Algebra Revision Notes — 9709 Paper 2

01

The modulus function

Syllabus requirement · §2.1

“

understand the meaning of |x|, sketch the graph of y = |ax + b|. Graphs of y = |f(x)| and y = f(|x|) for non-linear functions f are not included.

”

$|x|$ means "the size of $x$ , ignoring the sign": $|5| = 5$ and $|-5| = 5$ . So $|x|$ is never negative, and two different inputs can produce the same output — which is exactly why modulus equations have two cases.

Draw the line first, then fold everything below the x-axis upwards. The corner of the V sits where the expression inside the modulus is zero.

Sketching y = |ax + b|

1
Draw $y = ax + b$ lightly — the ordinary straight line.
2
Find where it crosses the $x$ -axis, at $x = -\dfrac{b}{a}$ . That point is the corner of the V.
3
Reflect the part below the axis upwards. Everything above stays put.
4
Mark the intercepts: the $y$ -intercept is $|b|$ , and the corner is on the $x$ -axis.
Sketch questions are usually 2 marks: one for the shape, one for the intercepts. Label them.

What is not examined

The syllabus is explicit: graphs of $y = |\mathrm{f}(x)|$ and $y = \mathrm{f}(|x|)$ for non-linear $\mathrm{f}$ are not included. In Paper 2 the thing inside the modulus is always linear.

Your turn

One with a positive x-coefficient, one with a negative one — the fold direction is the same either way, but the corner and intercept land in different places.

19709/22 F/M 2024 Q2(a)2 marks
Sketch the graph of $y = |3x-7|$ , stating the coordinates of the points where the graph meets the axes.
Stuck? Show hint
Find where 3x − 7 = 0 first — that is the corner. The y-intercept is just |{-7}|.
Show solution
y = |3x − 7|: the V-corner sits where 3x − 7 = 0, i.e. at x = ⁷⁄₃.
1. 1
  Draw the underlying line $y = 3x-7$ lightly: gradient $3$ , $y$ -intercept $-7$ .
2. 2
  Find where it crosses the $x$ -axis: $3x-7=0 \;\Longrightarrow\; x=\tfrac73$ That is the corner of the V.
3. 3
  Reflect the part of the line below the $x$ -axis upwards. Everything from $x=0$ to $x=\tfrac73$ was negative, so that whole stretch flips.
  Only the part below the axis moves — the line was already positive for x > 7/3, and that part stays exactly where it is.
4. 4
  Read off the axis crossings on the folded graph: the $x$ -intercept is still $\left(\tfrac73,0\right)$ , and the $y$ -intercept is now $\left(0,\left\lvert-7\right\rvert\right)=(0,7)$ — positive, because it has been reflected up.
Answer
V-shaped graph, corner at $\left(\tfrac73,0\right)$ , meeting the $y$ -axis at $(0,7)$
2
Sketch the graph of $y=|4-3x|$ , stating the coordinates of the points where the graph meets the axes.
Stuck? Show hint
The x-coefficient is negative this time — find the corner the same way regardless.
Show solution
1. 1
  Draw $y=4-3x$ lightly: gradient $-3$ , $y$ -intercept $4$ .
2. 2
  Find where it crosses the $x$ -axis: $4-3x=0 \;\Longrightarrow\; x=\tfrac43$
3. 3
  This line is positive for $x<\tfrac43$ and negative for $x>\tfrac43$ — the opposite way round from a positive-gradient line. Reflect the negative part, $x>\tfrac43$ , upwards.
  The sign of the coefficient decides which side of the corner gets folded, not whether a fold happens at all — every y = |ax+b| graph is a V with two rising arms.
4. 4
  Read off the intercepts on the folded graph: $x$ -intercept $\left(\tfrac43,0\right)$ , $y$ -intercept unchanged at $(0,4)$ since that part of the line was never negative.
Answer
V-shaped graph, corner at $\left(\tfrac43,0\right)$ , meeting the $y$ -axis at $(0,4)$

Practise sketching y = |ax + b|Real past-paper questions · Modulus function: graph of y = |ax + b| and solving modulus equations/inequalities

02

Solving modulus equations and inequalities

Syllabus requirement · §2.1

“

use relations such as |a| = |b| ⟺ a² = b² and |x − a| < b ⟺ a − b < x < a + b when solving equations and inequalities, e.g. |3x − 2| = |2x + 7|, 2x + 5 < |x + 1|.

