Factoring x² + bx + c

Tier 1 Foundations

1. (x + 3)(x + 8) and (y − 4)(y − 9). For x² + 11x + 24, find two numbers whose product is 24 and whose sum is 11, which are 3 and 8, giving (x + 3)(x + 8). For y² − 13y + 36, find two negative numbers whose product is 36 and whose sum is −13, which are −4 and −9, giving (y − 4)(y − 9). Multiply to check: (x + 3)(x + 8) = x² + 8x + 3x + 24 = x² + 11x + 24, and (y − 4)(y − 9) = y² − 9y − 4y + 36 = y² − 13y + 36.
2. Length is x + 9 meters and width is x + 5 meters; when x = 6, the length is 15 meters and the width is 11 meters. Find two numbers whose product is 45 and whose sum is 14, which are 9 and 5. Factor the area trinomial as (x + 9)(x + 5), so the length is x + 9 meters and the width is x + 5 meters. Substitute x = 6: the length is 6 + 9 = 15 meters and the width is 6 + 5 = 11 meters.
3. Multiplying (x − 7)(x − 4) gives x² − 11x + 28, which has −11x and +28 instead of −3x and −28; the correct factored form is (x − 7)(x + 4). Multiply Orla's binomials: (x − 7)(x − 4) = x² − 4x − 7x + 28 = x² − 11x + 28. Because the constant term −28 is negative, the two numbers must have opposite signs. The numbers −7 and +4 multiply to −28 and add to −3, giving (x − 7)(x + 4).
4. Rug A has dimensions x + 7 feet and x − 3 feet; Rug B has dimensions x − 6 feet and x + 2 feet; Rug A has the greater perimeter by 16 feet. Factor Rug A's area: x² + 4x − 21 = (x + 7)(x − 3), so its perimeter is 2(x + 7 + x − 3) = 4x + 8 feet. Factor Rug B's area: x² − 4x − 12 = (x − 6)(x + 2), so its perimeter is 2(x − 6 + x + 2) = 4x − 8 feet. Subtract the perimeters: (4x + 8) − (4x − 8) = 16 feet.
5. Length is x − 5 feet and width is x − 7 feet; when x = 12, the perimeter is 24 feet. Find two numbers whose product is +35 and whose sum is −12, which are −5 and −7. The factored area is (x − 5)(x − 7), so the length is x − 5 feet and the width is x − 7 feet. When x = 12, the length is 12 − 5 = 7 feet, the width is 12 − 7 = 5 feet, and the perimeter is 2(7 + 5) = 24 feet.

Tier 2 Intermediate

6. 4x(x + 2)(x + 5); when x = 3, the dimensions are 12 inches, 5 inches, and 8 inches. Factor out the greatest common factor 4x from all three terms: 4x(x² + 7x + 10). Factor x² + 7x + 10 into (x + 2)(x + 5), giving 4x(x + 2)(x + 5). Substitute x = 3 into each factor: 4(3) = 12 inches, 3 + 2 = 5 inches, and 3 + 5 = 8 inches.
7. −(x − 8)(x + 3) and (a − 8b)(a + 2b). Factor −1 out of −x² + 5x + 24 to get −(x² − 5x − 24). The numbers −8 and 3 multiply to −24 and add to −5, giving −(x − 8)(x + 3). For a² − 6ab − 16b², find two expressions in b that multiply to −16b² and add to −6b, which are −8b and 2b, giving (a − 8b)(a + 2b).
8. The binomial factor 3x + 9 still has a common factor of 3; the completely factored form is 3(x + 3)(x − 5), and the area when x = 7 is 60 square feet. In Lazaro's answer (3x + 9)(x − 5), both terms of 3x + 9 are divisible by 3, so it is not completely factored. Factoring out the greatest common factor 3 from 3x² − 6x − 45 gives 3(x² − 2x − 15) = 3(x + 3)(x − 5). At x = 7, the area is 3(7 + 3)(7 − 5) = 3(10)(2) = 60 square feet.
9. x² − 7x + 12 factors as (x − 3)(x − 4); x² − 7x − 12 is prime because no integer factor pair of −12 adds up to −7. For x² − 7x + 12, the numbers −3 and −4 multiply to +12 and add to −7, so it factors as (x − 3)(x − 4). For x² − 7x − 12, the integer factor pairs of −12 are 1 and −12, 2 and −6, 3 and −4, −1 and 12, −2 and 6, and −3 and 4. The sums of those pairs are −11, −4, −1, 11, 4, and 1; since none of the sums equal −7, x² − 7x − 12 is prime.
10. Length is x + 6y meters and width is x + 3y meters; when x = 10 and y = 2, the length is 22 meters and the width is 16 meters. Find two terms in y that multiply to 18y² and add to 9y, which are 6y and 3y. Factor the area as (x + 6y)(x + 3y); since y > 0, x + 6y is the length and x + 3y is the width. When x = 10 and y = 2, the length is 10 + 6(2) = 22 meters and the width is 10 + 3(2) = 16 meters.

