01 · A sign check
Is x + 3 a factor of x³ + 2x² − 5x − 6?
Hint
Evaluate at x = −3.
Worked solution
f(−3) = −27 + 18 + 15 − 6 = 0
Yes. In fact f(x) = (x + 3)(x + 1)(x − 2).
Understand · explore · practise
Factor theorem
Use the factor theorem to factorise cubics and quartics, solve polynomial equations and find unknown coefficients. Includes fractional roots, repeated roots and graph connections.
Before you startPolynomial division and solving quadratics
01 / The theorem
A root and a linear factor describe the same fact.
For a polynomial f(x), the factor theorem says:
f(p) = 0 ⇔ (x − p) is a factor of f(x)
Both directions matter. A known factor lets you calculate a zero; a known zero lets you extract a factor. A factor x + 2 corresponds to p = −2, not 2.
For f(x) = x³ − 2x² − 5x + 6, try the values in the model. A point on the x-axis identifies a factor. A non-zero result rules that candidate out.
Test a candidate rootExplore at your pace
f(x) = x³ − 2x² − 5x + 6. The graph crosses at −2, 1 and 3. At x = 0, f(0) = 6, so x is not a factor.
Watch roots identify linear factors
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Why it works
Substitute the divisor’s root into the division identity.
Division by x − p gives f(x) = (x − p)q(x) + r, with constant remainder r. Substitute x = p:
f(p) = (p − p)q(p) + r = r
This is the remainder theorem. If f(p) = 0, the remainder is zero, so x − p is a factor. Conversely, if x − p is a factor, substitution gives f(p) = 0.
A direct proof for any cubic
Let f(x) = ax³ + bx² + cx + d, with a ≠ 0, and suppose f(p) = 0. Subtract the zero f(p):
f(x) = f(x) − f(p)
= a(x³ − p³) + b(x² − p²) + c(x − p)
= (x − p)[a(x² + xp + p²)
+ b(x + p) + c]
This exhibits the factor directly. It does not rely on a numerical test of several x-values.
03 / Factorise fully
Find one root, divide, then factor the quotient.
For an integer-coefficient monic polynomial, integer roots must divide the constant term. This gives useful candidates, not a promise that an integer root exists.
For the main cubic, f(3) = 27 − 18 − 15 + 6 = 0. Divide by x − 3, then factor the resulting quadratic.
From one root to all threeWorked example
f(x) = x³ − 2x² − 5x + 6
We know x − 3 is a factor.
f(x) = (x − 3)(x² + x − 2)
Polynomial division supplies the quotient.
f(x) = (x − 3)(x + 2)(x − 1)
Factor the quadratic.
f(x) = 0 ⇒ x = −2, 1 or 3
All three factors are needed for the complete solution set.
04 / Fractional roots
For ax + b, substitute −b/a.
When a linear factor has leading coefficient a ≠ 0, solve ax + b = 0 to find the test value. In particular, test x = −1/2 for the factor 2x + 1.
g(x) = 2x³ − x² − 13x − 6
g(−1/2) = −1/4 − 1/4 + 13/2 − 6 = 0
g(x) = (2x + 1)(x² − x − 6)
= (2x + 1)(x − 3)(x + 2)
The roots are −1/2, 3 and −2. Test exact fractions; a rounded decimal may hide whether the remainder is truly zero.
Choosing rational candidates
If an integer-coefficient polynomial has a rational root p/q written in lowest terms, p divides its constant term and q divides its leading coefficient. This narrows trials. If the constant term is zero, first extract a factor x.
05 / Unknown coefficients
Each evaluation gives an equation in the coefficients.
If x − 1 and x + 2 are factors of f(x) = x³ + ax² + bx + 6, then f(1) = f(−2) = 0.
Two coefficients, two conditionsWorked example
1 + a + b + 6 = 0
a + b = −7
Use x = 1.
−8 + 4a − 2b + 6 = 0
2a − b = 1
Use x = −2 and simplify.
