Secondary 3 Additional Mathematics | Singapore G3
Polynomials
How One Remainder Can Tell You Whether a Factor Exists
If a polynomial P(x) is divided by x − a, the remainder is P(a). So one substitution can tell you what a full polynomial division would leave behind.
This is one of the first places in Additional Mathematics where a calculation becomes a theorem.
Instead of dividing an entire cubic by x − 2, you may be able to evaluate P(2). Instead of guessing whether x + 3 is a factor, you can test P(−3). Instead of attacking a cubic equation all at once, one known root can reduce it to a quadratic you already know how to solve.
The real topic is not “polynomial long division.” It is learning how the value of a polynomial at one input reveals divisibility structure.
The current Singapore G3 Additional Mathematics syllabus includes multiplication and division of polynomials, remainder and factor theorems, factorising polynomials, solving cubic equations, and the identities for sums and differences of cubes. Partial fractions sits in the same syllabus section, but it deserves its own teaching owner rather than being mixed into this article.
SEAB 2027 G3 Additional Mathematics syllabus (K341) →
The Topic Job
Use polynomial structure to move between values, factors, roots and lower-degree problems.
A strong student should be able to see four statements as connected:
- P(a) = 0;
- x − a is a factor of P(x);
- x = a is a root of P(x) = 0;
- the graph y = P(x) meets the x-axis at x = a.
These are different languages describing the same mathematical event.
What Is a Polynomial?
A polynomial in x is an expression built from non-negative integer powers of x with numerical coefficients.
Examples:
- 4x³ − 7x + 2;
- x⁵ + 3x² − 1;
- 6;
- 2x⁴ − x³ + 8x² + x − 9.
Not polynomials in x:
- 1/x, because x has power −1;
- √x, because x has power 1/2;
- 2ˣ, because x is in the exponent.
Degree Tells You the Highest Power
For:
5x⁴ − 2x³ + 7,
the degree is 4.
Degree matters because division, possible root counts and graph behaviour depend on it. Dividing a degree-4 polynomial by a degree-1 polynomial leaves a quotient of degree 3.
Polynomial Division Is Ordinary Division With Powers
Every division statement has the structure:
dividend = divisor × quotient + remainder.
For numbers:
17 = 5(3) + 2.
For polynomials:
P(x) = (x − a)Q(x) + r,
where r is a constant because the remainder must have lower degree than the linear divisor.
Where the Remainder Theorem Comes From
Start with:
P(x) = (x − a)Q(x) + r.
Now substitute x = a:
P(a) = (a − a)Q(a) + r.
Since a − a = 0:
P(a) = r.
That is the Remainder Theorem.
When P(x) is divided by x − a, the remainder is P(a).
The theorem is not magic. The divisor disappears because we deliberately substitute the value that makes it zero.
Worked Example 1 — Find a Remainder Without Dividing
Find the remainder when:
P(x) = 2x³ − 5x² + 4x − 7
is divided by x − 2.
Use P(2):
P(2) = 2(8) − 5(4) + 4(2) − 7
= 16 − 20 + 8 − 7
= −3.
The remainder is −3.
No long division was required.
Why x + 3 Means Substitute −3
The theorem is written for a divisor x − a.
If the divisor is:
x + 3,
rewrite mentally as:
x − (−3).
So substitute x = −3.
Do not memorise “opposite sign” without meaning. Ask:
What value makes the divisor equal to zero?
The Factor Theorem Is the Zero-Remainder Case
If P(a) = 0, then the remainder on division by x − a is zero.
Therefore x − a divides P(x) exactly.
x − a is a factor of P(x) if and only if P(a) = 0.
The Factor Theorem is therefore not a separate unrelated rule. It is the Remainder Theorem with remainder zero.
Worked Example 2 — Test a Factor
Show that x − 2 is a factor of:
P(x) = x³ − 3x² − 4x + 12.
Evaluate P(2):
8 − 12 − 8 + 12 = 0.
Therefore, by the Factor Theorem:
x − 2 is a factor.
A Known Factor Reduces a Cubic to a Quadratic
Once we know x − 2 is a factor, divide the cubic by x − 2:
x³ − 3x² − 4x + 12 = (x − 2)(x² − x − 6).
