The Factor Theorem says f(a) = 0 exactly when (x − a) is a factor: a root and a factor are one fact. The Rational Root Theorem lists every possible rational root, a factor of the constant over a factor of the leading coefficient. Test the list, divide out each hit, and the problem shrinks by one degree each time.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
The candidate list
For 2x³ − 3x² − 3x + 2: constant 2 gives ±1, ±2; leading 2 gives divisors 1, 2. Candidates ±1, ±2, ±1/2. The theorem finds nothing; it makes the search finite.
Testing
f(1) = 2 − 3 − 3 + 2 = −2, no. f(−1) = −2 − 3 + 3 + 2 = 0, yes. Test 1 and −1 first: each is only a sum of coefficients, with alternating signs for −1.
Synthetic division
Divide by (x + 1) using −1: bring down 2; 2 × (−1) = −2, so −3 − 2 = −5; −5 × (−1) = 5, so −3 + 5 = 2; 2 × (−1) = −2, so 2 − 2 = 0. Quotient 2x² − 5x + 2, remainder 0, as the Factor Theorem promised.
Finish the quadratic
2x² − 5x + 2 = (2x − 1)(x − 2). Roots 1/2 and 2. All three: −1, 1/2, 2. The 1/2 was on the candidate list because the leading coefficient was 2.
When the quadratic will not factor
x³ − 8 = (x − 2)(x² + 2x + 4). The quadratic has discriminant 4 − 16 = −12, so its roots are complex: x = −1 ± i√3. Three roots in total, one real, as a cubic must have.
A check that costs nothing
The roots of ax³ + bx² + cx + d sum to −b/a and multiply to −d/a. For 2x³ − 3x² − 3x + 2: sum 3/2 and −1 + 1/2 + 2 = 3/2; product −1 and (−1)(1/2)(2) = −1. Both agree.
Step 2: Try It Yourself
Tap and try it out.
Solve 2x³ − 3x² − 3x + 2 = 0
- 1candidates ±1, ±2, ±1/2factors of 2 over factors of 2
Step 0 of 4
Step 3: In Real Life
A box from a sheet
Cut squares of side x from a 20 by 30 sheet and fold a box of volume 1,000: x(20 − 2x)(30 − 2x) = 1,000. A cubic. Find one root, divide it out, and solve what is left.
Step 4: Watch an Example
One step at a time.
Watch Kofi Solve a Cubic
Kofi solves x³ − 4x² + x + 6 = 0.
- Step 1
The constant is 6 with leading coefficient 1, so the candidates are ±1, ±2, ±3 and ±6. Testing x = −1 gives −1 − 4 − 1 + 6 = 0, a root.
Step 5: Your Turn
Practice makes it stick.
The Candidates
Problem 1 of 2
For x³ + 2x² − 5x − 6, how many candidate rational roots does the theorem list, counting both signs?
The Test
Problem 2 of 2
f(2) = 0. Is (x − 2) a factor?
Hunt the Roots
1 of 8
x³ − x = 0. How many real roots?
2 of 8
x³ − 8 = 0. What is the real root?
3 of 8
f(3) = 0. Which factor does that give?
4 of 8
After dividing a cubic by a linear factor, what degree remains?
5 of 8
A cubic has roots 2, 3i and one more. Imaginary coefficient of the third?
6 of 8
x³ − 6x² + 11x − 6 factors as (x−1)(x−2)(x−3). Largest root?
7 of 8
Put the solving process in order.
- 1Test candidates until one gives zero.
- 2Divide the polynomial by that factor.
- 3Solve the smaller polynomial that remains.
- 4List the candidate rational roots.
8 of 8
x³ − 4x = 0. How many real roots?
Step 6: Quick Check
Show what you know.
Question 1 of 2
x³ − 9x = 0. How many real roots?
Question 2 of 2
What does the Rational Root Theorem actually provide?
What You Learned
- A root and a factor are the same fact, by the Factor Theorem.
- The Rational Root Theorem gives a finite list of candidates: constant factors over leading-coefficient factors.
- Divide out each root found and solve what remains; a non-factoring quadratic gives a complex pair.
- The roots sum to −b/a and multiply to ±d/a, a free check.