DWG NO. 2.4 — Lesson 4 of 6

Finding Real Zeros

Unit 2: Polynomial and Rational Functions · ~30 min

Objective Use the Rational Root Theorem to list possible rational zeros of a polynomial, then confirm and find them all with synthetic division.

The Rational Root Theorem

For a polynomial with integer coefficients, every possible rational zero has the form \(\frac{p}{q,}\) where p is a factor of the constant term and q is a factor of the leading coefficient. This doesn't guarantee any of them actually is a zero — it just narrows an infinite search down to a short, finite list worth testing.

Step 1: List all integer factors of the constant term (the p values, ±).
Step 2: List all integer factors of the leading coefficient (the q values, ±).
Step 3: Form every \(\frac{p}{q}\) combination — this is your list of candidates.
Step 4: Test candidates with synthetic division. A remainder of 0 confirms a zero (Factor Theorem, from 2.3) and hands you a lower-degree depressed polynomial to keep working with.
Step 5: Repeat on the depressed polynomial until it's down to a quadratic, then finish with factoring or the quadratic formula.

f(x) = x³ − 2x² − 5x + 6 — real zeros at x = −2, 1, 3, found using the Rational Root Theorem

Worked Example 1 · Full search, monic leading coefficient

ProblemFind all real zeros of f(x) = x³ − 2x² − 5x + 6.

1Constant term 6: factors ±1, 2, 3, 6. Leading coefficient 1: factors ±1. Candidates: ±1, ±2, ±3, ±6.

2Test x = 1: f(1) = 1−2−5+6 = 0. It's a zero. Synthetic division (c=1, coefficients 1,−2,−5,6) gives depressed polynomial x² − x − 6.

3Factor the quadratic: x² − x − 6 = (x−3)(x+2), giving zeros x = 3 and x = −2.

Real zeros: x = −2, 1, 3

Worked Example 2 · Non-monic leading coefficient

ProblemFind all real zeros of f(x) = 2x³ + 3x² − 8x + 3.

1Constant 3: factors ±1, 3. Leading coefficient 2: factors ±1, 2. Candidates: ±1, ±3, ±½, \(\frac{\pm3}{2.}\)

2Test x = 1: f(1) = 2+3−8+3 = 0. Synthetic division (c=1, coefficients 2,3,−8,3) gives depressed polynomial 2x² + 5x − 3.

3Factor: 2x² + 5x − 3 = (2x−1)(x+3), giving zeros x = ½ and x = −3.

Real zeros: x = −3, ½, 1

Worked Example 3 · Depressed polynomial with irrational zeros

ProblemFind all real zeros of f(x) = x³ + x² − 3x − 3.

1Constant 3: factors ±1, 3. Leading coefficient 1. Candidates: ±1, ±3.

2Test x = −1: f(−1) = −1+1+3−3 = 0. Synthetic division (c=−1, coefficients 1,1,−3,−3) gives depressed polynomial x² − 3.

3x² − 3 doesn't factor over the rationals — no more candidates from the list will work. Solve directly: x² = 3, so x = ±\(\sqrt{3}\).

Real zeros: x = −1, \(\sqrt{3}\), −\(\sqrt{3}\) — the last two are irrational, so the Rational Root Theorem could never have found them directly, only narrowed the search enough to expose them

Guided practice

1.List all possible rational zeros of f(x) = x³ − 4x² − 7x + 10 using the Rational Root Theorem.
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Constant 10: factors ±1,2,5,10. Leading coefficient 1: factors ±1. Candidates: ±1, ±2, ±5, ±10.
2.Confirm x = 1 is a zero of f(x) = x³ − 4x² − 7x + 10 using synthetic division, and state the depressed quadratic.
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c=1, coefficients 1,−4,−7,10: 1(1)+(−4)=−3; −3(1)+(−7)=−10; −10(1)+10=0. Remainder 0 confirms the zero. Depressed polynomial: x² − 3x − 10
3.Using your answer to problem 2, find all real zeros of f(x) = x³ − 4x² − 7x + 10.
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x² − 3x − 10 = (x−5)(x+2), giving zeros 5 and −2. Real zeros: x = −2, 1, 5
4.List all possible rational zeros of f(x) = 3x³ − 2x² − 7x − 2.
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Constant −2: factors ±1,2. Leading coefficient 3: factors ±1,3. Candidates: ±1, ±2, ±⅓, \(\frac{\pm^{2}}{3}\)
5.Find all real zeros of f(x) = 2x³ − x² − 8x + 4.
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Candidates include ±½. Testing x = ½: f(½) = 0.25−0.25−4+4 = 0. Synthetic division gives depressed polynomial 2x² − 8 = 2(x−2)(x+2). Real zeros: x = −2, ½, 2

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