For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: multiplication, division, remainder and factor theorems, cubic equations, and sum or difference of cubes form the main route.
School-sensitive extension: Vieta relations, cubic inequalities, and parameter-heavy factor problems may vary by school.
2027 national comparison: K341 Topic A4 covers polynomial multiplication and division, remainder and factor theorems, cubic equations, cube identities, and partial fractions.
Check your school: confirm whether synthetic division or Vieta methods are permitted or expected.
Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 09) Polynomials cover? A: Factor theorem, remainder theorem, and inequalities involving higher-degree polynomials for IP AMaths.
Polynomials underpin factorisation, curve sketching, and inequality solving. Memorise the theorems that convert substitution into proofs of factor status.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
The core idea is simple: Polynomial questions turn substitution into factor and remainder information.
Use it as a working check: Use f(a) = 0 to prove x - a is a factor, then factor fully before solving inequalities or finding roots.
Then go one layer deeper: Example: if f(2) = 0, divide by x - 2 first. The remaining quadratic often unlocks the full factorisation and the sign chart.
Choosing the polynomial route
Read the wording before expanding. Polynomial questions usually tell you whether substitution, division, coefficient comparison, or a sign chart should come first.
Question cue
Best first move
What to watch
"Show that x−a is a factor"
Substitute x=a and prove f(a)=0.
Do not divide until the zero remainder has been shown.
"Find the remainder when divided by x−a"
Evaluate f(a) directly.
The answer is a number, not a quotient.
A factor and a remainder condition are both given
Convert each condition into an equation, then solve simultaneously.
For x+a, substitute x=−a.
Roots are described but the full polynomial is unknown
Use Vieta relationships or build the factorised form first.
Check the leading coefficient before expanding.
A polynomial inequality is given
Factor fully, mark critical values, then test interval signs.
Repeated roots may touch the axis without changing sign.
Common trap:f(a)=0 proves x−a is a factor, not x+a. The sign inside the bracket is opposite to the substituted root.
1 Key results
Factor theorem: if f(a)=0, then (x−a) divides f(x).
Remainder theorem: dividing f(x) by x−a leaves remainder f(a).
For cubic ax3+bx2+cx+d, the sum of roots is −b/a, pairwise sum c/a
Multiply polynomials by distributing every term, and divide using long division or synthetic division where your school permits it.
Sum of cubes: a3+b3=(a+b)(a2−ab+b2)
Difference of cubes: a3−b3=(a−b)(a2+ab+b2)
The signs in the quadratic factor follow the pattern "same, opposite, positive": the first bracket uses the same sign as the cubes, the middle term uses the opposite sign, and the final square term is positive.
2 Worked example - Factorisation
Given f(x)=x3−4x2+x+6 and f(2)=0, factorise f(x).
Since f(2)=0, (x−2) is a factor.
Perform long division or equate: divide to obtain x2−2x−3.
Factor quadratic: x2−2x−3=(x−3)(x+1).
Hence f(x)=(x−2)(x−3)(x+1).
Quotient recovery checkpoint
Once a linear factor is known, recover the remaining polynomial before trying to solve the whole cubic. Coefficient comparison is a neat alternative to long division when the leading coefficient is simple.
Known information
Set up
Compare coefficients
Result
x−2 is a factor of x3−4x2+x+6
x3−4x2+x+6=(x−2)(x2+px+q)
Expand: x3+(p−2)x2+(q−2p)x−2q
Match constants: −2q=6, so q=−3.
Same setup
Use the x2 coefficient.
p−2=−4
p=−2
Check the middle coefficient
Substitute p=−2, q=−3.
q−2p=−3−2(−2)=1
Misconception check: f(2)=0 only proves the first factor. You still need division or coefficient comparison to find the remaining factor before listing all roots.
Sign-chart root behaviour checkpoint
Before testing intervals, mark each critical value with its root multiplicity. This tells you whether the graph crosses the axis or just touches it.
Factor pattern
Root behaviour
Sign-chart effect
Common trap
(x−a) appears once
The graph crosses the axis at x=a.
The sign changes across a.
Forgetting to switch signs after a simple root.
(x−a)2 appears
The graph touches the axis at x=a.
The sign usually stays the same across a.
Splitting the solution interval as if the sign changed.
(x−a)3 appears
The graph crosses with flattening at x=a.
The sign changes across a.
Treating every repeated root as a no-change root.
Inequality uses ≤ or ≥
Roots that make the expression zero may be included.
Use square brackets or include the exact root where appropriate.
Including roots for strict < or > inequalities.
Misconception check: the sign chart is not just alternating signs from left to right. Odd multiplicity changes sign; even multiplicity keeps the sign on both sides.
3 Worked example - Polynomial inequality
Solve x3−4x2+x+6≤0.
Using factorisation above, inequality is (x−2)(x−3)(x+1)≤0.
Critical points: x=−1,2,3.
Sign chart:
For x<−1, all factors negative → product negative → included.
Between −1 and 2, two factors negative, one positive → product positive → excluded.
Between 2 and 3
Include roots because inequality is non-strict.
Solution set: x≤−1 or 2≤x≤3.
4 Worked example - Using factor and remainder conditions
Let f(x)=x3−6x2+ax+b. Given that (x−2) is a factor and the remainder upon division by x+1 is 15, determine a and b, then factorise f(x).
