Factors of Polynomials

This is the shortest major topic in 0606 and the most reliably scoring: question patterns barely change between sessions, and a drilled student should target full marks here every time. The price of entry is care with signs.

The remainder theorem

When a polynomial $p(x)$ is divided by $(x - a)$, the remainder is $p(a)$, no division required. Divided by $(ax - b)$, the remainder is $p\left(\frac{b}{a}\right)$. The exam line that earns the method mark is the substitution statement itself:

$p(x) = 2x^3 - 3x^2 + 4x - 5$, divided by $(x - 2)$: remainder $= p(2) = 16 - 12 + 8 - 5 =$ $7$

Write “remainder $= p(2)$” before the arithmetic, that’s the M mark. Typical variants: finding unknown coefficients from given remainders (“when divided by $(x + 1)$ the remainder is $6$”), substitute, form an equation, solve. With two conditions you get simultaneous equations in the unknowns.

The factor theorem

The special case that powers everything: $(x - a)$ is a factor of $p(x) \iff p(a) = 0$. Uses:

• Show $(x - 3)$ is a factor: compute $p(3)$, show it’s $0$, and say so. ”$p(3) = 0$, therefore $(x - 3)$ is a factor” (the sentence is a mark; this is a show-that command)
• Find $k$ given a factor: set $p(a) = 0$, solve for $k$
• Factorise a cubic: find the first root by trial (test $x = \pm 1, \pm 2, \pm 3$, and factors of the constant term)

The sign trap: for factor $(x + 2)$, substitute $x = -2$. For $(2x - 1)$, substitute $x = \frac{1}{2}$. Slowing down on exactly this line is worth marks every session.

Factorising and solving cubics: the full routine

Solve $2x^3 - x^2 - 13x - 6 = 0$.

1. Find one root by trial. Try $x = -2$: $2(-8) - 4 + 26 - 6 = 0$ ✓ So $(x + 2)$ is a factor (state the theorem conclusion).
2. Extract the quadratic, by comparing coefficients (or long division if you prefer): $2x^3 - x^2 - 13x - 6 = (x + 2)(2x^2 + bx - 3)$. Expanding gives $2x^3 + (b + 4)x^2 + (2b - 3)x - 6$. Matching $x^2$ terms: $b + 4 = -1 \to$ $b = -5$. Verify with the $x$ term: $2(-5) - 3 = -13$ ✓ (always check the middle term, free error detection).
3. So $(x + 2)(2x^2 - 5x - 3) = 0 \to (x + 2)(2x + 1)(x - 3) = 0$
4. $x = -2$, $x = -\frac{1}{2}$, $x = 3$, all three roots stated.

Each numbered step is a marking point: root found with theorem cited (M, A), quadratic factor correct (M, A), final factorisation and all solutions (A). The verify-the-middle-term habit in step 2 catches nearly every slip before it costs anything.

Where this topic connects

Cubic factorising feeds cubic inequalities and graphical solutions; the “find unknown coefficients” pattern reuses quadratic and simultaneous-equation machinery; and on the non-calculator Paper 1 the arithmetic is hand-friendly by design, ugly numbers mean a wrong root.

Common mistakes in this topic

• Substituting $+a$ for factor $(x + a)$, the topic’s signature error
• Computing $p(a)$ correctly but never writing the conclusion sentence for “show that” questions
• Coefficient slips when extracting the quadratic, fix: verify against the middle term before moving on
• Stopping after factorising when the question said solve (all roots required)
• Trial roots chosen randomly instead of from factors of the constant term

A topic this mechanical should be a guaranteed 6-8 marks. If it isn’t yet, one focused session sorts it, free 1-hour trial class with Teacher Rig, booked on WhatsApp.