Then factorise the quadratic:
(x − 2)(x − 3)(x + 2).
So the cubic equation:
x³ − 3x² − 4x + 12 = 0
has roots:
x = 2, 3, −2.
The key move was not “solve a cubic.” It was:
find one root → extract one linear factor → reduce the remaining problem to a quadratic.
How Do You Find the First Root?
In school questions, a factor may be given, a condition may determine one, or a simple integer root may be discoverable from likely candidates.
If P(x) has integer coefficients and leading coefficient 1, any integer root must divide the constant term.
For:
x³ − 5x² − 2x + 24,
possible integer roots include ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24.
You should not test them randomly forever. Use signs, size and any clue in the question to narrow candidates.
When the Divisor Is ax + b
If the divisor is:
2x − 3,
set it equal to zero:
2x − 3 = 0 → x = 3/2.
The remainder is therefore P(3/2).
The theorem is not fundamentally about “x − a.” It is about substituting the root of the linear divisor.
Unknown Coefficients From Remainder Conditions
Suppose:
P(x) = x³ + ax² + bx + 6.
You are told:
- x − 1 is a factor;
- the remainder on division by x + 2 is 12.
Translate each statement.
Factor condition:
P(1) = 0.
Remainder condition:
P(−2) = 12.
Those become two simultaneous equations in a and b.
This is a central exam habit:
translate English divisibility information into substitution equations before doing algebra.
Polynomial Long Division: Keep Every Power in Place
Consider dividing:
x³ + 5x − 4
by x − 2.
The x² term is missing. Write the polynomial mentally as:
x³ + 0x² + 5x − 4.
This placeholder prevents columns from shifting.
The long-division cycle is:
- divide leading term by leading term;
- write the quotient term;
- multiply back;
- subtract the whole line;
- bring down the next term;
- repeat.
Most long-division errors are subtraction or alignment errors, not polynomial-concept errors.
Synthetic Division: Useful, but Know What It Is Doing
Some schools teach synthetic division as a compact method for division by x − a.
It can be efficient, but it should not become a mysterious table. Every step is a compressed version of polynomial long division using coefficients.
If the student cannot explain why the final number is the remainder, return to:
P(x) = (x − a)Q(x) + r.
The Sum and Difference of Cubes
The syllabus includes:
a³ + b³ = (a + b)(a² − ab + b²)
and:
a³ − b³ = (a − b)(a² + ab + b²).
Do not remember only the sign pattern. Verify by expansion.
Example:
x³ − 27 = x³ − 3³ = (x − 3)(x² + 3x + 9).
Notice that x² + 3x + 9 does not factorise over the real numbers because its discriminant is negative.
Why x³ + 8 Has a Factor x + 2
Use the Factor Theorem:
P(x) = x³ + 8.
P(−2) = −8 + 8 = 0.
So x + 2 is a factor.
The sum-of-cubes identity then tells us the complete factorisation:
x³ + 8 = (x + 2)(x² − 2x + 4).
The identity and Factor Theorem are consistent views of the same structure.
Repeated Factors
If:
P(x) = (x − 2)²(x + 1),
x = 2 is a repeated root.
The graph touches the x-axis there rather than simply crossing in the ordinary way.
At later levels, derivatives provide another test: repeated root a often satisfies both P(a) = 0 and P′(a) = 0. You do not need that method yet, but it shows how polynomial structure connects forward into calculus.
Remainders From Different Divisors Give Different Information
Knowing P(1) = 4 tells you the remainder when dividing by x − 1.
Knowing P(2) = 0 tells you x − 2 is a factor.
Knowing P(−3) = 7 tells you the remainder on division by x + 3 is 7.
Each value is a small receipt from the polynomial at one input. Several such receipts can determine unknown coefficients.
The Graph Connection
If P(a) = 0, then y = P(x) passes through (a, 0).
So:
root ↔ x-intercept ↔ zero of function ↔ linear factor.
This connection becomes useful when a graph is supplied and you are asked to infer factors, or when factorisation is used to sketch sign changes.