3a = −6 ⇒ a = −2
b = −5
Solve simultaneously.
f(x) = (x − 1)(x + 2)(x − 3)
Check both conditions and the complete factorisation.
06 / Real roots and graphs
A cubic need not have three distinct real roots.
After extracting a linear factor, examine the quadratic quotient. Use its discriminant if factorisation over the real numbers is unclear.
x³ − 2x² + 4x − 8
= (x − 2)(x² + 4)
x² + 4 is always positive for real x. The only real root is x = 2.
x³ + 3x² − 4 = (x + 2)²(x − 1)
There are two distinct real roots: −2 (repeated) and 1. The graph touches the axis at −2 and crosses at 1. Root multiplicity matters as well as the number of roots.
The main model’s three simple roots each give a crossing. Its y-intercept is f(0) = 6. Its positive cubic leading term puts the left end down and the right end up.
Exact surd roots
x³ − x² − 3x − 1
= (x + 1)(x² − 2x − 1)
= 0 ⇒ x = −1 or 1 ± √2
Only 1 + √2 is positive. A root need not be rational or an integer.
07 / Your turn
Find every root and check any conditions.
Give a factorisation that can be multiplied back. Keep repeated roots and exact surds clear.
02 · Factor and solve
x³ − 4x² + x + 6 = 0
Hint
x = 2 is a root. Divide by x − 2.
Worked solution
(x − 2)(x² − 2x − 3) = 0
(x − 2)(x − 3)(x + 1) = 0
x = −1, 2, 3
03 · A fractional root
Show 2x + 1 is a factor of 2x³ − x² − 13x − 6, then solve the cubic.
Hint
Use x = −1/2, then divide by 2x + 1.
Worked solution
f(−1/2) = 0
f(x) = (2x + 1)(x² − x − 6)
= (2x + 1)(x − 3)(x + 2)
Roots: −1/2, 3, −2.
04 · A non-zero evaluation
f(x) = x³ + px + q, f(−1) = 0 and f(2) = 15. Find p and q, then all real roots.
Hint
−1 − p + q = 0 and 8 + 2p + q = 15.
Worked solution
q − p = 1, 2p + q = 7
p = 2, q = 3
f(x) = (x + 1)(x² − x + 3)
The quadratic discriminant is 1 − 12 = −11. The only real root is −1.
05 · A quartic
Solve 2x⁴ − 5x³ − x² + 6x = 0.
Hint
Extract x; test x = 2 in the remaining cubic.
Worked solution
x(2x³ − 5x² − x + 6)
= x(x − 2)(2x² − x − 3)
= x(x − 2)(x + 1)(2x − 3)
Roots: 0, 2, −1, 3/2.
06 · A repeated root
Factor x³ + 3x² − 4 and state the distinct real roots.
Hint
x = 1 is a root. Divide and inspect the quotient.
Worked solution
(x − 1)(x² + 4x + 4)
= (x − 1)(x + 2)²
The distinct roots are 1 and −2; −2 has multiplicity two.
07 · Surd roots
Solve x³ − x² − 3x − 1 = 0 exactly.
Hint
x = −1 is a root.
Worked solution
(x + 1)(x² − 2x − 1) = 0
x = −1 or x = 1 ± √2
08 · Rule out further real roots
Show that x³ − 2x² + 4x − 8 = 0 has exactly one real root.
Hint
Test x = 2 and factor.
Worked solution
(x − 2)(x² + 4) = 0
Since x² + 4 ≥ 4 > 0 for real x, it cannot vanish. Thus x = 2 is the unique real root.
08 / Recap
A zero remainder gives a factor.
- f(p) = 0 exactly when x − p is a factor.
- For ax + b, test x = −b/a.
- Divide out one factor and continue with the quotient.
- Check the quadratic discriminant and repeated roots.
- Use each given evaluation to form an equation in unknown coefficients.
Section 1 of 8 · The theorem