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Quizzes > Mathematics > Algebra

Factoring with the AC Method Quiz: Ace Your Algebra Skills

Think you can master AC method quadratic factoring? Dive in now!

Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art shapes and algebraic symbols on dark blue background for AC method quadratic factoring quiz

Ready to master factoring with the AC method? Our free quiz challenge puts your algebra skills to the test and helps you conquer AC method quadratic factoring in style. Whether you're learning how to solve quadratic with ac method or reinforcing core concepts, this fun quiz offers a series of ac method practice problems designed for every level. Jump in, explore our detailed guide on factoring ax^2 + bx + c , and tackle engaging factorization questions that sharpen your technique. By the end, you'll confidently factor quadratic equations and unlock new math prowess. Are you up for the challenge? Start now!

Factor the quadratic expression x^2 + 5x + 6 using the AC method.
(x+3)(x+3)
(x-2)(x-3)
(x+1)(x+6)
(x+2)(x+3)
In the expression x^2 + 5x + 6, a = 1, b = 5, and c = 6. Multiply a and c (1*6 = 6) and find two numbers that multiply to 6 and add to b = 5, which are 2 and 3. Rewrite the middle term as 2x + 3x, factor by grouping, and you get (x+2)(x+3). This procedure is a direct application of the AC method. Learn more about the AC method
Factor x^2 - x - 6 using the AC method.
(x-2)(x+3)
(x-3)(x+2)
(x+3)(x-2)
(x-6)(x+1)
Here a = 1, b = -1, and c = -6, so ac = -6. We look for two numbers that multiply to -6 and add to -1, which are -3 and 2. Splitting the middle term gives x^2 - 3x + 2x - 6, and factoring by grouping yields (x-3)(x+2). The AC method simplifies this process. Further explanation
Factor 2x^2 + 7x + 3 using the AC method.
(2x-1)(x-3)
(2x+1)(x+3)
(2x+3)(x+1)
(x+1)(2x+3)
With a = 2, b = 7, c = 3, we get ac = 6. The numbers that multiply to 6 and add to 7 are 1 and 6. Rewrite 7x as 1x + 6x, group to factor: 2x^2 + 1x + 6x + 3 = x(2x+1) + 3(2x+1) = (2x+1)(x+3). The AC method makes handling a ? 1 straightforward. See steps here
Factor 3x^2 + 8x + 4 using the AC method.
(3x+1)(x+4)
(3x+4)(x+1)
(3x+2)(x+2)
(x+2)(3x+4)
Here a = 3, b = 8, c = 4 and ac = 12. We need two numbers that multiply to 12 and sum to 8, which are 6 and 2. Split the middle term: 3x^2 + 6x + 2x + 4, factor by grouping to get (3x+2)(x+2). This is exactly how the AC method works when a ? 1. More examples
Factor x^2 + 2x - 15 using the AC method.
(x+3)(x-5)
(x+5)(x-3)
(x+15)(x-1)
(x-5)(x-3)
We have a = 1, b = 2, c = -15 so ac = -15. Numbers that multiply to -15 and sum to 2 are 5 and -3. Rewriting yields x^2 + 5x - 3x - 15, then factor by grouping to get (x+5)(x-3). The AC method clarifies handling negative c. Detailed guide
Factor 4x^2 - 12x + 9 using the AC method.
(4x-3)(x-3)
(2x-1)(2x-9)
(2x-3)(2x-3)
(4x-9)(x-1)
With a = 4, b = -12, c = 9, we calculate ac = 36. We look for two numbers that multiply to 36 and add to -12, which are -6 and -6. Splitting gives 4x^2 - 6x - 6x + 9, and grouping yields (2x-3)(2x-3), or (2x-3)^2. This AC approach confirms a perfect square trinomial. Perfect squares explained
Factor 5x^2 + 5x - 30 using the AC method.
(5x+2)(x-15)
5(x+2)(x-15)
5(x-2)(x+3)
(x+2)(5x-15)
Here a = 5, b = 5, c = -30, so ac = -150. Numbers multiplying to -150 and summing to 5 are 15 and -10. Rewrite as 5x^2 +15x -10x -30, factor by grouping to get 5(x+3)(x-2). Factoring out the common 5 completes the AC method. Learn grouping
Factor 6x^2 - x - 2 using the AC method.
(3x+2)(2x-1)
(6x+1)(x-2)
(2x-1)(3x+2)
(2x+1)(3x-2)
With a = 6, b = -1, c = -2, compute ac = -12. You need two numbers that multiply to -12 and sum to -1, which are 3 and -4. Splitting the middle term gives 6x^2 + 3x - 4x - 2; grouping yields (3x+2)(2x-1). This demonstrates the efficiency of the AC method. Step-by-step guide
Factor 6x^2 + 11x + 3 using the AC method.
(6x+3)(x+1)
(6x+1)(x+3)
(3x+1)(2x+3)
(3x+3)(2x+1)
Here a = 6, b = 11, c = 3 so ac = 18. We look for two numbers that multiply to 18 and add to 11, which are 9 and 2. Rewrite the middle term: 6x^2 + 9x + 2x + 3, group to get (3x+1)(2x+3). The AC method streamlines factoring when a > 1. More practice
Factor x^2 + 13x + 36 using the AC method.
(x+6)(x+6)
(x+3)(x+12)
(x+2)(x+18)
(x+4)(x+9)
With a = 1, b = 13, c = 36, ac = 36. We need two numbers multiplying to 36 and summing to 13; 4 and 9 work. Write x^2 + 4x + 9x + 36, then factor to (x+4)(x+9). This is a straightforward AC method case when a = 1. AC method overview
