AMC 10 · 2025 · #8
Grade 11 algebraPick an answer.
The usual direction is polynomial first, roots second. Here I am handed a root and asked for the polynomial, so I run the machine backwards: turn the root into a factor, rebuild f(x) in factored form, then expand and read the coefficients off. The one gap in that plan is that a single root does not determine a cubic — I need all three. Organizing the given root differently (as an integer-coefficient quadratic it satisfies) hands me a second root for free, and naming the last root with a variable lets the known x² coefficient pin it down.
Turn the root into an equation
Clearing the radical gives an integer quadratic.
Squaring is the one move that makes a square root vanish, trading an awkward number for a clean integer equation.
11.N-RN.A.2Organize Information In More WaysShow the quadratic divides f
That quadratic is a factor of f.
A polynomial with integer coefficients cannot pick just one member of an irrational pair — the leftover would be a rational equation with an irrational solution.
A polynomial with integer coefficients cannot pick just one member of an irrational pair.
▸ Why?
Such roots arrive in matched pairs, so taking one drags the other along.
▸ Why?
A polynomial vanishing at a root carries that root's factor, so the whole pair's quadratic divides it.
Pin down the third root
Matching coefficients gives the third root -3.
The conjugate pair already contributes 8 toward a root total of 5, so the third root has to pull it back down by 3.
9.A-SSE.A.1Introduce A VariableExpand the factored form
Expanding reveals the whole cubic.
The factored form and the expanded form are the same polynomial in different clothing; expanding just puts the coefficients on display.
9.A-APR.A.1Organize Information In More WaysRead off a and b
a is -13 and b is 33, so the sum is 20.
Coefficient matching is legal because a polynomial has exactly one list of coefficients.
9.A-SSE.A.1Introduce A VariableA polynomial with integer coefficients can never take only one half of an irrational pair like 4±√(5) — it has to take both, and that free second root is enough to rebuild the whole polynomial.
- Turn the root into an equation
- Show the quadratic divides f
- Pin down the third root
- Expand the factored form
- Read off a and b