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.
Try it first
Someone has already found that x³ − 6x² + 5x + 12 = (x + 1)(x² − 7x + 12). Using that, solve x³ − 6x² + 5x + 12 = 0. What is the largest solution?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
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 Sam Solve x³ − 2x² − 5x + 6 = 0
A fresh cubic, solved the whole way: list, test, divide, finish the quadratic, then check.
- Step 1
The leading coefficient is 1 and the constant is 6, so every rational root has to be a factor of 6. With both signs, the candidates are ±1, ±2, ±3 and ±6: eight numbers, and nothing else is possible.
Watch Ravi Start, Then Finish It Yourself
Ravi solves x³ + 2x² − 5x − 6 = 0. He lists and tests; you divide, factor and check.
- Step 1
The constant is −6 and the leading coefficient is 1, so the candidates are ±1, ±2, ±3 and ±6.
Watch Noor Handle a Missing Term
Noor solves x³ − x² + 2 = 0. There is no x term, and the quadratic left over will not factor.
- Step 1
The cubic has no x term, so a 0 holds its place: the coefficients are 1, −1, 0, 2. Leaving the 0 out would slide every later number one place and give a quotient of the wrong degree.
Step 5: Your Turn
Practice makes it stick.
The Other Cut
Problem 1 of 17
A box is folded from a 20-inch by 30-inch sheet after a square of side x inches is cut from each corner. Its volume is x(20 − 2x)(30 − 2x) cubic inches, and it must hold 1,000 cubic inches. That equation tidies to x³ − 25x² + 150x − 250 = 0. A 5-inch cut works. What smaller cut also gives 1,000 cubic inches? Round to the nearest hundredth of an inch.
inches
The Storage Bin
Problem 2 of 17
A storage bin has a square base x feet on each side and is 2 feet taller than it is wide, so its volume is x²(x + 2) cubic feet. It must hold exactly 16 cubic feet. How long is each side of the base, in feet?
feet
Hunt the Roots
List, test, divide, then finish what is left.
Problem 3 of 17
f(2) = 0 for f(x) = x³ − 2x² − 9x + 18. Divide f by (x − 2). What is the quotient? Type it.
Problem 4 of 17
Solve x³ − 7x − 6 = 0. What is the largest solution?
Problem 5 of 17
For 3x³ − 2x² + 5x + 4, how many candidates does the Rational Root Theorem list, counting both signs?
Problem 6 of 17
Solve 2x³ − x² − 2x + 1 = 0. One solution is not a whole number. What is it? Give it as a decimal.
Problem 7 of 17
How many real solutions does x³ + 2x² + 2x + 1 = 0 have?
Whose Step?
One line in each solution is a slip. Find it, then fix it.
Problem 8 of 17
Ana solved x³ + 4x² + x − 6 = 0. One line is a slip. Which?
Ana's solution of x³ + 4x² + x − 6 = 0
- 1f(1) = 1 + 4 + 1 − 6 = 0
- 2synthetic division by 1: 1, 5, 6, remainder 0
- 3x² + 5x + 6 = (x + 2)(x + 3)
- Answerx = 1, 2, 3
Problem 9 of 17
What is the smallest solution of x³ + 4x² + x − 6 = 0, correctly?
Problem 10 of 17
Dev solved x³ − 13x + 12 = 0. One line is a slip. Which?
Dev's solution of x³ − 13x + 12 = 0
- 1f(1) = 1 − 13 + 12 = 0
- 21 | 1, −13, 12
- 3bottom row: 1, −12, remainder 0
- Answerx − 12 = 0, so x = 1, 12
Problem 11 of 17
What quotient should Dev get when he divides x³ − 13x + 12 by (x − 1)? Type it.
Problem 12 of 17
Maya looked for rational roots of 2x³ + x² + 2x + 1 = 0. One line is a slip. Which?
Maya's search for rational roots of 2x³ + x² + 2x + 1 = 0
- 1candidates ±1
- 2f(1) = 2 + 1 + 2 + 1 = 6
- 3f(−1) = −2 + 1 − 2 + 1 = −2
- Answerno rational roots
Problem 13 of 17
2x³ + x² + 2x + 1 = 0 does have one real root. What is it? Give it as a decimal.
Pick the Move
Decide the move before doing any of the work.
Problem 14 of 17
Leo wants to solve x³ − 6x² + 9x = 0. Which first move gets him furthest fastest?
Problem 15 of 17
Priya has found that f(2) = 0 for f(x) = x³ − 2x² − x + 2. Which move shrinks the cubic to a quadratic?
Same Root?
Judge without solving all the way.
Problem 16 of 17
Select every polynomial that shares the root x = 1 with x³ − 2x² − 5x + 6.
Problem 17 of 17
Select every polynomial with exactly the same roots as (x − 1)(x + 2)(x − 3).
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?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
For which number k is x = 2 a root of x³ + kx² − 4x + 4?
Stretch 2 of 2
A cubic x³ + bx² + cx + d has 2 and 3 among its roots and no x² term at all. Write it out in expanded form. Type the whole polynomial.
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.
For the grown-up
The misconception this lesson targets is reading a root off a factor with the sign kept: (x + 2) looks like "2", so the root "must" be 2, when it is −2. It sits under the Factor Theorem in both directions, so a student who finds f(3) = 0 then divides by (x + 3). Two smaller slips travel with it: a missing term written without its 0, which wrecks synthetic division silently, and a candidate list built from the constant alone, which loses every fractional root and leads to "no rational roots" when there is one. The sum-and-product check catches all three in a line, which is why the lesson ends every example with it.