➕ Algebra 1 · Quadratics

Solving Quadratics by Factoring practice questions

Questions start easy and step up. Before you answer, rate how sure you are. It trains you to know what you really know.

Answer key with explanations

1. Solve x² − 10x + 21 = 0.

   Answer: x = 3 or x = 7. Factor: two integers with product 21 and sum −10 are −3 and −7, so (x − 3)(x − 7) = 0. Then x − 3 = 0 or x − 7 = 0, so x = 3 or x = 7. Check: 9 − 30 + 21 = 0 and 49 − 70 + 21 = 0. The choice x = −3 or x = −7 uses the factor pair itself instead of solving each factor. The choice x = 3 or x = −7 gets only one sign right. The choice x = 10 or x = 21 just uses the numbers in the equation.

2. What is the first step in solving x² + 3x = 28 by factoring?

   Answer: Subtract 28 from both sides to get x² + 3x − 28 = 0. The zero-product property works only when a product equals 0, so first move the 28 to the left: x² + 3x − 28 = 0. Then factor: (x + 7)(x − 4) = 0, so x = −7 or x = 4. Setting the factors of x(x + 3) equal to 28 gives x = 28 or x = 25, and neither works: 28(31) ≠ 28 and 25(28) ≠ 28. Dividing by x would lose solutions. Taking square roots works neatly only for an equation like x² = a number, and here the left side also has a 3x term.

3. How many solutions does x² − 16x + 64 = 0 have, and what are they?

   Answer: One solution, x = 8 (a double root). Factor: two integers with product 64 and sum −16 are −8 and −8, so x² − 16x + 64 = (x − 8)² = 0. The factor x − 8 appears twice and gives only x = 8. Check: 64 − 128 + 64 = 0. The value x = −8 does not work: 64 + 128 + 64 = 256. The value x = 64 does not work either: 4096 − 1024 + 64 = 3136. And x = 8 is a solution, so it is not the no-solution case.

4. Solve 2x² − 18x = 0.

   Answer: x = 0 or x = 9. Factor out the GCF 2x: 2x(x − 9) = 0. The factor 2 is never 0. The factor x gives x = 0 and the factor x − 9 gives x = 9. Check: 0 = 0 and 2(81) − 18(9) = 162 − 162 = 0. The choice “x = 9 only” comes from dividing by x and losing x = 0. The choice “x = 2 or x = 9” treats the number 2 as a solution, but a constant factor can never be 0. The choice “x = 0 or x = −9” has the wrong sign.

5. Solve (x − 3)(x + 2) = 14.

   Answer: x = 5 or x = −4. The product is 14, not 0, so first multiply out and move the 14: x² − x − 6 = 14, so x² − x − 20 = 0. Factor: (x − 5)(x + 4) = 0, so x = 5 or x = −4. Check: (5 − 3)(5 + 2) = 2 × 7 = 14 and (−4 − 3)(−4 + 2) = (−7)(−2) = 14. The choice x = 17 or x = 12 sets each factor equal to 14, which does not work: (17 − 3)(17 + 2) = 266. The choice x = 3 or x = −2 solves the equation = 0 instead of 14. The choice x = 5 only misses the second solution.

6. A student solves x² = 11x by dividing both sides by x and writes x = 11. What did the student miss?

   Answer: The solution x = 0, because dividing by x is only allowed when x ≠ 0. Dividing both sides by x assumes x ≠ 0, but x = 0 makes the equation true: 0² = 11(0). Factoring avoids the problem: x² − 11x = 0, so x(x − 11) = 0, so x = 0 or x = 11. The value x = −11 does not work: 121 ≠ −121. Not every quadratic has one positive and one negative solution. And “x = 11 only” is wrong because x = 0 also works. Dividing by x is allowed only when x ≠ 0.

7. Which of these equations cannot be solved by factoring with integers?

   Answer: x² + 5x + 5 = 0. To factor x² + 5x + c, you need two integers with sum 5 and product c. For c = 6 the pair is 2 and 3, for c = 4 the pair is 1 and 4, and for c = 0 the pair is 0 and 5. For c = 5, the only pairs of integers with product 5 are 1 and 5 (sum 6) and −1 and −5 (sum −6), so no pair adds to 5. This equation does have solutions, but factoring with integers cannot find them.

8. Solve 9x² − 16 = 0.

   Answer: x = 4/3 or x = −4/3. This is a difference of squares: 9x² − 16 = (3x)² − 4² = (3x − 4)(3x + 4). Then 3x − 4 = 0 gives x = 4/3, and 3x + 4 = 0 gives x = −4/3. Check: 9(16/9) − 16 = 0. The choice x = ±4 forgets the 9: 9(16) − 16 = 128. The choice 4/3 only misses the negative solution. The choice ±3/4 inverts the fraction: 9(9/16) − 16 ≠ 0.

9. The length of a rectangle is 4 feet more than its width. The area is 60 square feet. What is the width?

   Answer: 6 feet. Let w be the width in feet. The length is w + 4, so w(w + 4) = 60, which gives w² + 4w − 60 = 0. Factor: (w + 10)(w − 6) = 0, so w = −10 or w = 6. A width cannot be negative, so reject w = −10 (it solves the equation, since (−10)(−6) = 60, but not the story). The width is 6 feet and the length is 10 feet: 6 × 10 = 60. The choice 10 is the length. The choice 15 is 60 ÷ 4, which is not a width that works: 15 × 19 = 285.

10. At which points does the graph of y = x² + 6x − 16 cross the x-axis?

    Answer: (−8, 0) and (2, 0). The graph crosses the x-axis where y = 0, so solve x² + 6x − 16 = 0. Factor: (x + 8)(x − 2) = 0, so x = −8 or x = 2, giving the points (−8, 0) and (2, 0). The choice (8, 0) and (−2, 0) copies the signs from the factors. The point (0, −16) is where the graph crosses the y-axis, and (−3, −25) is the vertex.