Solving Quadratics by Factoring

Tier 1 Foundations

1. x = 16 and x = −3. Find two numbers whose product is −48 and whose sum is −13. The numbers are −16 and 3, so the equation factors as (x − 16)(x + 3) = 0. Setting each factor equal to 0 gives x = 16 or x = −3. Checking x = 16 gives 16² − 13(16) − 48 = 256 − 208 − 48 = 0, and checking x = −3 gives (−3)² − 13(−3) − 48 = 9 + 39 − 48 = 0.
2. The solutions are t = 4 and t = −1; the flare hits the ground after 4 seconds. Set h = 0 to get −16(t − 4)(t + 1) = 0. By the zero product property, t − 4 = 0 or t + 1 = 0, so t = 4 or t = −1. Time after launch cannot be negative, so the flare hits the ground at 4 seconds.
3. 2 seconds. Set h = 0 to get −16t² + 64 = 0. Factor out −16 to get −16(t² − 4) = 0, then factor the difference of squares as −16(t − 2)(t + 2) = 0. Setting each variable factor to 0 gives t = 2 or t = −2, and rejecting negative time leaves 2 seconds.
4. Dividing by x is invalid when x = 0 and loses the solution x = 0; the complete solutions are x = 0 and x = 9. Dividing both sides by x assumes that x is not 0, which throws away the case where x = 0. Instead, subtract 9x from both sides to get x² − 9x = 0. Factor out x to get x(x − 9) = 0, which gives x = 0 or x = 9.
5. Equation A has more distinct solutions and also has the larger greatest solution; Equation A's greatest solution is 7 and Equation B's greatest solution is 5. For Equation A, setting each factor to 0 gives x = 7 and x = −2, so it has 2 distinct solutions and a greatest solution of 7. For Equation B, x² − 10x + 25 factors as (x − 5)² = 0, which has only 1 distinct solution, x = 5. Comparing them shows Equation A has 2 solutions versus 1, and 7 is greater than 5.

Tier 2 Intermediate

6. −3. Let x be the negative number, so x² = 5x + 24. Subtract 5x and 24 from both sides to get x² − 5x − 24 = 0. Factor as (x − 8)(x + 3) = 0, which gives x = 8 or x = −3. Since the number is negative, the answer is −3.
7. The width is 5 meters; the solution w = −9 makes no sense because the width of a dog run cannot be negative. The length is w + 4, so the area equation is w(w + 4) = 45, or w² + 4w = 45. Subtract 45 from both sides to get w² + 4w − 45 = 0, which factors as (w + 9)(w − 5) = 0. The solutions are w = −9 and w = 5. Since a physical length cannot be negative, w = −9 is rejected and the width is 5 meters.
8. 4 seconds. Set h = 0 to get −16t² + 60t + 16 = 0. Factor out the common factor −4 to get −4(4t² − 15t − 4) = 0. Factor the trinomial as −4(4t + 1)(t − 4) = 0, giving t = −1/4 or t = 4. Since time cannot be negative, the ball hits the ground after 4 seconds.
9. The zero product property only works when the product equals 0, not 14; the correct solutions are x = 5 and x = −4. Two numbers can multiply to 14 without either number being 14, so setting each factor equal to 14 is invalid. Expand the left side to get x² − x − 6 = 14, then subtract 14 from both sides to get x² − x − 20 = 0. Factoring gives (x − 5)(x + 4) = 0, so x = 5 or x = −4.
10. Ball B lands later, by 1 second; Ball A lands at 3 seconds and Ball B at 4 seconds. Ball A: −16t² + 32t + 48 = 0 gives −16(t² − 2t − 3) = −16(t − 3)(t + 1) = 0, so t = 3 or t = −1. Ball B: −16t² + 48t + 64 = 0 gives −16(t² − 3t − 4) = −16(t − 4)(t + 1) = 0, so t = 4 or t = −1. Time can't be negative, so Ball A lands at 3 seconds and Ball B at 4 seconds: Ball B lands 1 second later.

