Math 2 Quiz: Strategic Distributive Property
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Strategic Distributive PropertyQuestion 1 of 20

Which expression shows the correct strategic factoring of 2x3+6x2−4x−122x^3 + 6x^2 - 4x - 12 using the distributive property in reverse?

(2x2−4)(x+3)(2x^2 - 4)(x + 3)
2x(x2+3x−2)−122x(x^2 + 3x - 2) - 12
2(x+3)(x2−2)2(x + 3)(x^2 - 2)
2(x3+3x2−2x−6)2(x^3 + 3x^2 - 2x - 6)
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Math 2 Quiz

Math 2 Quiz: Strategic Distributive Property

Practice Strategic Distributive Property in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Strategic Distributive Property, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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.

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Question 1

Which expression shows the correct strategic factoring of 2x3+6x2−4x−122x^3 + 6x^2 - 4x - 12 using the distributive property in reverse?

  1. (2x2−4)(x+3)(2x^2 - 4)(x + 3)
  2. 2x(x2+3x−2)−122x(x^2 + 3x - 2) - 12
  3. 2(x+3)(x2−2)2(x + 3)(x^2 - 2) (correct answer)
  4. 2(x3+3x2−2x−6)2(x^3 + 3x^2 - 2x - 6)
Explanation: When you encounter a polynomial factoring problem asking for "strategic factoring using the distributive property in reverse," you're looking for a complete factorization that reveals the polynomial's structure through grouping. To factor 2x3+6x2−4x−122x^3 + 6x^2 - 4x - 12, start by looking for a greatest common factor. You can factor out 2 from all terms: 2(x3+3x2−2x−6)2(x^3 + 3x^2 - 2x - 6). Now use grouping on the expression inside the parentheses. Group the first two terms and last two terms: x3+3x2−2x−6=x2(x+3)−2(x+3)x^3 + 3x^2 - 2x - 6 = x^2(x + 3) - 2(x + 3). Notice both groups contain the factor (x+3)(x + 3), so you can factor it out: (x+3)(x2−2)(x + 3)(x^2 - 2). Including the initial factor of 2, the complete factorization is 2(x+3)(x2−2)2(x + 3)(x^2 - 2), which is choice C. Choice A gives (2x2−4)(x+3)(2x^2 - 4)(x + 3), but when expanded this yields 2x3+6x2−4x−122x^3 + 6x^2 - 4x - 12, which matches the original. However, this isn't fully factored since 2x2−4=2(x2−2)2x^2 - 4 = 2(x^2 - 2). Choice B shows 2x(x2+3x−2)−122x(x^2 + 3x - 2) - 12, which isn't even in factored form—it's still a sum of terms. Choice D gives 2(x3+3x2−2x−6)2(x^3 + 3x^2 - 2x - 6), which only shows the first step of factoring out the GCF but doesn't complete the grouping process. Remember: complete factoring means breaking down until no further common factors exist. Always check if your factored form can be simplified further.

Question 2

If 3(x−k)+2(x+k)=5x−73(x - k) + 2(x + k) = 5x - 7 for all values of xx, what is the value of kk?

  1. k=75k = \frac{7}{5}
  2. k=−7k = -7
  3. k=7k = 7 (correct answer)
  4. k=−75k = -\frac{7}{5}
Explanation: When you encounter an equation that's supposed to be true "for all values of x," you're dealing with an algebraic identity. This means the left and right sides must be equivalent expressions, so their coefficients must match exactly. Start by expanding the left side: 3(x−k)+2(x+k)=3x−3k+2x+2k=5x−k3(x - k) + 2(x + k) = 3x - 3k + 2x + 2k = 5x - k. Now you have 5x−k=5x−75x - k = 5x - 7. Since this equation must hold for all values of xx, the coefficients of like terms on both sides must be equal. The coefficients of xx already match (both are 5), so you need the constant terms to match as well. This gives you −k=−7-k = -7, which means k=7k = 7. Looking at the wrong answers: Choice A (k=75k = \frac{7}{5}) might result from incorrectly dividing by the coefficient of xx, but you don't divide when matching coefficients in an identity. Choice B (k=−7k = -7) comes from the common error of setting k=−7k = -7 instead of −k=−7-k = -7. Choice D (k=−75k = -\frac{7}{5}) combines both mistakes—the sign error and the unnecessary division. You can verify: substituting k=7k = 7 gives 3(x−7)+2(x+7)=3x−21+2x+14=5x−73(x - 7) + 2(x + 7) = 3x - 21 + 2x + 14 = 5x - 7 ✓ Study tip: When an equation holds "for all values" of a variable, expand both sides completely and match coefficients of like terms. Don't solve for the variable—solve for the parameter that makes the expressions identical.