”

Because the modulus hides a sign, every equation containing one has to be dealt with in a way that covers both possibilities. The syllabus names two tools, and each suits a different shape of question.

The two relations named in the syllabus

|a| = |b| \iff a^2 = b^2

Squaring removes both moduli at once — use it when BOTH sides have one

|x-a| < b \iff a-b < x < a+b

Reads as “x is within b of a” — use it for a simple inequality

Two V-shaped graphs. Their crossings are the solutions, which is why an equation with a modulus on each side normally has exactly two.

The two-case method

1
Case 1: assume the inside is positive. Drop the modulus bars and solve.
2
Case 2: assume the inside is negative. Replace the contents with its negative, then solve.
3
Check each answer back in the original equation.
“Check each solution against its case condition” is a recorded mark-scheme technique — squaring and case-splitting can both manufacture solutions that do not work.

Worked example9709/22 F/M 2022 Q13 marks

Solve the equation $|5x - 2| = |4x + 9|$ .

Show full working

1
Case 1 — the insides are equal. Drop both moduli: $5x - 2 = 4x + 9 \;\Rightarrow\; x = 11$
2
Case 2 — the insides are opposite in sign. Negate one side: $5x - 2 = -(4x + 9) \;\Rightarrow\; 9x = -7 \;\Rightarrow\; x = -\tfrac79$
The mark scheme's second mark is exactly this: “attempt solution of linear equation where signs of 5x and 4x are different”.
3
Or square instead, which handles both cases at once: $(5x-2)^2 = (4x+9)^2 \;\Rightarrow\; 9x^2 - 92x - 77 = 0$ and factorising gives the same two roots.

Answer

$x = 11$ and $x = -\dfrac{7}{9}$

The published mark scheme accepts either route in full. Squaring is usually quicker when there is a modulus on both sides; the two-case method is safer when only one side has one.

Worked example9709/21 O/N 2022 Q14 marks

Solve the inequality $|2x - 5| > x$ .

Show full working

1
Find the critical values first, by solving the corresponding equation $|2x-5| = x$ .
The mark scheme awards its first two marks for the two linear solutions, before any inequality reasoning.
2
Case 1: $2x - 5 = x$ , giving $x = 5$ .
3
Case 2: $2x - 5 = -x$ , giving $3x = 5$ and $x = \tfrac53$ .
4
Now decide which side of each critical value works. The V-shaped graph of $y = |2x-5|$ sits above the line $y = x$ outside the two crossing points, so $x < \tfrac53 \quad\text{or}\quad x > 5$
Sketching the V and the line takes seconds and settles the direction without any case bookkeeping.

Answer

$x < \dfrac{5}{3}$ or $x > 5$

The mark scheme is explicit: "must be 2 separate inequalities". Writing $\tfrac53 > x > 5$ — which is impossible — or joining them with "and" loses the final mark.

Squaring is always safe when both sides are moduli

For $|3x-2| = |2x+7|$ above, squaring works because both sides are already $\geqslant 0$ — a modulus can never be negative, so squaring cannot flip the inequality or manufacture a false direction.

The same is true for a strict inequality between two moduli: $|5x+7| > |2x-3|$ can be squared immediately, with no case-by-case justification needed, because $\left(\left\lvert A\right\rvert\right)^2 = A^2$ always. That is a stronger guarantee than " $2x+5 < |x+1|$ " gets, where only one side is a modulus and the other, $2x+5$ , is not known to be non-negative — there, squaring first requires checking the sign, or the case method is safer.

