Four worked problems on multiplicity, parameters, and complex roots
The graphs of four third-degree polynomials are shown below. Each example states the polynomial and asks a specific question about its x-intercepts, multiplicities, or unknown parameters. Try each one before revealing the solution.
Consider the polynomial \( f(x) = x^3 \).
Question: Why does the graph cross the x-axis at exactly one point, and why does it look "flat" there?
Single x-intercept. Solving \( x^3 = 0 \) gives the only zero at \( x = 0 \). Because \( f(x) = x^3 \) is the product of three identical factors, the zero at \( x = 0 \) has multiplicity 3.
Why it crosses (not touches). An odd multiplicity means the graph crosses the x-axis (unlike an even multiplicity, which touches and turns around). Multiplicity 3 is odd, so the graph crosses at \( x = 0 \).
Why it looks flat. At multiplicity 3, the graph is tangent to the x-axis at the crossing — it has an inflection there, so it lingers near \( y = 0 \) for a stretch before and after crossing. That "flat" appearance is the visual signature of a higher odd multiplicity.
Why only one intercept. A cubic has 3 complex roots counting multiplicity. Here all three roots coincide at \( x = 0 \), so exactly one real x-intercept appears on the graph.
Consider the polynomial \( f(x) = (x + 2)(x^2 + 3x + 1) \), which is a cubic.
Question: The graph crosses at \( x = -2 \). What are the coordinates of the other two x-intercepts?
Step 1 — Factor out the known linear root. Since \( x = -2 \) is an x-intercept, \( (x + 2) \) is a factor. The given factorization is already in that form:
\[ f(x) = (x + 2)(x^2 + 3x + 1) \]Step 2 — Solve the remaining quadratic. Set the quadratic factor to zero:
\[ x^2 + 3x + 1 = 0 \]The discriminant is \( D = 9 - 4 = 5 > 0 \), so there are two distinct real roots:
\[ x = \frac{-3 \pm \sqrt{5}}{2} \]Answer. The other two x-intercepts are exactly:
\[ \left( \frac{-3 - \sqrt{5}}{2},\; 0 \right) \quad \text{and} \quad \left( \frac{-3 + \sqrt{5}}{2},\; 0 \right) \]Numerically, these are approximately \( (-2.618,\; 0) \) and \( (-0.382,\; 0) \).
A cubic \( f(x) \) has x-intercepts at \( x = 1 \) and at the roots of \( x^2 - 3x + 1 \). Its y-intercept is \( (0, 1) \). Find the leading coefficient \( a \) and the coordinates of the other two x-intercepts.
Question: What are the two other x-intercepts, and what is the value of the leading coefficient?
Step 1 — Write the general factored form. With \( x = 1 \) as one factor and \( x^2 - 3x + 1 \) as the other, write:
\[ f(x) = a(x - 1)(x^2 - 3x + 1) \]Step 2 — Use the y-intercept \( (0, 1) \) to find \( a \).
\[ f(0) = a(0 - 1)(0^2 - 3(0) + 1) = a(-1)(1) = -a \]Since the y-intercept is 1, we have:
\[ -a = 1 \implies a = -1 \]So the cubic is \( f(x) = -(x - 1)(x^2 - 3x + 1) \).
Step 3 — Solve the quadratic for the other two x-intercepts. The quadratic \( x^2 - 3x + 1 = 0 \) has \( D = 9 - 4 = 5 > 0 \), so:
\[ x = \frac{3 \pm \sqrt{5}}{2} \]Answer.
Check: \( f(0) = -(-1)(1) = 1 \) ✓, and \( f(1) = -(0)(\ldots) = 0 \) ✓.
Consider the polynomial \( f(x) = (x + 1)(x^2 + x + 1) \), which is a cubic.
Question: The graph crosses the x-axis at \( x = -1 \). Why is this the only real x-intercept?
Step 1 — Factor out the known linear root. Since \( x = -1 \) is a real root, \( (x + 1) \) is a factor:
\[ f(x) = (x + 1)(x^2 + x + 1) \]Step 2 — Check whether the quadratic has real roots. The discriminant of \( x^2 + x + 1 \) is:
\[ D = b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3 \]Since \( D < 0 \), the quadratic has no real roots — its two roots are a complex-conjugate pair:
\[ x = \frac{-1 \pm i\sqrt{3}}{2} \]Conclusion. The cubic has one real root (\( x = -1 \)) and two non-real roots. Only the real root appears on the real Cartesian graph, so the graph crosses the x-axis exactly once.
Try each problem before revealing the answer.
The cubic \( f(x) = (x - 2)^3 \) crosses the x-axis at \( x = 2 \). What is the multiplicity of the zero? Does the graph cross or touch?
Multiplicity 3 (odd). The graph crosses the x-axis, with an inflection — flatter than a multiplicity-1 crossing but not a "touch."
A cubic has x-intercepts at \( x = -1, x = 1, x = 3 \) and y-intercept \( (0, 6) \). Find its leading coefficient and equation.
Form: \( f(x) = a(x + 1)(x - 1)(x - 3) \). At \( x = 0 \):
\[ f(0) = a(1)(-1)(-3) = 3a = 6 \implies a = 2 \]So \( f(x) = 2(x + 1)(x - 1)(x - 3) \).
The cubic \( f(x) = (x - 1)(x^2 + 4) \) has a root at \( x = 1 \). Does it have any other real roots? Explain.
The quadratic \( x^2 + 4 = 0 \) has discriminant \( D = 0 - 16 = -16 < 0 \), so no real roots — its roots are \( \pm 2i \).
So the cubic has exactly one real root at \( x = 1 \).
A cubic has three complex roots counting multiplicity. If one root is real and the other two form a complex-conjugate pair (with nonzero imaginary part), only one x-intercept appears on the real graph. For example, \(f(x) = (x + 1)(x^2 + x + 1)\) has one real root at \(x = -1\) and two complex roots.
The graph crosses the x-axis with an inflection — it's tangent to the axis at that point but continues through. It's flatter there than a simple (multiplicity-1) crossing.
Write the cubic as \(f(x) = a(x - r_1)(x - r_2)(x - r_3)\) using the x-intercepts. Substitute an additional point — usually the y-intercept \((0, f(0))\) — and solve for \(a\).
Use the discriminant \(D = b^2 - 4ac\). \(D > 0\) → two distinct real roots. \(D = 0\) → one double real root. \(D < 0\) → no real roots (complex conjugate pair).
Last updated: October 17, 2026