Question 3

Which expression is equivalent to 3x(2x−5)−2(x2−3x+1)3x(2x - 5) - 2(x^2 - 3x + 1) after applying the distributive property and combining like terms?

  1. 4x2−9x−24x^2 - 9x - 2 (correct answer)
  2. 4x2−21x−24x^2 - 21x - 2
  3. 6x2−21x−26x^2 - 21x - 2
  4. 6x2−9x−26x^2 - 9x - 2
Explanation: First distribute: 3x(2x−5)=6x2−15x3x(2x - 5) = 6x^2 - 15x and −2(x2−3x+1)=−2x2+6x−2-2(x^2 - 3x + 1) = -2x^2 + 6x - 2. Then combine: (6x2−15x)+(−2x2+6x−2)=4x2−9x−2(6x^2 - 15x) + (-2x^2 + 6x - 2) = 4x^2 - 9x - 2. Choice B incorrectly combines −15x+6x=−21x-15x + 6x = -21x. Choice C forgets to distribute the negative to x2x^2. Choice D makes both the error in B and fails to distribute properly.

Question 4

To factor 6x2+9x−156x^2 + 9x - 15, which represents the most strategic first step using the distributive property in reverse?

  1. Factor out 3: 3(2x2+3x−5)3(2x^2 + 3x - 5), then factor the quadratic if possible (correct answer)
  2. Look for two numbers that multiply to 6(−15)=−906(-15) = -90 and add to 9
  3. Rewrite as 6x2+15x−6x−156x^2 + 15x - 6x - 15 and factor by grouping
  4. Use the quadratic formula since this doesn't factor nicely
Explanation: The GCF of 6, 9, and 15 is 3, so factoring out 3 first gives 3(2x2+3x−5)3(2x^2 + 3x - 5). The quadratic 2x2+3x−52x^2 + 3x - 5 factors as (2x+5)(x−1)(2x + 5)(x - 1), giving the complete factorization 3(2x+5)(x−1)3(2x + 5)(x - 1). Choice B is a valid approach but more complex. Choice C uses incorrect splitting. Choice D is wrong since the expression does factor.

Question 5

To simplify 2(3x+1)−4(x−2)+3(2x+5)2(3x + 1) - 4(x - 2) + 3(2x + 5), after applying the distributive property, what is the coefficient of xx?

  1. 8 (correct answer)
  2. 2
  3. 10
  4. 6
Explanation: Distributing: 2(3x+1)=6x+22(3x + 1) = 6x + 2, −4(x−2)=−4x+8-4(x - 2) = -4x + 8, and 3(2x+5)=6x+153(2x + 5) = 6x + 15. The xx terms are 6x−4x+6x=8x6x - 4x + 6x = 8x, so the coefficient is 8. Choice B results from 6x−4x=2x6x - 4x = 2x (forgetting the third term). Choice C comes from 6x+4x=10x6x + 4x = 10x (sign error on the second term). Choice D comes from 6x+6x=12x6x + 6x = 12x then incorrectly subtracting instead of adding.

Question 6

Which approach using the distributive property would be most strategic for expanding (2x−3)(x+4)(x−1)(2x - 3)(x + 4)(x - 1)?