Do Not Confuse a Factor With a Root
x − 4 is a factor.
x = 4 is a root.
They correspond, but they are different mathematical objects.
This matters in explanations. “4 is a factor” is incorrect language.
The Earliest Weak Link
| What you see | Likely weak link |
|---|---|
| Uses P(3) for divisor x + 3 | does not solve divisor = 0 |
| Says P(2) = 0 but cannot state the factor | root–factor translation weak |
| Long division columns collapse | missing zero-coefficient placeholders / subtraction control |
| Finds one cubic root and stops | does not reduce remaining polynomial |
| Tests many integer roots blindly | no candidate-selection strategy |
| Uses remainder theorem but cannot explain it | does not understand dividend = divisor × quotient + remainder |
Common Mistakes to Repair
- Wrong substitution sign. Solve the divisor equal to zero.
- Confusing factor and root. x − a is the factor; a is the root.
- Dropping missing powers. Use zero coefficients during division.
- Subtracting only the first term in long division. Subtract the entire product line.
- Stopping after one factor. Continue until the remaining polynomial is solved or irreducible at the required level.
- Forgetting the factor theorem requires remainder zero. P(a) = 5 means x − a is not a factor.
- Memorising cube identities with mixed signs. Expand to verify.
Retrieval Check
- State the Remainder Theorem.
- State the Factor Theorem.
- What value should be substituted for divisor x + 5?
- Find the remainder when x³ − 4x + 7 is divided by x − 2.
- Show that x + 1 is a factor of x³ − 4x² + x + 6.
- Factorise x³ − 64.
Transfer Set
- P(x) = 2x³ + kx² − 5x + 6. Given that x − 2 is a factor, find k.
- P(x) = x³ + ax + b. The remainder when divided by x − 1 is 4 and x + 2 is a factor. Find a and b.
- Solve x³ − 2x² − 5x + 6 = 0, given that x − 1 is a factor.
- Find the remainder when 3x⁴ − x² + 5 is divided by 2x + 1.
- Factorise completely over the real numbers: 8x³ − 27.
Answer outline — open only after attempting
- P(2)=16+4k−10+6=0 → 12+4k=0 → k=−3.
- P(1)=1+a+b=4 → a+b=3. P(−2)=−8−2a+b=0 → b−2a=8. Solve: a=−5/3, b=14/3.
- Divide by x−1: quotient x²−x−6=(x−3)(x+2). Roots 1, 3, −2.
- 2x+1=0 → x=−1/2. Remainder P(−1/2)=3/16−1/4+5=79/16.
- (2x)³−3³=(2x−3)(4x²+6x+9). Quadratic discriminant 36−144<0, so complete over reals.
Independent Checks
- After factorising, multiply factors back.
- After finding a root, substitute it into the original polynomial.
- After division, verify P(x) = divisor × quotient + remainder.
- If a factor condition was used to find a parameter, re-evaluate P(a) at the end.
A check is strongest when it reconstructs the original claim in a different direction.
For Parents and Tutors — What This Topic Is Really Testing
Polynomial questions reveal whether a student can translate between forms of information.
A student may be perfectly able to substitute numbers but still fail because they do not translate:
- “x − 3 is a factor” → P(3) = 0;
- “remainder is 5 when divided by x + 2” → P(−2) = 5;
- “x = 4 is a root” → x − 4 is a factor;
- “one root is known” → divide and reduce the degree.
The diagnostic target is not whether the child remembers the theorem name. It is whether they can move from sentence → substitution → factor structure → reduced problem.
If long division repeatedly fails, repair alignment and subtraction separately. If theorem questions fail, repair the meaning of the linear divisor before assigning more polynomial practice.
Where This Connects Next
- Secondary 3 Additional Mathematics Topics Explained: The Complete A-Math Map
- Additional Mathematics Directory
- Secondary Math Tuition | Sec 3 Additional Mathematics Tutor
- Additional Mathematics | A Difficult Question Is Often Several Easy Ideas Joined Together
- Partial Fractions
Bukit Timah Tutor Mathematics
One substitution can reveal one remainder. One zero can reveal one factor. One factor can turn a cubic into a problem you already know how to solve.