Your turn

One equation by squaring, one inequality with a modulus on both sides (safe to square outright), and one inequality with only one side a modulus (where a case-derived value can turn out spurious).

19709/25 O/N 2025 Q3(a)3 marks
Solve the equation $|2x-3| = |5x+2|$ .
Stuck? Show hint
Two cases: the insides equal, or the insides opposite in sign.
Show solution
1. 1
  Case 1 — the insides are equal. Drop both moduli: $2x-3 = 5x+2 \;\Longrightarrow\; -3x=5 \;\Longrightarrow\; x=-\tfrac53$
2. 2
  Case 2 — the insides are opposite in sign. Negate one side: $2x-3 = -(5x+2) \;\Longrightarrow\; 2x-3=-5x-2 \;\Longrightarrow\; 7x=1 \;\Longrightarrow\; x=\tfrac17$
3. 3
  Check both back in the original equation — both are genuine here.
Answer
$x=-\dfrac53$ and $x=\dfrac17$
29709/22 M/J 2024 Q14 marks
Solve the inequality $|5x+7| > |2x-3|$ .
Stuck? Show hint
Both sides are moduli, so it is safe to square immediately — no case-by-case justification needed first.
Show solution
1. 1
  Both sides are moduli, so squaring is safe outright: $\left(5x+7\right)^2 > \left(2x-3\right)^2$
  A modulus is never negative, so squaring either side cannot change the direction of the inequality — the caution needed when only one side is a modulus does not apply here.
2. 2
  Expand both sides separately. Left: $\left(5x+7\right)^2 = 25x^2+70x+49$
3. 3
  Right: $\left(2x-3\right)^2 = 4x^2-12x+9$
4. 4
  Bring everything to one side: $25x^2+70x+49 - \left(4x^2-12x+9\right) > 0 \;\Longrightarrow\; 21x^2+82x+40>0$
5. 5
  Factorise: $\left(3x+10\right)\left(7x+4\right)>0$
  Check: 3×7=21 ✓, 10×4=40 ✓, and the cross terms 3(4)+10(7)=12+70=82 ✓.
6. 6
  A product of two linear factors is positive outside its roots. The roots are $x=-\tfrac{10}{3}$ and $x=-\tfrac47$ : $x<-\tfrac{10}{3} \quad\text{or}\quad x>-\tfrac47$
  Sketch the upward parabola through these two roots — it is above the x-axis (positive) on the two outer regions, which is exactly what “> 0” asks for. Note only one side being a modulus (like 2x + 5 < |x + 1|, from the syllabus's own example) would need the case method or a sign check first — with two moduli you can always square straight away.
Answer
$x<-\dfrac{10}{3}$ or $x>-\dfrac47$
39709/23 O/N 2024 Q24 marks
Solve the inequality $|x-7| > 4x+3$ .
Stuck? Show hint
Find both case-derived critical values, then test a point in each region — one of the two values may not actually be a genuine boundary.
Show solution
1. 1
  Case 1 — the inside is positive. Drop the modulus: $x-7 > 4x+3 \;\Longrightarrow\; -3x>10 \;\Longrightarrow\; x<-\tfrac{10}{3}$
  Dividing by −3 flips the inequality sign — easy to miss when the case algebra is compressed.
2. 2
  Case 2 — the inside is negative. Negate it: $-(x-7) > 4x+3 \;\Longrightarrow\; -x+7>4x+3 \;\Longrightarrow\; 4>5x \;\Longrightarrow\; x<\tfrac45$
3. 3
  Test a point from each candidate region against the original inequality — do not trust the case algebra alone.
4. 4
  Test $x=0$ (satisfies $x<\tfrac45$ ): $\;\left\lvert-7\right\rvert=7$ , $4(0)+3=3$ . Is $7>3$ ? Yes.
5. 5
  Test $x=-4$ (satisfies $x<-\tfrac{10}{3}$ , and also $x<\tfrac45$ ): $\;\left\lvert-11\right\rvert=11$ , $4(-4)+3=-13$ . Is $11>-13$ ? Yes — but this point is also inside $x<\tfrac45$ , so it does not distinguish the two regions.
6. 6
  Test $x=1$ (satisfies neither $x<-\tfrac{10}{3}$ nor $x<\tfrac45$ ): $\;\left\lvert-6\right\rvert=6$ , $4(1)+3=7$ . Is $6>7$ ? No.
  This shows x < 4/5 is the true boundary and x < −10/3 was never a separate region at all — it sits entirely inside x < 4/5, so it adds nothing. Not every value the case method produces survives as a genuine boundary — this is the inequality version of checking a squared equation's roots, and it is why testing a point beats just reporting both case answers.
Answer
$x<\dfrac45$

