Factors of Polynomials – Remainder and Factor Theorems
Learning Objectives
- State the Remainder Theorem and the Factor Theorem exactly as they appear in the Cambridge IGCSE Additional Mathematics (0606) syllabus.
- Apply each theorem to:
- Find the remainder when a polynomial is divided by a linear divisor.
- Test whether a linear expression is a factor of a polynomial.
- Factorise higher‑degree polynomials (cubic and quartic).
- Use synthetic (Horner’s) division efficiently to obtain both the remainder and the quotient when dividing by a linear factor.
- Identify and avoid the most common errors when working with these theorems.
3.1 Remainder Theorem – Formal Statement
Remainder Theorem (Cambridge wording):
If a polynomial f(x) is divided by the linear divisor x − a, then the remainder is f(a).
In symbolic form:
\[ f(x) = (x-a)\,q(x) + f(a) \]
where q(x) is the quotient polynomial.
Note: The theorem applies only when the divisor is written in the form x − a. If you are given a divisor such as a − x or 2 − x, rewrite it as x − a (e.g., a − x = -(x − a)) before using the theorem.
Check‑your‑understanding:
If f(x)=3x³‑4x²+5 is divided by x‑2, what is the remainder without performing any division?
3.2 Factor Theorem – Formal Statement
Factor Theorem (Cambridge wording):
The linear expression x − a is a factor of the polynomial f(x) iff f(a)=0.
Consequently, x − a is a factor ⇔ the remainder on division by x − a is zero.
How to use the Factor Theorem – Checklist
- Write the divisor in the form x − a (adjust the sign if necessary).
- Substitute a into the polynomial to evaluate f(a).
- If f(a)=0, then x − a is a factor; otherwise it is not.
Check‑your‑understanding:
Explain why x + 4 is a factor of g(x)=x³+2x²‑8x‑16 (show the single calculation required).
3.3 Rational Root Theorem (IGCSE level)
Statement:
For a polynomial with integer coefficients, any rational root written in lowest terms as \(\displaystyle \frac{p}{q}\) must satisfy:
- \(p\) divides the constant term.
- \(q\) divides the leading coefficient.
Hence the list of possible rational roots can be obtained by forming all fractions \(\pm\frac{p}{q}\) from those divisors.
In many IGCSE problems the leading coefficient is 1, so the possible rational roots are simply the factors of the constant term.
3.4 Synthetic Division (Horner’s Method)
Synthetic division provides a quick way to compute both the remainder f(a) and the quotient q(x) when dividing by x − a.
- Write the coefficients of f(x) in descending order of degree, inserting a zero for any missing term.
- Place the value a (the number that makes the divisor zero) to the left of a vertical bar.
- Bring the leading coefficient straight down – this becomes the first coefficient of the quotient.
- Multiply the number just written down by a, write the product under the next coefficient, and add vertically.
- Repeat step 4 until the last coefficient is processed.
- The final number on the right is the remainder f(a). All numbers above it (except the leading coefficient) are the coefficients of the quotient q(x).
Typical layout (vertical bar and horizontal line):
a | c_n c_{n-1} c_{n-2} … c_0
|
|_______________________________
| d_1 d_2 d_3 … R
|
Where the row d₁, d₂, … gives the coefficients of the quotient and R is the remainder.
Check‑your‑understanding:
Perform synthetic division for the polynomial f(x)=2x³‑3x²+4x‑5 with divisor x‑1. State the quotient and remainder.
3.5 Worked Example – Using the Remainder Theorem
Problem: Find the remainder when f(x)=2x³‑5x²+3x‑7 is divided by x‑2.
According to the Remainder Theorem, the remainder is f(2):
\[
\begin{aligned}
f(2) &= 2(2)^{3} - 5(2)^{2} + 3(2) - 7 \\
&= 2\cdot8 - 5\cdot4 + 6 - 7 \\
&= 16 - 20 + 6 - 7 \\
&= -5 .
\end{aligned}
\]
Hence the remainder is −5.
3.6 Worked Example – Using the Factor Theorem
Problem: Determine whether x + 3 is a factor of g(x)=x⁴+2x³−x²−6x−9.
Rewrite the divisor as x − (‑3) so that a = –3. Evaluate:
\[
\begin{aligned}
g(-3) &= (-3)^{4}+2(-3)^{3}-(-3)^{2}-6(-3)-9 \\
&= 81 - 54 - 9 + 18 - 9 \\
&= 27 .