Tier 3 Advanced

11. 2.5 inches, or 5/2 inches. The outer width is 2x + 8 and the outer height is 2x + 12, so (2x + 8)(2x + 12) = 221. Expanding gives 4x² + 40x + 96 = 221, and subtracting 221 gives 4x² + 40x − 125 = 0. Factoring gives (2x − 5)(2x + 25) = 0, so x = 5/2 or x = −25/2. Rejecting the negative width leaves 2.5 inches.
12. 1 second and 4 seconds. Set h = 96 to get −16t² + 80t + 32 = 96. Subtract 96 from both sides to get −16t² + 80t − 64 = 0, then factor out −16 to get −16(t² − 5t + 4) = 0. Factoring the trinomial gives −16(t − 1)(t − 4) = 0, so the rocket is at 96 feet at t = 1 second on the way up and t = 4 seconds on the way down.
13. −16t² + 40t = 0; the initial upward velocity is 40 feet per second. With solutions t = 0 and t = 5/2 and leading coefficient −16, the factored equation is −16t(t − 5/2) = 0. Distributing −16t gives −16t² + 40t = 0. Matching −16t² + 40t = 0 to −16t² + vt + s = 0 shows that s = 0 and v = 40 feet per second.
14. Dividing by (x − 4) is invalid when x = 4 and loses the solution x = 4; the solutions are x = 1 and x = 4. Dividing both sides by (x − 4) assumes x − 4 is not 0, which erases the solution x = 4. Instead, subtract 5(x − 4) from both sides to get (x − 4)(2x + 3) − 5(x − 4) = 0. Factor out the common binomial (x − 4) to get (x − 4)(2x − 2) = 0, which gives x = 4 or x = 1.
15. Equation B has more distinct solutions, while Equation A has the larger greatest solution; Equation A's only solution is 3/2, and Equation B's solutions are 3/4 and 1. Equation A factors as the perfect square (2x − 3)² = 0, so its only solution is x = 3/2. For Equation B, subtract 2x² + 12x − 7 from both sides to get 4x² − 7x + 3 = 0, which factors as (4x − 3)(x − 1) = 0 and gives x = 3/4 and x = 1. Equation B has 2 distinct solutions compared to 1 for Equation A, while Equation A's greatest solution of 3/2 is greater than Equation B's greatest solution of 1.

Tier 4 Challenge

16. The path width is 2 meters; x = 12 makes no sense because two path widths would total 24 meters, which is wider than the 12-meter and 16-meter garden dimensions. The inner planting bed has width 12 − 2x and length 16 − 2x, so (12 − 2x)(16 − 2x) = 96. Expanding gives 192 − 56x + 4x² = 96, and subtracting 96 gives 4x² − 56x + 96 = 0. Factoring out 4 gives 4(x² − 14x + 24) = 0, or 4(x − 2)(x − 12) = 0, so x = 2 or x = 12. If x = 12, the inner bed dimensions 12 − 2x and 16 − 2x would be −12 meters and −8 meters, which is impossible, so x = 2 meters.
17. 2 seconds. Find when the flare is at 96 feet by setting −16t² + 64t + 48 = 96. Subtract 96 from both sides to get −16t² + 64t − 48 = 0, and factor out −16 to get −16(t² − 4t + 3) = 0, or −16(t − 1)(t − 3) = 0. The flare reaches 96 feet at t = 1 second on the way up and passes 96 feet again at t = 3 seconds on the way down. The total time spent at or above 96 feet is 3 − 1 = 2 seconds.
18. Vikram lost the solution x = 5 by canceling x − 5, and replaced x − 5 on the right side with 0 instead of 1; the true solutions are x = 5 and x = −1. Canceling x − 5 divides by 0 when x = 5, losing that root, and dividing x − 5 by itself leaves 1 on the right side rather than 0. Subtract (x − 5) from both sides to get (x − 5)(2x + 3) − 1(x − 5) = 0. Factor out (x − 5) to get (x − 5)(2x + 3 − 1) = 0, or (x − 5)(2x + 2) = 0, which gives x = 5 or x = −1.
19. Equation B has more distinct solutions and also has the larger greatest solution; Equation A's greatest solution is k and Equation B's greatest solution is 2k. Equation A is a perfect square trinomial that factors as (x − k)² = 0, so it has only 1 distinct solution, x = k. Equation B factors as (x − 2k)(x + k) = 0, which gives x = 2k and x = −k. Since k is a positive integer, 2k and −k are distinct and 2k is greater than k, so Equation B has 2 distinct solutions and the larger greatest solution, 2k.
20. Factored form: (x − 2)(x + 2)(x + 3i)(x − 3i) = 0; solutions: x = 2, x = −2, x = 3i, and x = −3i. Treat x⁴ + 5x² − 36 = 0 as a quadratic in x² and factor it as (x² − 4)(x² + 9) = 0. Factor x² − 4 as (x − 2)(x + 2). Rewrite x² + 9 as x² − 9i² = x² − (3i)² and factor it as (x + 3i)(x − 3i), giving (x − 2)(x + 2)(x + 3i)(x − 3i) = 0 and solutions x = 2, x = −2, x = −3i, and x = 3i.