Common mistakes

$|3x-2| = |2x+7| \Rightarrow 3x - 2 = 2x + 7$ only
Also $3x - 2 = -(2x+7)$
One equation cannot capture both signs; you lose one of the two solutions.
Squaring $|x-4| < 3x$ straight away
Splitting into cases, because $3x$ is not known to be positive
Squaring an inequality preserves it only when both sides are non-negative.
Keeping every root that comes out of the squared equation
Substituting each back into the original
Squaring can create roots that satisfy the squared equation but not the modulus one.

In the exam

69 parts · 2.6 marks each · the most-tagged sub-topic in Paper 2

Modulus questions are short and frequent — usually one 2- or 3-mark part, often a sketch followed by a solve. The two recorded techniques that dominate are "solve each resulting linear equation" and "check each solution against its case condition". The checking is worth real marks.

Practise modulus questionsReal past-paper questions · Modulus function: graph of y = |ax + b| and solving modulus equations/inequalities

03

Polynomial division

Syllabus requirement · §2.1

“

divide a polynomial, of degree not exceeding 4, by a linear or quadratic polynomial, and identify the quotient and remainder (which may be zero).

”

Dividing $2x^3 - 3x^2 - 11x + 6$ by $x - 3$ is long division with letters. The layout is the same as with numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat.

What division produces: the original polynomial rewritten as divisor × quotient + remainder. Setting x = a kills the quotient term, which is where the remainder theorem comes from.

Long division, one cycle at a time

1
Divide the leading terms. $2x^3 \div x = 2x^2$ — that is the first term of the quotient.
2
Multiply the whole divisor by it and write the result underneath: $2x^2(x-3) = 2x^3 - 6x^2$ .
3
Subtract, and bring down the next term. $-3x^2 - (-6x^2) = 3x^2$ , then bring down $-11x$ .
Subtracting a negative is where nearly every long-division error happens. Write the signs out.
4
Repeat until what is left has a lower degree than the divisor. Whatever remains is the remainder.
5
Check by multiplying back: divisor × quotient + remainder should rebuild the original.

The comparing-coefficients shortcut

Instead of dividing, write down what the answer must look like and match coefficients. For $2x^3 - 3x^2 - 11x + 6 \equiv (x-3)(ax^2+bx+c)$ :

$x^3$ : $a = 2$
constant: $-3c = 6$ , so $c = -2$
$x^2$ : $b - 3a = -3$ , so $b = 3$

Three lines, no subtraction, and it is fully accepted. It is usually faster once you have practised it.

Your turn

One divisor with a non-unit x-coefficient, and one quadratic divisor — the syllabus explicitly includes both.