  1. First multiply (2x−3)(x+4)(2x - 3)(x + 4), then multiply the result by (x−1)(x - 1)
  2. First multiply (x+4)(x−1)(x + 4)(x - 1), then multiply the result by (2x−3)(2x - 3) (correct answer)
  3. Distribute (2x−3)(2x - 3) to both (x+4)(x + 4) and (x−1)(x - 1) simultaneously
  4. Use the distributive property to rewrite as 2x(x+4)(x−1)−3(x+4)(x−1)2x(x + 4)(x - 1) - 3(x + 4)(x - 1)
Explanation: Multiplying (x+4)(x−1)(x + 4)(x - 1) first gives (x2+3x−4)(x^2 + 3x - 4), which is simpler to multiply by (2x−3)(2x - 3) than the result from choice A. Choice A works but creates a more complex intermediate step. Choice C is not a valid application of the distributive property. Choice D is technically correct but unnecessarily complicates the process by separating the first factor.

Question 7

Before solving 5(2x−3)+7=4(x+1)−2(3x−5)5(2x - 3) + 7 = 4(x + 1) - 2(3x - 5), Maria wants to verify her distributive property work. After distributing, which equation should she have?

  1. 10x−15+7=4x+4+6x−1010x - 15 + 7 = 4x + 4 + 6x - 10
  2. 10x−15+7=4x+4−6x+1010x - 15 + 7 = 4x + 4 - 6x + 10 (correct answer)
  3. 10x+15+7=4x−4−6x+1010x + 15 + 7 = 4x - 4 - 6x + 10
  4. 10x−8=4x+4−6x+1010x - 8 = 4x + 4 - 6x + 10
Explanation: When you encounter an equation with multiple terms in parentheses, the distributive property is your first step. You need to multiply each term inside the parentheses by the coefficient outside, paying careful attention to positive and negative signs. Let's work through each part of 5(2x−3)+7=4(x+1)−2(3x−5)5(2x - 3) + 7 = 4(x + 1) - 2(3x - 5): On the left side: 5(2x−3)=5⋅2x+5⋅(−3)=10x−155(2x - 3) = 5 \cdot 2x + 5 \cdot (-3) = 10x - 15. The +7+7 stays as is. On the right side: 4(x+1)=4x+44(x + 1) = 4x + 4, and −2(3x−5)=−2⋅3x+(−2)⋅(−5)=−6x+10-2(3x - 5) = -2 \cdot 3x + (-2) \cdot (-5) = -6x + 10. Note that multiplying two negatives gives a positive. This gives us: 10x−15+7=4x+4−6x+1010x - 15 + 7 = 4x + 4 - 6x + 10, which matches answer choice B. Answer choice A incorrectly shows +6x+6x instead of −6x-6x in the final term, missing that −2×3x=−6x-2 \times 3x = -6x. Answer choice C makes two errors: it shows +15+15 instead of −15-15 (forgetting that 5×(−3)=−155 \times (-3) = -15) and −4-4 instead of +4+4 (forgetting that 4×1=+44 \times 1 = +4). Answer choice D appears to have already combined like terms on the left side, which isn't what the question asks for—it wants to see the equation immediately after distribution, before any combining. Remember: when distributing with negative coefficients, track your signs carefully. Each multiplication follows the same rules you learned for multiplying signed numbers.

Question 8

When preparing to solve 12(4x−6)=34(2x+8)−5\frac{1}{2}(4x - 6) = \frac{3}{4}(2x + 8) - 5, which form would be most convenient to work with after applying the distributive property?

  1. 2x−3=3x2+6−52x - 3 = \frac{3x}{2} + 6 - 5
  2. 2x−3=3x+122−52x - 3 = \frac{3x + 12}{2} - 5
  3. 4x−6=3x+12−104x - 6 = 3x + 12 - 10
  4. 8x−12=6x+24−208x - 12 = 6x + 24 - 20 (correct answer)
Explanation: The most convenient approach is to clear fractions by multiplying everything by 4, then distribute: 4⋅12(4x−6)=4⋅34(2x+8)−4⋅54 \cdot \frac{1}{2}(4x - 6) = 4 \cdot \frac{3}{4}(2x + 8) - 4 \cdot 5 gives 2(4x−6)=3(2x+8)−202(4x - 6) = 3(2x + 8) - 20, which becomes 8x−12=6x+24−208x - 12 = 6x + 24 - 20. This eliminates fractions entirely. Choices A and B retain fractions unnecessarily. Choice C only partially clears fractions.