Factor 4x^2 - x - 3 using the AC method.
(2x+3)(2x-1)
(4x+3)(x-1)
(2x-1)(2x+3)
(4x-3)(x+1)
Here a = 4, b = -1, c = -3 so ac = -12. Two numbers that multiply to -12 and sum to -1 are 3 and -4. Splitting gives 4x^2 + 3x - 4x - 3; grouping yields (4x+3)(x-1). The AC method makes splitting clear. Grouping tutorial
Factor 3x^2 - x - 4 using the AC method.
(3x+4)(x-1)
(3x+1)(x-4)
(x-4)(3x+1)
(3x-4)(x+1)
For 3x^2 - x - 4, a = 3, b = -1, c = -4 so ac = -12. Numbers that multiply to -12 and add to -1 are 3 and -4. Rewriting yields 3x^2 +3x -4x -4 and grouping gives (3x-4)(x+1). The AC method highlights the splitting step. See detailed steps
Factor 2x^2 + x - 6 using the AC method.
(2x+1)(x-6)
(x-3)(2x+2)
(2x+3)(x-2)
(2x-3)(x+2)
With a = 2, b = 1, c = -6, we have ac = -12. The pair of numbers that multiply to -12 and sum to 1 are 4 and -3. Rewrite as 2x^2 + 4x - 3x - 6, group to get (2x-3)(x+2). This shows the power of the AC method for non-unit leading coefficients. AC practice
Factor 5x^2 - 14x + 3 using the AC method.
(5x-3)(x-1)
(5x-1)(x-3)
(5x-3)(x+1)
(5x-1)(x-3)
Here a = 5, b = -14, c = 3 gives ac = 15. We look for factors of 15 summing to -14, which are -15 and 1. Rewriting: 5x^2 - 15x + x + 3 and grouping leads to (5x-3)(x-1). This is a clear AC method application. See grouping
Factor 8x^2 + 10x - 3 using the AC method.
(8x-3)(x+1)
(4x-1)(2x+3)
(8x+1)(x-3)
(2x-1)(4x+3)
For 8x^2 + 10x - 3, a = 8, b = 10, c = -3, so ac = -24. The numbers 12 and -2 multiply to -24 and sum to 10. Splitting gives 8x^2 + 12x - 2x - 3; grouping yields (4x-1)(2x+3). This highlights the AC splitting process. Further reading
Factor 6x^2 - 7x - 3 using the AC method.
(6x+1)(x-3)
(3x-3)(2x+1)
(2x-3)(3x+1)
(2x+3)(3x-1)
Here a = 6, b = -7, c = -3 so ac = -18. The pair 2 and -9 multiply to -18 and sum to -7. Splitting yields 6x^2 + 2x - 9x - 3; grouping gives (2x-3)(3x+1). The AC method identifies the right pair for splitting. AC explanation
Factor 6x^2 + 13x - 5 using the AC method.
(3x+1)(2x-5)
(6x+5)(x-1)
(6x-1)(x+5)
(3x-1)(2x+5)
With a = 6, b = 13, c = -5, ac = -30. The numbers 15 and -2 multiply to -30 and sum to 13. Split into 6x^2 +15x -2x -5, group to get (3x-1)(2x+5). This showcases the AC method for larger coefficients. Advanced AC method
Factor 12x^2 - 5x - 2 using the AC method.
(4x-1)(3x+2)
(3x-2)(4x+1)
(6x+1)(2x-2)
(12x+1)(x-2)
Here a = 12, b = -5, c = -2, so ac = -24. The numbers 3 and -8 multiply to -24 and sum to -5. Rewriting gives 12x^2 +3x -8x -2, grouping yields (3x-2)(4x+1). The AC method is effective even with a = 12. Read more
Factor 9x^2 + 30x + 25 using the AC method.
(9x+1)(x+25)
(9x+5)(x+5)
(3x+25)(3x+1)
(3x+5)(3x+5)
With a = 9, b = 30, c = 25, ac = 225. The pair 15 and 15 multiply to 225 and sum to 30, so split into 9x^2 +15x +15x +25. Grouping produces (3x+5)(3x+5), a perfect square. The AC method confirms this special case. Perfect square trinomials
Factor 10x^2 - 23x + 12 using the AC method.
(5x-4)(2x-3)
(10x-3)(x-4)
(5x-3)(2x-4)
(10x-4)(x-3)
Here a = 10, b = -23, c = 12 gives ac = 120. The pair -15 and -8 multiply to 120 and sum to -23. Splitting yields 10x^2 - 15x - 8x + 12, grouping leads to (5x-4)(2x-3). The AC method handles negative sums smoothly. Negative factors
Factor 14x^2 + 9x - 10 using the AC method.
(14x+5)(x-2)
(7x+2)(2x-5)
(14x-2)(x+5)
(7x-2)(2x+5)
With a = 14, b = 9, c = -10, ac = -140. The pair 20 and -7 multiply to -140 and add to 9. Splitting gives 14x^2 +20x -7x -10; grouping yields (7x-2)(2x+5). AC method works for mixed signs. Mixing signs
Factor 9x^2 - 24x + 16 using the AC method.
(3x-4)(3x-4)
(3x-2)(3x-8)
(3x-1)(3x-16)
(9x-4)(x-4)
For 9x^2 - 24x + 16, a = 9, b = -24, c = 16 and ac = 144. The pair -12 and -12 multiply to 144 and sum to -24, so split to get a perfect square: (3x-4)^2. This confirms that the trinomial is a perfect square. Perfect squares
Factor 15x^2 - 13x - 2 using the AC method.
(x-2)(15x+1)
(15x-2)(x+1)
(5x-2)(3x+1)
(3x-2)(5x+1)
Here a = 15, b = -13, c = -2, so ac = -30. The pair -15 and 2 multiply to -30 and sum to -13. Splitting yields 15x^2 - 15x + 2x - 2; grouping gives (15x+1)(x-2). AC method makes splitting systematic. Detailed grouping
Factor 24x^2 + 38x + 15 using the AC method.
(4x+5)(6x+3)
(12x+5)(2x+3)
(8x+5)(3x+3)
(4x+3)(6x+5)
In 24x^2 + 38x + 15, a = 24, b = 38, c = 15 so ac = 360. The pair 20 and 18 multiply to 360 and sum to 38. Splitting yields 24x^2 + 20x + 18x + 15; grouping gives (4x+3)(6x+5). The AC method empowers you to handle large coefficients reliably. Advanced examples
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Study Outcomes