19709/22 F/M 2025 Q6(a)3 marks
Find the quotient and remainder when $18x^3-6x^2-30x+4$ is divided by $(3x-1)$ .
Stuck? Show hint
Divide the leading terms first: 18x³ ÷ 3x, not 18x³ ÷ x — the coefficient of x in the divisor is not 1.
Show solution
1. 1
  Divide the leading terms. $18x^3 \div 3x = 6x^2$ — the first term of the quotient.
2. 2
  Multiply the whole divisor by it: $6x^2(3x-1) = 18x^3-6x^2$ .
3. 3
  Subtract, and bring down the next term: $\left(18x^3-6x^2-30x+4\right) - \left(18x^3-6x^2\right) = -30x+4$
4. 4
  Divide again. $-30x \div 3x = -10$ — the second term of the quotient.
5. 5
  Multiply the divisor by it: $-10(3x-1) = -30x+10$ .
6. 6
  Subtract to find what is left: $\left(-30x+4\right) - \left(-30x+10\right) = -6$ This has degree $0$ , lower than the divisor's degree $1$ , so the division stops here.
Answer
quotient $6x^2-10$ , remainder $-6$
29709/22 O/N 2025 Q4(a)3 marks
The polynomial $\mathrm{p}(x) = x^4-10x^3+20x^2-30x+40$ . Find the quotient when $\mathrm{p}(x)$ is divided by $\left(x^2+3\right)$ and show that the remainder is $-11$ .
Stuck? Show hint
The divisor is quadratic, so each subtraction removes two terms of degree at a time — the method is otherwise identical.
Show solution
1. 1
  Divide the leading terms. $x^4 \div x^2 = x^2$ — the first term of the quotient.
2. 2
  Multiply the whole divisor by it: $x^2\left(x^2+3\right) = x^4+3x^2$ .
3. 3
  Subtract, bringing down the remaining terms: $\left(x^4-10x^3+20x^2-30x+40\right) - \left(x^4+3x^2\right) = -10x^3+17x^2-30x+40$
  20x² − 3x² = 17x², and the −10x³, −30x and +40 terms had nothing to subtract against, so they come straight down.
4. 4
  Divide again. $-10x^3 \div x^2 = -10x$ — the second term of the quotient.
5. 5
  Multiply the divisor by it: $-10x\left(x^2+3\right) = -10x^3-30x$ .
6. 6
  Subtract: $\left(-10x^3+17x^2-30x+40\right) - \left(-10x^3-30x\right) = 17x^2+40$
7. 7
  Divide a third time. $17x^2 \div x^2 = 17$ — the third term of the quotient.
8. 8
  Multiply the divisor by it: $17\left(x^2+3\right) = 17x^2+51$ .
9. 9
  Subtract to find what is left: $\left(17x^2+40\right) - \left(17x^2+51\right) = -11$ This is a constant, degree $0$ , lower than the divisor's degree $2$ — so it is the remainder, confirming $-11$ . $\;\blacksquare$
Answer
quotient $x^2-10x+17$ , remainder $-11$ (shown)

Practise polynomial divisionReal past-paper questions · Polynomial division by linear or quadratic polynomial

04

The factor and remainder theorems

Syllabus requirement · §2.1

“

use the factor theorem and the remainder theorem, e.g. to find factors and remainders, solve polynomial equations or evaluate unknown coefficients. Including factors of the form (ax + b) in which the coefficient of x is not unity, and including calculation of remainders.

”

Both theorems come from one line. Division says

\mathrm{f}(x) \equiv (x - a)\,\mathrm{Q}(x) + R

Now put $x = a$ . The first term becomes zero whatever $\mathrm{Q}$ is, so $\mathrm{f}(a) = R$ . That is the whole idea: the remainder is just the function evaluated at the root of the divisor.

The algebra and the graph are the same fact: p(a) = 0 at every x-axis crossing, and the y-intercept is the remainder on division by x — read p(k) off the curve at any point k and you have the remainder for dividing by (x − k).

\mathrm{f}(a) = R

Remainder theorem — dividing by (x − a)

\mathrm{f}(a) = 0 \iff (x-a) \text{ is a factor}

Factor theorem — the zero-remainder case

\mathrm{f}\!\left(-\tfrac{b}{a}\right) = 0 \iff (ax+b) \text{ is a factor}

…when the x coefficient is not 1

Getting the substitution right

For a divisor $(x - 3)$ you substitute $x = +3$ : the sign flips. For $(2x + 1)$ you substitute $x = -\tfrac12$ , because that is what makes the divisor zero. Set the divisor to zero and solve — do not guess the sign.