Question 9

To solve 2(3x−1)−4(x−2)=3(x+1)2(3x - 1) - 4(x - 2) = 3(x + 1), Alex distributes and gets 6x−2−4x+8=3x+36x - 2 - 4x + 8 = 3x + 3. What should Alex do next to most efficiently continue solving?

  1. Move all terms with xx to the left side before combining like terms
  2. Combine like terms on the left to get 2x+6=3x+32x + 6 = 3x + 3, then solve (correct answer)
  3. Factor the left side before moving terms to isolate xx
  4. Substitute specific values of xx to check which ones satisfy the equation
Explanation: When solving linear equations, the most efficient approach after distributing is to simplify each side by combining like terms before moving variables around. This reduces complexity and minimizes the chance of algebraic errors. Looking at Alex's work: 6x−2−4x+8=3x+36x - 2 - 4x + 8 = 3x + 3, the left side has like terms that can be combined. The xx terms are 6x−4x=2x6x - 4x = 2x, and the constants are −2+8=6-2 + 8 = 6. This gives us 2x+6=3x+32x + 6 = 3x + 3, which is much cleaner to work with. From here, you can easily subtract 2x2x from both sides to get 6=x+36 = x + 3, then subtract 3 to find x=3x = 3. Choice A suggests moving variables first, but this creates unnecessary complexity. You'd have 6x−4x−3x=3+2−86x - 4x - 3x = 3 + 2 - 8, which requires you to track more terms and increases error risk. Choice C mentions factoring, but there's nothing meaningful to factor on the left side. The terms 6x−2−4x+86x - 2 - 4x + 8 don't share common factors that would simplify the equation. Choice D suggests substitution, but this is a guess-and-check approach that's inefficient and doesn't demonstrate algebraic understanding. It also doesn't guarantee you'll find the solution systematically. Remember: always combine like terms on each side of an equation before moving terms across the equals sign. This "clean up first" strategy keeps your work organized and reduces computational errors.

Question 10

When expanding (x+3)2−(x−2)2(x + 3)^2 - (x - 2)^2, which strategic application of the distributive property leads to the simplest form?

  1. Expand each square separately: (x2+6x+9)−(x2−4x+4)=10x+5(x^2 + 6x + 9) - (x^2 - 4x + 4) = 10x + 5
  2. Use difference of squares: [(x+3)+(x−2)][(x+3)−(x−2)]=(2x+1)(5)=10x+5[(x + 3) + (x - 2)][(x + 3) - (x - 2)] = (2x + 1)(5) = 10x + 5 (correct answer)
  3. Factor first: (x+3+x−2)(x+3−x+2)=(2x+1)(5)=10x+5(x + 3 + x - 2)(x + 3 - x + 2) = (2x + 1)(5) = 10x + 5
  4. Distribute the negative: (x+3)2+(−1)(x−2)2(x + 3)^2 + (-1)(x - 2)^2, then expand both terms
Explanation: Recognizing this as a difference of squares a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b) where a=x+3a = x + 3 and b=x−2b = x - 2 is most efficient. This gives (2x+1)(5)=10x+5(2x + 1)(5) = 10x + 5 directly. Choice A works but requires more computation. Choice C has the right idea but incorrect notation. Choice D doesn't simplify the process.

Question 11

When solving 4(x−3)−2(x+1)=3(x−2)+54(x - 3) - 2(x + 1) = 3(x - 2) + 5, which equation correctly results after applying the distributive property to both sides?

