AMC 10 · 2025 · #17
Grade 11 algebrageometry-2dPick an answer.
The three factors are evenly spaced around the middle one, z + 2i. That symmetry is the whole opening: introduce a new variable w = z + 2i so the middle factor becomes just w and the outer two become w - i and w + i, a matched pair. Their product collapses by difference of squares, killing the quadratic term and leaving a bare cubic w³ + w + 10 that is easy to factor. From there the work splits into clean subproblems: find one root by the Remainder Theorem, divide out to get a quadratic, solve it (its roots are complex), then plot all three w-roots and read off the area. Because z = w - 2i only slides every point straight down, the triangle in the w-picture is congruent to the one in the z-picture, so its area is the answer.
Re-center on the middle factor
The shifts are evenly spaced with 2i in the middle, so set w = z + 2i: the middle factor is w and the outer pair becomes w - i and w + i.
Naming the center of the symmetry as the new variable makes the two outer factors mirror images of each other.
9.A-SSE.A.2Introduce A VariableCollapse the matched pair
Multiplying the conjugate pair, (w - i)(w + i) = w² + 1, so the product is w(w² + 1) + 10 = w³ + w + 10 — a tidy cubic with no w² term.
A number times its conjugate wipes out the imaginary part, turning i-laden factors into a plain real quadratic.
A number times its mirror image wipes out the imaginary part, turning two complex factors into one real quadratic.
▸ Why?
Nonreal factors of a real polynomial come in matched mirror pairs, so the pairing is always available.
▸ Why?
Multiplying a sum by its matching difference gives one square minus the other, which is what erases the imaginary term.
Find one root by testing divisors
A rational root must divide 10, so test ±1, ±2, ±5, ±10. Plugging in w = -2 gives -8 - 2 + 10 = 0, so it is a root and (w + 2) is a factor.
A rational root can only be built from divisors of the constant term, so a short list of guesses is guaranteed to catch it.
11.A-APR.B.2Identify SubproblemsDivide out the known factor
Dividing w³ + w + 10 by (w + 2) gives quotient w² - 2w + 5, so the cubic factors completely as (w + 2)(w² - 2w + 5).
Peeling off a known linear factor drops the cubic to a quadratic you can finish with the usual formula.
11.A-APR.D.6Identify SubproblemsSolve the quadratic (complex roots)
For w² - 2w + 5 = 0 the discriminant is 4 - 20 = -16, so w = 1 ± 2i; the three w-roots are -2, 1 + 2i, and 1 - 2i.
A negative discriminant means the roots leave the real line and become a conjugate pair straddling it.
11.N-CN.C.7Introduce A VariablePlot the roots and read the area
Points (-2,0), (1,2), (1,-2): vertical side 4, height 3, so area = ½ · 4 · 3 = 6; translation to z preserves area, so (A).
A conjugate pair gives a vertical side, so the far root's horizontal distance to that side is the height and the area falls right out.
10.G-GPE.B.7Draw A DiagramWhen factors are evenly spaced, rename the middle one as your variable so the outsides become a matched plus/minus pair that collapses, then plot the roots and let the conjugate pair hand you a vertical side.
- Re-center on the middle factor
- Collapse the matched pair
- Find one root by testing divisors
- Divide out the known factor
- Solve the quadratic (complex roots)
- Plot the roots and read the area