Worked example9709/22 M/J 2022 Q57 marks

The polynomial $\mathrm{p}(x)$ is defined by $\mathrm{p}(x) = 2x^3 + ax^2 - 3x - 4,$ where $a$ is a constant. It is given that $(x-4)$ is a factor of $\mathrm{p}(x)$ .

(a) Find the value of $a$ and hence factorise $\mathrm{p}(x)$ . [4]
(b) Show that the equation $\mathrm{p}\!\left(\mathrm{e}^{3y}\right) = 0$ has only one real root, and find its exact value. [3]

Show full working

1
(a) $(x-4)$ is a factor, so $\mathrm{p}(4) = 0$ : $128 + 16a - 12 - 4 = 0 \;\Rightarrow\; 16a = -112 \;\Rightarrow\; a = -7$
2
So $\mathrm{p}(x) = 2x^3 - 7x^2 - 3x - 4$ . Divide by $(x-4)$ , or compare coefficients: $\mathrm{p}(x) = (x-4)\left(2x^2 + x + 1\right)$
By inspection: leading 2, constant −4 ÷ −4 = 1, then the x² coefficient fixes the middle term.
3
(b) Replacing $x$ by $\mathrm{e}^{3y}$ gives $\left(\mathrm{e}^{3y} - 4\right)\left(2\mathrm{e}^{6y} + \mathrm{e}^{3y} + 1\right) = 0$ .
4
The quadratic factor, read as a quadratic in $\mathrm{e}^{3y}$ , has discriminant $1 - 8 = -7 < 0$ , so it contributes no real roots.
This is the “show that there is only one root” mark — the discriminant is the evidence, and the scheme demands it explicitly.
5
That leaves $\mathrm{e}^{3y} = 4$ . Take logarithms: $3y = \ln 4$ , so $y = \tfrac13\ln 4$

Answer

(a) $a = -7$ , $\mathrm{p}(x) = (x-4)\left(2x^2+x+1\right)$ (b) $y = \tfrac13\ln 4$

Part (b) is this topic in its natural habitat: a factorised cubic feeding a logarithm question. Paper 2 chains topics together far more than Paper 1 does — the factor theorem is rarely the whole question.

Solving a cubic

1
Find one root by trial, testing the factors of the constant term: $\pm1$ , $\pm2$ , $\pm3$ , … until $\mathrm{f}(a) = 0$ .
Any whole-number root has to divide the constant term, so the list of candidates is short.
2
That gives one factor $(x - a)$ . Divide, or compare coefficients, to get the quadratic factor.
3
Solve the quadratic by factorising or the formula.
4
Report all the roots — a cubic has up to three, and the quadratic may have none.

Two unknowns need two conditions

So far each example has had one unknown coefficient, found from one factor. A quartic or a cubic with two unknown coefficients needs two separate conditions — usually two given factors, occasionally a factor and a stated remainder — each turned into its own equation, then solved as simultaneous equations.

Two unknowns from two factors

9709/22 M/J 2025 Q5(a)4 marks

The polynomial $\mathrm{p}(x)$ is defined by $\mathrm{p}(x) = ax^4+bx^3+13x^2-35x+15,$ where $a$ and $b$ are constants. It is given that $(2x-1)$ and $(x-3)$ are factors of $\mathrm{p}(x)$ . Find the values of $a$ and $b$ .