  1. 4x−3−2x−1=3x−2+54x - 3 - 2x - 1 = 3x - 2 + 5
  2. 4x−12−2x+2=3x−6+54x - 12 - 2x + 2 = 3x - 6 + 5
  3. 4x−12−2x−2=3x−6+54x - 12 - 2x - 2 = 3x - 6 + 5 (correct answer)
  4. 4x−12−2x−2=3x+6+54x - 12 - 2x - 2 = 3x + 6 + 5
Explanation: When you encounter equations with parentheses, the distributive property is your first step. You need to multiply each term outside the parentheses by every term inside, paying careful attention to signs. Let's work through the left side: 4(x−3)−2(x+1)4(x - 3) - 2(x + 1). For the first term, 4(x−3)=4x−124(x - 3) = 4x - 12. For the second term, −2(x+1)=−2x−2-2(x + 1) = -2x - 2 (notice the negative sign distributes to both terms). This gives us 4x−12−2x−24x - 12 - 2x - 2. For the right side: 3(x−2)+53(x - 2) + 5. Here, 3(x−2)=3x−63(x - 2) = 3x - 6, so we get 3x−6+53x - 6 + 5. The complete equation after distribution is 4x−12−2x−2=3x−6+54x - 12 - 2x - 2 = 3x - 6 + 5, which is answer choice C. Looking at the wrong answers: Choice A incorrectly shows 4(x−3)4(x - 3) as 4x−34x - 3 instead of 4x−124x - 12, forgetting to multiply 4 by both terms. Choice B shows −2(x+1)-2(x + 1) as −2x+2-2x + 2, incorrectly making the last term positive when it should be negative. Choice D shows 3(x−2)3(x - 2) as 3x+63x + 6, changing the minus sign to plus when distributing. Remember: when distributing a negative sign, it affects every term inside the parentheses. Double-check your signs by working slowly through each multiplication, especially when negative numbers are involved.

Question 12

To factor x3−4x2+4x−16x^3 - 4x^2 + 4x - 16 by grouping, which initial application of the distributive property in reverse is most strategic?

  1. Group as x3−4x2+(4x−16)x^3 - 4x^2 + (4x - 16), then factor each part separately
  2. Group as x(x2−4x+4)−16x(x^2 - 4x + 4) - 16, then factor the quadratic
  3. Factor out xx first: x(x2−4x+4)−16x(x^2 - 4x + 4) - 16
  4. Group as x2(x−4)+4(x−4)x^2(x - 4) + 4(x - 4), then factor out (x−4)(x - 4) (correct answer)
Explanation: When you encounter a four-term polynomial like x3−4x2+4x−16x^3 - 4x^2 + 4x - 16, factoring by grouping means strategically pairing terms so each group shares a common factor, ultimately revealing a shared binomial factor. The key insight is to group the first two terms and last two terms: x2(x−4)+4(x−4)x^2(x - 4) + 4(x - 4). From x3−4x2x^3 - 4x^2, you factor out x2x^2 to get x2(x−4)x^2(x - 4). From 4x−164x - 16, you factor out 44 to get 4(x−4)4(x - 4). Notice both groups contain the factor (x−4)(x - 4), so you can factor it out: (x−4)(x2+4)(x - 4)(x^2 + 4). This is exactly what option D describes. Option A groups as (x3−4x2)+(4x−16)(x^3 - 4x^2) + (4x - 16) but doesn't specify the crucial step of ensuring both groups yield the same binomial factor. Option B starts by factoring xx from only the first three terms, which doesn't create the balanced grouping needed for this method—you'd get x(x2−4x+4)−16x(x^2 - 4x + 4) - 16, where the quadratic doesn't factor nicely with the remaining −16-16. Option C is essentially the same flawed approach as B, prematurely factoring xx instead of grouping strategically. The most effective strategy for factoring by grouping is to look for ways to pair terms so that each pair, when factored, reveals the same binomial factor. Always check that your grouping will actually lead to a common factor before proceeding.

Question 13

Which expression demonstrates the most strategic use of the distributive property to simplify 6x+93−4x−82\frac{6x + 9}{3} - \frac{4x - 8}{2}?

