 6
- GG. 2
- HH. 1
- JJ. 3
Question 19
- AA. 0 only
- BB. 0 or 7
- CC. 7 only
- DD. no solution
Question 20
- FF. 0, −4, 4
- GG. −4, 4
- HH. 4, 0
- JJ. −4, 0
▸ Answers and Explanations
Show the answers and explanations
1: A. Factor out x: x(x² − x − 6) = x(x − 3)(x + 2) = 0, giving x = 0, 3, or −2. Choice B reverses the sign of the two nonzero roots.
2: J. Rewrite 16 as a power of 2: 2^(x+3) = 2⁴. Set the exponents equal: x + 3 = 4, giving x = 1. Choice F reports the exponent value 4 instead of solving for x.
3: B. Group: x²(x + 2) − 9(x + 2) = (x + 2)(x² − 9) = (x + 2)(x − 3)(x + 3) = 0, giving x = −2, 3, or −3. Choice A reverses the sign of all three roots.
4: F. Substitute x = −1 and set the result equal to 0: −1 + 4 − k + 2 = 0, so 5 − k = 0, giving k = 5. Choice G reports the sign-reversed value, k = −5.
5: A. Rewrite both sides with base 3: (3³)^x = (3²)^(x+1), so 3^(3x) = 3^(2x+2). Set the exponents equal: 3x = 2x + 2, giving x = 2. Choice B reports the sign-reversed value, x = −2.
6: G. Setting the exponents equal: 2x = x + 6, so x = 6. Then y = x² − 4 = 36 − 4 = 32. Choice C reports x itself instead of the requested value of y.
7: B. Group: x²(x + 5) − 4(x + 5) = (x + 5)(x² − 4) = (x + 5)(x − 2)(x + 2) = 0, giving x = −5, 2, or −2. Choice A reverses the sign of all three roots.
8: H. Rewrite 64 as a power of 4: 4^(x−1) = 4³. Set the exponents equal: x − 1 = 3, giving x = 4. Choice F reports the exponent value 3 instead of solving for x.
9: C. Substitute x = 3 and set the result equal to 0: 27 − 18 + 3k − 15 = 0, so 3k − 6 = 0, giving k = 2. Choice A reports the sign-reversed value, k = −2.
10: G. Rewrite 27 as a power of 3: 3^(2x) = 3³. Set the exponents equal: 2x = 3, giving x = 1.5. Choice F subtracts instead of dividing, treating 2x = 3 as x = 3 − 2.
11: B. Factor out x: x(x² − 9) = x(x − 3)(x + 3) = 0, giving x = 0, 3, or −3. Choice A omits the root x = 0.
12: H. Squaring gives 3x + 1 = x² − 2x + 1, so x² − 5x = 0, giving x = 0 or x = 5. Checking x = 0: √1 = 1, but 0 − 1 = −1; 1 ≠ −1, so x = 0 is extraneous. Checking x = 5: √16 = 4, and 5 − 1 = 4, which matches. Choice G reports both candidate solutions without checking for the extraneous root.
13: D. Rewrite 1/16 as a power of 2: 2^(x+3) = 2⁻⁴. Set the exponents equal: x + 3 = −4, giving x = −7. Choice C reports the sign-reversed value, x = 7.
14: G. Rewrite 125 as a power of 5: 5^(x−2) = 5³. Set the exponents equal: x − 2 = 3, giving x = 5. The requested value is x² + 1, not x² alone: 25 + 1 = 26. Choice F reports x² without adding 1.
15: A. Testing small integer values shows x = 2 is a root; dividing gives x² − 2x − 15 = (x − 5)(x + 3). The full factorization is (x − 2)(x − 5)(x + 3) = 0, giving x = 2, 5, or −3. Choice D reverses the sign of the root x = −3 only.
16: F. Rewrite 125 as a power of 5: 5^(x+2) = 5³. Set the exponents equal: x + 2 = 3, giving x = 1. Choice G reports the exponent value 3 instead of solving for x.
17: D. Substitute x = 2 and set the result equal to 0: 8 + 4k − 10 − 6 = 0, so 4k − 8 = 0, giving k = 2. Choice A reports the sign-reversed value, k = −2.
18: J. Rewrite 8 as a power of 2: (2³)^x = 2^(x+6), so 2^(3x) = 2^(x+6). Set the exponents equal: 3x = x + 6, so 2x = 6, giving x = 3. Choice G divides by 3 instead of 2 when isolating x, giving 6/3 = 2 instead of 6/2 = 3.
19: C. Squaring gives 3x + 4 = x² − 4x + 4, so x² − 7x = 0, giving x = 0 or x = 7. Checking x = 0: √4 = 2, but 0 − 2 = −2; 2 ≠ −2, so x = 0 is extraneous. Checking x = 7: √25 = 5, and 7 − 2 = 5, which matches. Choice B reports both candidate solutions without checking for the extraneous root.
20: F. Factor out x: x(x² − 16) = x(x − 4)(x + 4) = 0, giving x = 0, 4, or −4. Choice J omits the root x = 4.