Show full working

1
Turn the first factor into an equation. $(2x-1)$ is a factor, so $\mathrm{p}\!\left(\tfrac12\right)=0$ : $a\left(\tfrac12\right)^4 + b\left(\tfrac12\right)^3 + 13\left(\tfrac12\right)^2 - 35\left(\tfrac12\right) + 15 = 0$
2x − 1 = 0 gives x = ½ — set the divisor to zero and solve, exactly as the callout above says.
2
Simplify each term: $\tfrac{1}{16}a + \tfrac18 b + \tfrac{13}{4} - \tfrac{35}{2} + 15 = 0$ . Multiply every term by $16$ to clear the fractions: $a + 2b + 52 - 280 + 240 = 0 \;\Longrightarrow\; a+2b+12=0$
3
Turn the second factor into a separate equation. $(x-3)$ is a factor, so $\mathrm{p}(3)=0$ : $a(81) + b(27) + 13(9) - 35(3) + 15 = 0$
This is a genuinely separate condition — do not try to combine it with the first equation before both are written down cleanly.
4
Simplify: $81a+27b+117-105+15=0 \;\Longrightarrow\; 81a+27b+27=0$ . Divide every term by $27$ : $3a+b+1=0$
5
Now solve the two equations simultaneously. From the second, $b=-1-3a$ .
6
Substitute into the first: $a + 2\left(-1-3a\right) + 12 = 0 \;\Longrightarrow\; a - 2 - 6a + 12 = 0 \;\Longrightarrow\; -5a+10=0$
7
So $a=2$ . Substitute back: $b = -1-3(2) = -7$

Answer

$a=2$ , $b=-7$

Keep the two equations in the simplest whole-number form before combining them — a+2b+12=0 and 3a+b+1=0 are far easier to eliminate between than the fractional versions they started as.

Your turn

One single-unknown factor-theorem problem with a non-unit x-coefficient, and one two-unknown problem like the example above.

1
The polynomial $\mathrm{p}(x) = 4x^3+kx^2-x-3$ , where $k$ is a constant. Given that $(2x-1)$ is a factor of $\mathrm{p}(x)$ , find the value of $k$ , and hence factorise $\mathrm{p}(x)$ completely.
Stuck? Show hint
2x − 1 = 0 gives x = ½ — substitute that into p(x) and set the result to zero.
Show solution
1. 1
  $(2x-1)$ is a factor, so $\mathrm{p}\!\left(\tfrac12\right)=0$ : $4\left(\tfrac12\right)^3 + k\left(\tfrac12\right)^2 - \tfrac12 - 3 = 0$
2. 2
  Simplify each term: $4\left(\tfrac18\right) + k\left(\tfrac14\right) - \tfrac12 - 3 = 0 \;\Longrightarrow\; \tfrac12 + \tfrac14k - \tfrac72 = 0$
3. 3
  Solve for $k$ : $\tfrac14k = 3 \;\Longrightarrow\; k=12$
  ½ − 7/2 = −3, so ¼k must cancel that: ¼k = 3.
4. 4
  So $\mathrm{p}(x) = 4x^3+12x^2-x-3$ . Divide by $(2x-1)$ : leading term $4x^3\div2x=2x^2$ , giving $2x^2(2x-1)=4x^3-2x^2$ to subtract, leaving $14x^2-x$ ; next term $14x^2\div2x=7x$ , giving $7x(2x-1)=14x^2-7x$ to subtract, leaving $6x-3$ ; last term $6x\div2x=3$ , giving $3(2x-1)=6x-3$ , leaving $0$ : $\mathrm{p}(x) = (2x-1)\left(2x^2+7x+3\right)$
5. 5
  Factorise the quadratic: $2x^2+7x+3 = (2x+1)(x+3)$
  Check by expanding back: 2x·x + 2x·3 + 1·x + 1·3 = 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓.
Answer
$k=12$ , $\mathrm{p}(x) = (2x-1)(2x+1)(x+3)$
2
The polynomial $\mathrm{p}(x) = ax^3+bx^2-11x+6$ , where $a$ and $b$ are constants. It is given that $(x-2)$ and $(x+3)$ are factors of $\mathrm{p}(x)$ . Find the values of $a$ and $b$ .
Stuck? Show hint
Two factors, two substitutions, two simultaneous equations — the same pattern as the worked example.
Show solution
1. 1
  $(x-2)$ is a factor, so $\mathrm{p}(2)=0$ : $8a+4b-22+6=0 \;\Longrightarrow\; 8a+4b-16=0$ Divide by $4$ : $2a+b-4=0$
2. 2
  $(x+3)$ is a factor, so $\mathrm{p}(-3)=0$ : $-27a+9b+33+6=0 \;\Longrightarrow\; -27a+9b+39=0$ Divide by $3$ : $-9a+3b+13=0$
3. 3
  Solve simultaneously. From the first, $b=4-2a$ . Substitute into the second: $-9a+3\left(4-2a\right)+13=0 \;\Longrightarrow\; -9a+12-6a+13=0 \;\Longrightarrow\; -15a+25=0$
4. 4
  So $a=\tfrac53$ . Substitute back: $b=4-2\left(\tfrac53\right) = 4-\tfrac{10}{3} = \tfrac23$
  Non-integer coefficients are unusual for this topic but not impossible — trust the algebra rather than assuming a mistake because the numbers are not whole.
Answer
$a=\dfrac53$ , $b=\dfrac23$

