🔤 Algebra

OCR FSMQ Additional Maths · Overview of the whole topic

Level 3 · Ages 15–16
1 The Big Idea

Algebra is the section that underpins every other part of this qualification. The specification describes the course as four "pure" topics plus two numerical topics, all underpinned by an Algebra section — so the fluency you build here is spent everywhere else.

At GCSE you learned to do algebra. At Level 3 you are expected to choose the algebra. A question rarely says "factorise this" — it gives you a situation, and you decide that factorising is the way in.
What is new compared with GCSE
Cubics and the factor theorem · polynomial division
Quadratic inequalities · regions in two variables
Recurrence relationships · algebraic fractions with surds
Calculators are allowed in the FSMQ examination — but read the command words carefully. If a question says "In this question you must show detailed reasoning", a bare answer from a solver earns nothing. You must write down a complete analytical method.
2 The Language You Need

Statement AL1 asks you to know and use algebraic vocabulary and notation. The specification gives a complete list, so these words are examinable in their own right.

WordMeaningExample
ConstantA fixed number that does not vary.The $7$ in $3x + 7$
CoefficientThe number multiplying a variable.The $3$ in $3x^2$
ExpressionA collection of terms with no equals sign.$3x^2 - 5x$
EquationTwo expressions set equal; true for particular values.$x^2 = 9$
IdentityTrue for every value; written with $\equiv$.$(x+1)^2 \equiv x^2 + 2x + 1$
IndexA power. Plural: indices.The $5$ in $x^5$
VariableA quantity that is free to change.$t$ in a distance–time model
UnknownA fixed value you are trying to find.$x$ in $2x + 1 = 9$
$\mathbf{f}(x)$Function notation — the output when the input is $x$.$\mathrm{f}(x) = x^3 - 2x$
Equation or identity? $x^2 = 9$ is an equation: it is only true when $x = \pm 3$. But $x^2 - 9 \equiv (x-3)(x+3)$ is an identity: it is true for every $x$. Examiners award marks for using $\equiv$ correctly when completing the square, because AL5 is written as an identity.
3 The Six Core Techniques
1 — Factorising a quadratic
$ax^2 + bx + c$: find two numbers multiplying to $ac$ and adding to $b$
2 — The quadratic formula
$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,  discriminant $\Delta = b^2 - 4ac$
3 — Completing the square (AL5)
$ax^2 + bx + c \equiv a(x + p)^2 + q$
4 — The factor theorem (AL4)
If $\mathrm{f}(a) = 0$ then $(x - a)$ is a factor of $\mathrm{f}(x)$
5 — Solving a quadratic inequality (AL8)
Solve the equation, sketch the parabola, then read off the sign you need
6 — A recurrence relationship (AL10)
$x_{n+1} = \mathrm{f}(x_n)$ together with a starting value $x_1$
Worked Example — Several techniques in one question

$\mathrm{f}(x) = x^3 - 4x^2 + x + 6$. Show that $(x + 1)$ is a factor, hence solve $\mathrm{f}(x) = 0$, and then solve $\mathrm{f}(x) \geqslant 0$.

①$\mathrm{f}(-1) = -1 - 4 - 1 + 6 = 0$, so by the factor theorem $(x+1)$ is a factor.
②Dividing: $x^3 - 4x^2 + x + 6 = (x+1)(x^2 - 5x + 6)$.
③$x^2 - 5x + 6 = (x-2)(x-3)$, so $\mathrm{f}(x) = (x+1)(x-2)(x-3)$.
④$\mathrm{f}(x) = 0$ when $x = -1,\; 2,\; 3$.
⑤A positive cubic starts low and ends high, crossing at $-1, 2, 3$. It is above the axis between $-1$ and $2$, and again after $3$.
⑥$\mathrm{f}(x) \geqslant 0$ for $-1 \leqslant x \leqslant 2$ or $x \geqslant 3$.
This single question uses AL4, AL3, AL6 and AL8. The specification says explicitly that items "will require learners to use two or more content statements" — expect it.
4 The Algebra Subtopics

Every specification statement in the Algebra section, and the page that covers it.

5 Common Mistakes to Avoid
Dividing an inequality by a negative. $-2x < 6$ gives $x > -3$, not $x < -3$. The sign flips.
Treating a quadratic inequality like an equation. $x^2 > 9$ does not give $x > 3$. It gives $x > 3$ or $x < -3$. Always sketch.
Cancelling across a $+$ sign. $\dfrac{x + 3}{3}$ is not $x$. You may only cancel factors of the whole numerator.
Forgetting a root. A cubic has up to three roots. If you find one with the factor theorem and stop, you have thrown away most of the marks.
Losing the "$\equiv$". When completing the square you are rewriting an expression, not solving anything. Do not introduce an $= 0$ that was never there.
6 Quick Reference

Identity

True for all $x$; use $\equiv$.