  1. 6(x+1.5)3−4(x−2)2=2(x+1.5)−2(x−2)=7\frac{6(x + 1.5)}{3} - \frac{4(x - 2)}{2} = 2(x + 1.5) - 2(x - 2) = 7
  2. 6x3+93−4x2+82=2x+3−2x+4=7\frac{6x}{3} + \frac{9}{3} - \frac{4x}{2} + \frac{8}{2} = 2x + 3 - 2x + 4 = 7
  3. 2x+3−2x+4=72x + 3 - 2x + 4 = 7 by direct division of each term
  4. 3(2x+3)3−2(2x−4)2=(2x+3)−(2x−4)=7\frac{3(2x + 3)}{3} - \frac{2(2x - 4)}{2} = (2x + 3) - (2x - 4) = 7 (correct answer)
Explanation: When you encounter algebraic expressions with fractions that need simplification, look for opportunities to factor out common terms from numerators before dividing. This strategic approach often reveals the most elegant path to the solution. The most effective method here is shown in choice D: 3(2x+3)3−2(2x−4)2=(2x+3)−(2x−4)=7\frac{3(2x + 3)}{3} - \frac{2(2x - 4)}{2} = (2x + 3) - (2x - 4) = 7. This approach factors the greatest common factor from each numerator first. In the first fraction, factoring out 3 gives us 3(2x+3)3(2x + 3), which then cancels perfectly with the denominator 3. Similarly, factoring out 2 from the second numerator gives us 2(2x−4)2(2x - 4), which cancels with the denominator 2. This leaves us with (2x+3)−(2x−4)=2x+3−2x+4=7(2x + 3) - (2x - 4) = 2x + 3 - 2x + 4 = 7. Choice A incorrectly factors the first term as 6(x+1.5)6(x + 1.5) instead of recognizing that 6x+9=3(2x+3)6x + 9 = 3(2x + 3). Choice B splits each fraction into separate terms, which works but doesn't demonstrate strategic use of the distributive property as requested. Choice C mentions "direct division" but doesn't show the factoring step that makes the distributive property evident. Remember: when simplifying algebraic fractions, always look for common factors in the numerator that match the denominator. Factoring first often leads to clean cancellations and reveals the underlying structure of the expression more clearly than term-by-term division.

Question 14

To solve the equation 2x+34+x−16=5\frac{2x + 3}{4} + \frac{x - 1}{6} = 5, which expression represents the most strategic first step using the distributive property?

  1. Multiply the entire equation by 12 to get 3(2x+3)+2(x−1)=603(2x + 3) + 2(x - 1) = 60 (correct answer)
  2. Multiply the entire equation by 24 to get 6(2x+3)+4(x−1)=1206(2x + 3) + 4(x - 1) = 120
  3. Distribute within fractions first: 2x4+34+x6−16=5\frac{2x}{4} + \frac{3}{4} + \frac{x}{6} - \frac{1}{6} = 5
  4. Multiply the entire equation by 10 to get 10(2x+3)4+10(x−1)6=50\frac{10(2x + 3)}{4} + \frac{10(x - 1)}{6} = 50
Explanation: The LCD of 4 and 6 is 12, so multiplying by 12 clears all denominators efficiently, giving 3(2x+3)+2(x−1)=603(2x + 3) + 2(x - 1) = 60. Choice B uses 24 which works but creates unnecessarily large numbers. Choice C distributes prematurely without clearing denominators. Choice D multiplies by 10, which doesn't clear the denominators completely.

Question 15

To solve x+13−x−24=56\frac{x + 1}{3} - \frac{x - 2}{4} = \frac{5}{6}, after clearing denominators by multiplying by 12, what equation results from applying the distributive property?

