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?
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.
Check‑your‑understanding:
Explain why x + 4 is a factor of g(x)=x³+2x²‑8x‑16 (show the single calculation required).
Statement:
For a polynomial with integer coefficients, any rational root written in lowest terms as \(\displaystyle \frac{p}{q}\) must satisfy:
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.
Synthetic division provides a quick way to compute both the remainder f(a) and the quotient q(x) when dividing by x − a.
Typical layout (vertical bar and horizontal line):
a | cn c{n-1} c{n-2} … c0 | |_ | d1 d2 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.
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.
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.
Problem: Factorise h(x)=x³‑6x²+11x‑6 completely.
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:
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
\[
h(x)=(x-1)(x-2)(x-3).
\]
| 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. |
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:
(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 \((x‑2)^{2}(x‑1)(x+1)\).)