\end{aligned}
\]
Since g(‑3) ≠ 0, x + 3 is **not** a factor.
3.7 Factoring a Cubic Using the Factor Theorem
Problem: Factorise h(x)=x³‑6x²+11x‑6 completely.
- Find a possible root.
The constant term is ‑6 and the leading coefficient is 1, so by the Rational Root Theorem the only possible rational roots are
\(\displaystyle \pm1,\;\pm2,\;\pm3,\;\pm6\).
Test them:
- h(1)=1‑6+11‑6=0 ⇒ x‑1 is a factor.
- Perform synthetic division with a = 1.
1 | 1 -6 11 -6
|
|_________________________
| 1 -5 6 0
|
Quotient: x²‑5x+6. The remainder is 0, confirming that x‑1 is indeed a factor.
Factor the quadratic:
\[
x^{2}-5x+6=(x-2)(x-3).
\]
Therefore
\[
h(x)=(x-1)(x-2)(x-3).
\]
3.8 Common Mistakes to Avoid
- Divisor sign. The theorems apply only to a divisor written as x − a. If you start with a − x or 2‑x, rewrite it first.
- Missing zero coefficients. When a term is absent (e.g., no x³ term), insert a zero in the synthetic‑division table; otherwise the alignment of coefficients will be wrong.
- Sign errors in evaluation. Substitute carefully; a misplaced minus sign gives an incorrect remainder.
- Assuming only integers can be factors. Rational roots such as ±½, ±⅓ etc. are also possible – use the Rational Root Theorem to list all candidates.
3.9 Summary Table
| Theorem |
Formal Statement (Cambridge) |
How to Apply |
| Remainder Theorem |
If f(x) is divided by x‑a, the remainder is f(a). |
Write the divisor as x‑a and substitute a into the polynomial (or use synthetic division). |
| Factor Theorem |
x‑a is a factor of f(x) iff f(a)=0. |
1) Write divisor as x‑a. 2) Compute f(a). 3) If the result is 0, the divisor is a factor; otherwise it is not. |
3.10 Practice Questions
- Find the remainder when p(x)=4x⁴‑3x³+2x‑5 is divided by x+1.
- Show that x‑2 is a factor of q(x)=x³‑4x²+5x‑2 and factorise q(x) completely.
- Use synthetic division to divide r(x)=2x³+7x²‑x+6 by x‑3. State the quotient and remainder.
- Determine all linear factors of s(x)=x⁴‑5x³+6x²+4x‑8 (use the Factor Theorem and synthetic division). Express the final factorisation.
3.11 Answers to Practice Questions
-
Remainder:
\[
p(-1)=4(-1)^{4}-3(-1)^{3}+2(-1)-5=4+3-2-5=0.
\]
The remainder is **0**, so x+1 is a factor.
-
Verification:
\[
q(2)=2^{3}-4\cdot2^{2}+5\cdot2-2=8-16+10-2=0,
\]
therefore x‑2 is a factor.
Synthetic division (a=2):
2 | 1 -4 5 -2
|
|_______________________
| 2 -4 2 0
|
Quotient: x²‑2x+1 = \((x‑1)^{2}\).
Complete factorisation: \[
q(x)=(x‑2)(x‑1)^{2}.
\]
-
Synthetic division (a=3):
3 | 2 7 -1 6
|
|_______________________
| 6 39 114 120
|
Quotient: 2x²+13x+38 Remainder: 120.
Hence
\[
r(x)=(x‑3)(2x^{2}+13x+38)+120.
\]
-
Possible rational roots: ±1, ±2, ±4, ±8.
Evaluations:
\[
s(2)=0,\qquad s(-1)=0,\qquad s(4)=0.
\]
Perform successive synthetic divisions:
- Divide by x‑2 (a=2) → quotient \(x^{3}-3x^{2}+0x-4\).
- Divide the new cubic by x+1 (a=‑1) → quotient \(x^{2}-4x+4\).
- Factor the quadratic: \(x^{2}-4x+4=(x‑2)^{2}\).
(Alternatively, divide once more by x‑4 (a=4) to obtain the linear factor x‑4 and the remaining factor x‑1.)
Final factorisation (distinct linear factors as required by the syllabus):
\[
s(x)=(x‑2)(x+1)(x‑4)(x‑1).
\]
(If multiplicities are needed, the factorisation can be written as \((x‑2)^{2}(x‑1)(x+1)\).)