  1. 3(x+1)−4(x−2)=103(x + 1) - 4(x - 2) = 10
  2. 4(x+1)−3(x−2)=604(x + 1) - 3(x - 2) = 60
  3. 4(x+1)−3(x−2)=104(x + 1) - 3(x - 2) = 10 (correct answer)
  4. 12(x+1)−12(x−2)=512(x + 1) - 12(x - 2) = 5
Explanation: When solving equations with fractions, you'll often need to clear denominators by multiplying through by the least common multiple (LCM). This transforms the equation into one without fractions, making it easier to solve. To clear the denominators in x+13−x−24=56\frac{x + 1}{3} - \frac{x - 2}{4} = \frac{5}{6}, you multiply every term by 12 (the LCM of 3, 4, and 6). This gives you: 12⋅x+13−12⋅x−24=12⋅5612 \cdot \frac{x + 1}{3} - 12 \cdot \frac{x - 2}{4} = 12 \cdot \frac{5}{6} Simplifying each term: 123=4\frac{12}{3} = 4, 124=3\frac{12}{4} = 3, and 12⋅56=10\frac{12 \cdot 5}{6} = 10 This results in: 4(x+1)−3(x−2)=104(x + 1) - 3(x - 2) = 10 Looking at the wrong answers: Choice A incorrectly uses coefficients 3 and 4 instead of 4 and 3 – this comes from confusing which denominator cancels with which part of 12. Choice B has the right coefficients but calculates the right side as 60 instead of 10, likely from multiplying 12×5=6012 \times 5 = 60 without dividing by 6. Choice D keeps all the 12s without simplifying the fractions, missing the key step of reducing 123\frac{12}{3} and 124\frac{12}{4}. The correct answer is C: 4(x+1)−3(x−2)=104(x + 1) - 3(x - 2) = 10. Study tip: When clearing denominators, always double-check that you've simplified each fraction completely. Write out 123=4\frac{12}{3} = 4 and 124=3\frac{12}{4} = 3 explicitly to avoid coefficient mix-ups.

Question 16

When solving 5(x+2)=3(x−4)+225(x + 2) = 3(x - 4) + 22, which step correctly applies the distributive property and sets up for the most efficient solution?

  1. 5x+10=3x+12+225x + 10 = 3x + 12 + 22, then 5x+10=3x+345x + 10 = 3x + 34
  2. 5x+10=3x−12+225x + 10 = 3x - 12 + 22, then 5x+10=3x−345x + 10 = 3x - 34
  3. 5x+2=3x−4+225x + 2 = 3x - 4 + 22, then 5x+2=3x+185x + 2 = 3x + 18
  4. 5x+10=3x−12+225x + 10 = 3x - 12 + 22, then 5x+10=3x+105x + 10 = 3x + 10 (correct answer)
Explanation: When you encounter an equation with parentheses on both sides, the distributive property is your first tool. You need to multiply each term inside the parentheses by the coefficient outside, paying careful attention to signs. Let's work through 5(x+2)=3(x−4)+225(x + 2) = 3(x - 4) + 22 step by step. On the left side, 5(x+2)=5x+105(x + 2) = 5x + 10. On the right side, 3(x−4)=3x−123(x - 4) = 3x - 12 (notice the negative sign is distributed). This gives us 5x+10=3x−12+225x + 10 = 3x - 12 + 22. Combining the constants on the right: −12+22=10-12 + 22 = 10, so we get 5x+10=3x+105x + 10 = 3x + 10. Choice D correctly shows this process: 5x+10=3x−12+225x + 10 = 3x - 12 + 22, then 5x+10=3x+105x + 10 = 3x + 10. Choice A makes a sign error when distributing 3(x−4)3(x - 4), incorrectly getting 3x+123x + 12 instead of 3x−123x - 12. This is a common mistake when students forget that the negative sign must be distributed to each term. Choice B correctly applies the distributive property initially but makes an error when combining constants: −12+22=10-12 + 22 = 10, not −34-34. Choice C fails to distribute properly on the left side, writing 5x+25x + 2 instead of 5x+105x + 10. This happens when students add the coefficient to the constant instead of multiplying. Study tip: When distributing, always multiply the outside number by every term inside the parentheses, and be extra careful with negative signs. Double-check your arithmetic when combining like terms—it's where most errors occur.

Question 17

When factoring 12x3+18x2−30x12x^3 + 18x^2 - 30x, what is the most efficient first application of the distributive property in reverse?

  1. Factor out 6x6x to get 6x(2x2+3x−5)6x(2x^2 + 3x - 5) (correct answer)
  2. Factor out 3x3x to get 3x(4x2+6x−10)3x(4x^2 + 6x - 10)
  3. Factor out 2x2x to get 2x(6x2+9x−15)2x(6x^2 + 9x - 15)
  4. Factor out xx to get x(12x2+18x−30)x(12x^2 + 18x - 30)
Explanation: The GCF of 12, 18, and 30 is 6, and all terms contain at least one factor of xx, so 6x6x is the greatest common factor. Factoring out 6x6x gives the simplest remaining expression. Choice B factors out only 3x3x (not the complete GCF). Choice C factors out only 2x2x. Choice D factors out only xx, leaving larger coefficients that could be simplified further.

