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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: A-SSE.A.1, A-SSE.A.2, A-CED.A.1, A-CED.A.2, A-CED.A.3, A-CED.A.4, A-REI.D.11, A-APR.C.5, N-CN.C.8, N-CN.C.9 · Level: Grade 11 Algebra 2
Problem 1 Interpret a model
500 is the starting amount in dollars (the value when t = 0). The factor 1.04 means the balance is multiplied by 1.04 each year, a 4% yearly increase. 500 = initial deposit; 1.04 = yearly growth factor (4%).
Problem 2 Difference of squares twice
x4 − 625 = (x2 + 25)(x2 − 25) = (x2 + 25)(x + 5)(x − 5). (x2 + 25)(x + 5)(x − 5).
Problem 3 Factor with a coefficient
16x4 = (4x2)2 and 81 = 92, so the expression is (4x2 + 9)(4x2 − 9) = (4x2 + 9)(2x + 3)(2x − 3). (4x2 + 9)(2x + 3)(2x − 3).
Problem 4 Rearrange a volume formula
Multiply by 3: 3V = πr2h. Divide by πr2. h = 3V/(πr2).
Problem 5 Rearrange a trapezoid area formula
Multiply by 2: 2A = h(b1 + b2). Divide by h: 2A/h = b1 + b2. Subtract b1. b2 = 2A/h − b1.
Problem 6 Expand with Pascal’s triangle
Row 4 is 1, 4, 6, 4, 1 (each entry is the sum of the two above). Then (x + 2)4 = x4 + 4(2)x3 + 6(4)x2 + 4(8)x + 16. x4 + 8x3 + 24x2 + 32x + 16.
Problem 7 Expand with a negative term
Row 3 is 1, 3, 3, 1. (2x)3 + 3(2x)2(−1) + 3(2x)(−1)2 + (−1)3 = 8x3 − 12x2 + 6x − 1. 8x3 − 12x2 + 6x − 1.
Problem 8 A single coefficient
The term is C(6, 2) · x2 · 24 = 15 · 16 · x2. 240.
Problem 9 Factor over the complex numbers
x2 + 25 = x2 − (5i)2 = (x + 5i)(x − 5i). Check: x2 − 25i2 = x2 + 25. (x + 5i)(x − 5i).
Problem 10 Zeros and factors of a quadratic
x = (4 ± √(16 − 52))/2 = (4 ± 6i)/2 = 2 ± 3i. (x − 2 − 3i)(x − 2 + 3i). Check: (x − 2)2 + 9 = x2 − 4x + 13.
Problem 11 Generate a Pythagorean triple
Right side: x4 − 2x2y2 + y4 + 4x2y2 = x4 + 2x2y2 + y4, the left side. For (4, 1): 172 = 152 + 82, since 289 = 225 + 64. Triple 8, 15, 17.
Problem 12 Solve f(x) = g(x) from a graph
The curves cross at x = 2 and x = 5. Algebraically, x2 − 6x + 7 = x − 3 gives x2 − 7x + 10 = 0, so (x − 2)(x − 5) = 0. x = 2 or x = 5.
Problem 13 Solve with a table
Compare columns: at x = 2 both are 10, and at x = 3 both are 15. In the other columns they differ. x = 2 and x = 3.
Problem 14 Garden constraint
The length is 12 − w, so the area is w(12 − w) ≥ 20. Then w2 − 12w + 20 ≤ 0, so (w − 2)(w − 10) ≤ 0. 2 ≤ w ≤ 10 feet (every width in this interval gives area at least 20).
Problem 15 Depreciation model
V(t) = 24000(0.88)t. Set 24000(0.88)t = 15000: 0.88t = 0.625, so t = ln 0.625 ÷ ln 0.88 = 3.677. V(t) = 24000(0.88)t; ≈ 3.7 years.
Problem 16 Mixture as a rational equation
Solve (4 + x)/(20 + x) = 0.5. Cross-multiply: 4 + x = 10 + 0.5x, so 0.5x = 6. x = 12 liters. Check: 16/32 = 0.5.
Problem 17 Ticket sales system
Let x be adult tickets and y child tickets: x + y = 150 and 12x + 7y = 1330. Substitute y = 150 − x: 12x + 1050 − 7x = 1330, so 5x = 280. 56 adult and 94 child tickets. Check: 672 + 658 = 1330.
Problem 18 Budget constraints
Let x = notebooks and y = pens: 4x + 1.5y ≤ 60, x ≥ 6, y ≥ 0. Test (6, 20): 4(6) + 1.5(20) = 24 + 30 = 54 ≤ 60 and 6 ≥ 6. Yes, (6, 20) is allowed.
Problem 19 Binomial Theorem with complex numbers
Row 4 is 1, 4, 6, 4, 1: 1 + 4i + 6i2 + 4i3 + i4 = 1 + 4i − 6 − 4i + 1. −4.
Problem 20 Identities for mental arithmetic
(144 + 25)(12 + 5)(12 − 5) = 169 · 17 · 7 = 169 · 119 = 20,111. Check: 20,736 − 625 = 20,111. 20,111.