Common mistakes

$(x - 3)$ is a factor, so test $\mathrm{f}(-3)$
Test $\mathrm{f}(3)$
Substitute the value that makes the divisor zero. (x − 3) is zero at x = +3.
$(2x+1)$ is a factor, so test $\mathrm{f}(-1)$
Test $\mathrm{f}\left(-\tfrac12\right)$
2x + 1 = 0 gives x = −½. The syllabus explicitly includes factors where the x coefficient is not 1.
Stopping at $(x-3)(2x^2-x+2)$ without comment
Checking the quadratic's discriminant and saying it does not factorise
“Completely” is an instruction; the last mark is for showing you finished.
Dividing when the question only asks for the remainder
Evaluating $\mathrm{f}(a)$
The remainder theorem exists precisely to save you the division.

In the exam

68 parts on the theorems · 47 on division · 3.2 and 3.6 marks each

These are the longest-scoring parts in the topic, and "apply the factor theorem" is the single most recorded technique in Paper 2 algebra. The standard question gives you two conditions on unknown coefficients, then asks you to factorise or solve the resulting cubic — a structure that has barely changed in five years.

Practise the factor and remainder theoremsReal past-paper questions · Factor theorem and remainder theorem

Everything on one page

|x| = \begin{cases} x & x \geqslant 0 \\ -x & x < 0 \end{cases}

Definition of the modulus

|a| = |b| \iff a^2 = b^2

Squaring removes both moduli

|x-a| < b \iff a-b < x < a+b

“x is within b of a”

\mathrm{f}(x) \equiv (x-a)\mathrm{Q}(x) + R

The division identity

R = \mathrm{f}(a)

Remainder theorem

\mathrm{f}(a) = 0 \iff (x-a) \text{ is a factor}

Factor theorem

\mathrm{f}\!\left(-\tfrac{b}{a}\right) = 0 \iff (ax+b) \text{ is a factor}

…for a non-unit x coefficient

Can you do all of these?

Sketch $y = |2x - 3|$ with both intercepts labelled
Solve $|3x-2| = |2x+7|$ by squaring, and check both roots
Solve $|x-4| < 3x$ by splitting into cases
Say why squaring is safe for an equation but risky for an inequality
Divide $2x^3 - 3x^2 - 11x + 6$ by $(x-3)$ and state the quotient and remainder
Get the same quotient by comparing coefficients instead
Find the remainder when a cubic is divided by $(x+2)$ without dividing
Decide whether $(2x+1)$ is a factor of a given cubic
Factorise a cubic completely and say when the quadratic factor cannot be factorised
Turn two given factors into two equations and solve them simultaneously for two unknown coefficients