Question 18

To solve the equation 4(2x−3)=5x−(3x−8)4(2x - 3) = 5x - (3x - 8), Maya first applies the distributive property to both sides. What equation does she obtain?

  1. 8x−12=2x+88x - 12 = 2x + 8 (correct answer)
  2. 8x−12=5x−3x−88x - 12 = 5x - 3x - 8
  3. 8x−12=8x−88x - 12 = 8x - 8
  4. 8x−12=5x+3x+88x - 12 = 5x + 3x + 8
Explanation: Applying distributive property: 4(2x−3)=8x−124(2x - 3) = 8x - 12 and 5x−(3x−8)=5x−3x+8=2x+85x - (3x - 8) = 5x - 3x + 8 = 2x + 8. Choice B shows the right side before combining like terms. Choice C incorrectly treats 5x−(3x−8)5x - (3x - 8) as 5x+3x−85x + 3x - 8. Choice D incorrectly distributes the negative sign as 5x+3x+85x + 3x + 8.

Question 19

If a(x+2)−b(x−1)=7x+5a(x + 2) - b(x - 1) = 7x + 5 is an identity true for all values of xx, what are the values of aa and bb?

  1. a=4,b=−3a = 4, b = -3 (correct answer)
  2. a=3,b=−4a = 3, b = -4
  3. a=5,b=2a = 5, b = 2
  4. a=2,b=5a = 2, b = 5
Explanation: Expanding the left side: a(x+2)−b(x−1)=ax+2a−bx+b=(a−b)x+(2a+b)a(x + 2) - b(x - 1) = ax + 2a - bx + b = (a - b)x + (2a + b). For this to equal 7x+57x + 5, we need a−b=7a - b = 7 and 2a+b=52a + b = 5. Solving: From the first equation, b=a−7b = a - 7. Substituting: 2a+(a−7)=52a + (a - 7) = 5, so 3a=123a = 12 and a=4a = 4. Then b=4−7=−3b = 4 - 7 = -3. Choices B, C, and D represent common algebraic errors in solving this system.

Question 20

Which strategy would be most efficient for solving 7(x−4)+3(2x+1)=5(x+2)−187(x - 4) + 3(2x + 1) = 5(x + 2) - 18?

  1. Divide both sides by the sum of all coefficients before applying distributive property
  2. Move all parenthetical expressions to one side before distributing to minimize computation
  3. Factor out the greatest common factor from both sides before distributing
  4. Distribute all terms first, then collect like terms and solve the resulting linear equation (correct answer)
Explanation: When solving linear equations with multiple parenthetical expressions, your goal is to transform the equation into its simplest form systematically. The most straightforward approach is to apply the distributive property first, then combine like terms. Starting with 7(x−4)+3(2x+1)=5(x+2)−187(x - 4) + 3(2x + 1) = 5(x + 2) - 18, distributing gives you 7x−28+6x+3=5x+10−187x - 28 + 6x + 3 = 5x + 10 - 18. Then combine like terms: 13x−25=5x−813x - 25 = 5x - 8. Finally, solve: 8x=178x = 17, so x=178x = \frac{17}{8}. This direct method requires no complex mental gymnastics and follows the standard order of operations. Option A is problematic because there's no meaningful "sum of all coefficients" to divide by, and dividing before distributing would create fractions unnecessarily. Option B suggests moving parenthetical expressions to one side, but this actually complicates the algebra since you'd still need to distribute eventually, and rearranging first adds extra steps. Option C recommends factoring out a greatest common factor, but there's no common factor among all terms on either side that would simplify the equation meaningfully. Option D correctly identifies the most efficient path: distribute first, then solve the resulting linear equation systematically. Remember that with multi-step equations involving parentheses, the distribute-then-simplify approach is almost always your most reliable strategy. Don't overcomplicate by looking for shortcuts that aren't there—sometimes the straightforward method is genuinely the best.