  1. Understand the AC Method -

    Explain the steps involved in factoring with the AC method to confidently decompose quadratic expressions.

  2. Identify Factor Pairs -

    Analyze quadratic coefficients (A, B, C) to determine the correct factor combinations for efficient AC method quadratic factoring.

  3. Apply the AC Method to Quadratics -

    Use the AC method to split the middle term and factor quadratic equations accurately.

  4. Solve Quadratic Equations -

    Execute factoring quadratic equations techniques to find and verify solutions in practice problems.

  5. Evaluate Problem-Solving Strategies -

    Compare the AC method with other factoring approaches to decide when it's the most effective tool.

  6. Build Exam Confidence -

    Complete ac method practice problems in a free quiz format to strengthen skills and prepare for algebra assessments.

Cheat Sheet

  1. Breaking Down ax² + bx + c with the AC Method -

    Begin by computing a·c to combine the leading coefficient and constant term; for 6x² + 11x + 3, a·c = 18. Then find two numbers that multiply to 18 and add to b (11), which are 9 and 2, before splitting the middle term. Drawing from MIT OpenCourseWare, this structured approach makes factoring with the ac method systematic and reliable.

  2. Splitting the Middle Term Strategically -

    In AC method quadratic factoring, you rewrite bx as mx + nx where m + n = b and m·n = a·c; for example, x² + 5x + 6 becomes x² + 3x + 2x + 6. A handy mnemonic is "Find Product, then Partition" to recall these two steps. Khan Academy's algebra resources emphasize this split as the core of factoring quadratic equations.

  3. Factoring by Grouping Technique -

    Once you've split the middle term, group the four-term polynomial into pairs and factor common terms: x² + 3x + 2x + 6 → x(x + 3) + 2(x + 3). Then factor out the shared binomial (x + 3) to get (x + 2)(x + 3). The University of Minnesota's math department highlights grouping as a universal method for factoring quadratic equations.

  4. Verifying Solutions with the Zero Product Property -

    After factoring, set each factor equal to zero - if (2x - 1)(3x + 4) = 0, then 2x - 1 = 0 or 3x + 4 = 0 - to solve quadratic equations neatly. This check confirms your factor pairs are correct and aligns with techniques taught at the University of Texas at Austin. Always plug your roots back into the original equation for a final validation.

  5. Building Confidence with AC Method Practice Problems -

    Regularly tackle ac method practice problems that vary a, b, and c to strengthen pattern recognition; start with a = 1 before advancing to higher values. Research in the American Mathematical Monthly shows spaced repetition and varied exercises boost proficiency in solving quadratics with the ac method. Track progress by timing yourself and reviewing errors to master each step under quiz conditions.

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