8 Rational Zeros

💡 Learning Objectives

By the end of this lesson you will be able to (AP CED 1.8.A):

• Determine the zeros of a rational function.
• Explain why a zero of the numerator is only a zero of r(x) when the x-value is in the domain.
• Use the zeros and undefined points of r to solve rational inequalities such as r(x) > 0 or r(x) ≤ 0.

💡 Definitions

The real zeros of r(x) = N(x)/D(x) are the real values of x for which N(x) = 0 AND D(x) ≠ 0.

• If N(a) = 0 and D(a) ≠ 0, then a is a zero of r.
• If N(a) = 0 and D(a) = 0, then a is NOT automatically a zero of r — we must simplify and check (see Topic 1.10 on holes).

💡 Quick Check

Test your instinct: which x-values are zeros of r(x) = (x − 3)(x + 2) / (x − 5)?

• Candidates from the numerator: x = 3, x = −2.
• Is the denominator nonzero there? Yes — D(3) = −2 ≠ 0, D(−2) = −7 ≠ 0.

So both are zeros. x = 5 is excluded (denominator is zero there).

📘 Example

Example 1 — Finding zeros

Find the real zeros of r(x) = (x² − 4) / (x² − 1).

Numerator factors: x² − 4 = (x − 2)(x + 2), so potential zeros are x = 2 and x = −2.

Denominator: x² − 1 = (x − 1)(x + 1), zero at x = 1 and x = −1.

Neither candidate (2 or −2) makes the denominator zero, so both are zeros of r. x = 1 and x = −1 are excluded from the domain (vertical asymptotes, as we'll see in 1.9).

⚠️ Common Mistake

Setting the WHOLE rational expression equal to zero and 'cross-multiplying' is an easy way to forget the domain restriction. Always state the excluded values first, then solve N(x) = 0, then check each solution is in the domain.

📘 Example

Example 2 — Solving a rational inequality with a sign chart

Solve (x − 1)(x + 2) / (x − 3) > 0.

Boundary points: x = 1, x = −2 (zeros of numerator), and x = 3 (excluded). These split the real line into four intervals: (−∞, −2), (−2, 1), (1, 3), (3, ∞).

Test each interval with a sample point: x = −3 gives (−)(−)/(−) = negative; x = 0 gives (−)(+)/(−) = positive; x = 2 gives (+)(+)/(−) = negative; x = 4 gives (+)(+)/(+) = positive.

Solution: (−2, 1) ∪ (3, ∞). Note x = 3 is NOT included (undefined), and x = −2, 1 are NOT included (strict inequality).

🎯 AP Tip

For a non-strict inequality r(x) ≥ 0, INCLUDE the zeros of the numerator (where r equals zero) but NEVER include points where the denominator is zero. The domain always wins.

📘 Try This

Find the zeros of r(x) = (x² − 9) / (x² − 2x − 3).

• Factor numerator: (x − 3)(x + 3).
• Factor denominator: (x − 3)(x + 1).
• Candidates from numerator: x = 3, x = −3.
• Check denominator: at x = 3, denom is also zero — exclude for now. At x = −3, denom = 6 ≠ 0 — keep.

Zero: x = −3 only. (x = 3 is a HOLE, covered in 1.10.)