Tier 3 Advanced

11. k = 21 gives (x + 1)(x + 20); k = 12 gives (x + 2)(x + 10); k = 9 gives (x + 4)(x + 5). List all pairs of positive whole numbers whose product is 20: 1 and 20, 2 and 10, and 4 and 5. Add each pair to find the possible values of the middle coefficient k: 1 + 20 = 21, 2 + 10 = 12, and 4 + 5 = 9. The three factored forms are (x + 1)(x + 20), (x + 2)(x + 10), and (x + 4)(x + 5).
12. −16(t − 5)(t + 1); 128 feet at t = 3 seconds. Factor out the negative greatest common factor −16 from all three terms: −16(t² − 4t − 5). Factor t² − 4t − 5 as (t − 5)(t + 1), giving −16(t − 5)(t + 1). At t = 3, the height is −16(3 − 5)(3 + 1) = −16(−2)(4) = 128 feet.
13. (x + 11)(x + 3). Multiply the binomials: (x + 10)(x + 4) = x² + 14x + 40. Subtract 7: x² + 14x + 40 − 7 = x² + 14x + 33. Factor x² + 14x + 33 into (x + 11)(x + 3).
14. When factoring out −2, he did not flip the signs of +12x and +32 to get −2(x² − 6x − 16), and (x + 8)(x − 2) multiplies to x² + 6x − 16 rather than x² + 6x + 16; the correct factored form is −2(x − 8)(x + 2). Divide each term of −2x² + 12x + 32 by −2: −2x² ÷ (−2) = x², 12x ÷ (−2) = −6x, and 32 ÷ (−2) = −16, giving −2(x² − 6x − 16). Find two numbers that multiply to −16 and add to −6, which are −8 and +2. Write the completely factored expression: −2(x − 8)(x + 2).
15. The shared wall has length x − 3 feet; Room A is wider along its other dimension by 11 feet. Factor Room A's area: x² + 3x − 18 = (x + 6)(x − 3). Factor Room B's area: x² − 8x + 15 = (x − 5)(x − 3). Both rooms share the factor x − 3 feet. Compare their other dimensions by subtracting: (x + 6) − (x − 5) = 11 feet, so Room A is wider by 11 feet.

Tier 4 Challenge

16. k = 7 gives (x + 1)(x + 7); k = 12 gives (x + 2)(x + 6); k = 15 gives (x + 3)(x + 5); k = 16 gives (x + 4)(x + 4). In (x + a)(x + b) = x² + 8x + k, the positive whole numbers a and b must add to 8, and k = ab. List all pairs of positive whole numbers that add to 8: 1 and 7, 2 and 6, 3 and 5, and 4 and 4. Multiply each pair to find k: 1 × 7 = 7, 2 × 6 = 12, 3 × 5 = 15, and 4 × 4 = 16.
17. Even though 17 is prime, −17 still has the integer factor pairs 17 and −1, and −17 and 1; k = 16 gives (x + 17)(x − 1), and k = −16 gives (x − 17)(x + 1). Every nonzero integer has at least the factor pairs formed with 1 and −1. For −17, the integer factor pairs are 17 and −1, which add to 16, and −17 and 1, which add to −16. Therefore, x² + 16x − 17 factors as (x + 17)(x − 1), and x² − 16x − 17 factors as (x − 17)(x + 1).
18. Bin A has base dimensions x − 3y centimeters and x + 2y centimeters; Bin B has base dimensions x − 3y centimeters and x + y centimeters; Bin A has the greater base perimeter by 2y centimeters. Factor Bin A's volume: 3x(x² − xy − 6y²) = 3x(x − 3y)(x + 2y). Its base perimeter is 2(x − 3y + x + 2y) = 4x − 2y centimeters. Factor Bin B's volume: 3x(x² − 2xy − 3y²) = 3x(x − 3y)(x + y). Its base perimeter is 2(x − 3y + x + y) = 4x − 4y centimeters. Subtract the base perimeters: (4x − 2y) − (4x − 4y) = 2y centimeters.
19. Largest possible perimeter is 4x − 20 feet and smallest possible perimeter is 4x − 50 feet; the difference is 30 feet. Since (x − a)(x − b) = x² − (a + b)x + 24, we have ab = 24, and the perimeter is 2(x − a + x − b) = 4x − 2(a + b) feet. The positive whole-number factor pairs of 24 are 1 and 24 with sum 25, 2 and 12 with sum 14, 3 and 8 with sum 11, and 4 and 6 with sum 10. Subtracting the smallest sum gives the largest perimeter 4x − 2(10) = 4x − 20 feet, and subtracting the largest sum gives the smallest perimeter 4x − 2(25) = 4x − 50 feet; their difference is (4x − 20) − (4x − 50) = 30 feet.
20. Factored form is y = 2x(x − 2)(x − 5); the zeroes are x = 0, x = 2, and x = 5 meters; at x = 3 meters, the glider is 12 meters below the ridge. Factor out the greatest common factor 2x: y = 2x(x² − 7x + 10) = 2x(x − 2)(x − 5). Set each factor equal to 0 to find the zeroes: 2x = 0 gives x = 0, x − 2 = 0 gives x = 2, and x − 5 = 0 gives x = 5 meters. At x = 3, y = 2(3)(3 − 2)(3 − 5) = 6(1)(−2) = −12 meters, meaning 12 meters below the ridge.