Discriminant

$b^2-4ac$: positive $\to$ 2 roots, zero $\to$ 1, negative $\to$ none.

Factor theorem

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

Completed square

$a(x+p)^2+q$ gives the vertex at $(-p,\,q)$.

Negative multiplier

Multiplying or dividing an inequality by a negative reverses it.

Quadratic inequality

Solve $=0$, sketch, read the interval.

Regions

Solid line for $\leqslant$ or $\geqslant$, dashed for $<$ or $>$.

Recurrence

$x_{n+1}=\mathrm{f}(x_n)$ needs a starting term to be useful.

Surds

Rationalise by multiplying by the conjugate.

7 Practice Questions

Ten questions drawing on the whole Algebra section. Work each one through before opening the solution.

Question 1

Write down the coefficient of $x$ and the constant term in $5 - 7x + 2x^2$.

▶ Show solution

Coefficient of $x$ is $-7$ — the sign belongs to the coefficient.

Constant term is $5$.

Question 2

Simplify $\dfrac{1}{x} - \dfrac{2}{x+1}$ as a single fraction.

▶ Show solution

Common denominator $x(x+1)$:

$\dfrac{(x+1) - 2x}{x(x+1)} = \dfrac{1 - x}{x(x+1)}$

Question 3

Express $\sqrt{12} + \sqrt{27}$ in the form $k\sqrt{3}$.

▶ Show solution

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ and $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$.

Sum $= 5\sqrt{3}$, so $k = 5$.

Question 4

Express $x^2 - 6x + 11$ in the form $(x+p)^2 + q$, and hence state the minimum value of the expression.

▶ Show solution

Half of $-6$ is $-3$, and $(x-3)^2 = x^2 - 6x + 9$.

$x^2 - 6x + 11 \equiv (x-3)^2 + 2$

A square is never negative, so the minimum value is $2$, when $x = 3$.

Question 5

Solve the simultaneous equations $3x + 2y = 12$ and $y = x - 1$.

▶ Show solution

Substitute: $3x + 2(x-1) = 12$

$5x - 2 = 12$, so $5x = 14$ and $x = 2.8$.

Then $y = 2.8 - 1 = 1.8$.

Question 6

Show that $(x - 2)$ is a factor of $\mathrm{f}(x) = x^3 + x^2 - 8x + 4$, and find the remaining quadratic factor.

▶ Show solution

$\mathrm{f}(2) = 8 + 4 - 16 + 4 = 0$, so $(x-2)$ is a factor.

Dividing gives $x^3 + x^2 - 8x + 4 = (x-2)(x^2 + 3x - 2)$.

Check: $x^2 + 3x - 2$ has discriminant $9 + 8 = 17$, which is not a perfect square, so it does not factorise further over the integers.

Question 7

Solve the inequality $2x^2 - 5x - 3 < 0$.

▶ Show solution

Factorise: $2x^2 - 5x - 3 = (2x + 1)(x - 3)$.

Roots at $x = -\tfrac12$ and $x = 3$.

The coefficient of $x^2$ is positive, so the parabola is below the axis between the roots.

$-\dfrac{1}{2} < x < 3$

Question 8

A sequence is defined by $x_{n+1} = 3x_n - 4$ with $x_1 = 3$. Find $x_2$, $x_3$ and $x_4$, and describe the behaviour of the sequence.

▶ Show solution

$x_2 = 3(3) - 4 = 5$

$x_3 = 3(5) - 4 = 11$

$x_4 = 3(11) - 4 = 29$

The terms increase without limit — the sequence diverges. (The fixed point is $x = 2$, since $3(2)-4=2$; starting above it, the terms run away from it.)

Question 9

A rectangular field has a perimeter of $100$ m and an area of $600$ m$^2$. Set up and solve equations to find its dimensions.

▶ Show solution

Let the sides be $x$ and $y$. Then $2x + 2y = 100$, so $y = 50 - x$.

Area: $x(50 - x) = 600$

$50x - x^2 = 600 \;\Rightarrow\; x^2 - 50x + 600 = 0$

$(x - 20)(x - 30) = 0$, so $x = 20$ or $x = 30$.

Either way the field is $20$ m by $30$ m.

Question 10

£$2000$ is invested at $4\%$ compound interest per year, and a further £$500$ is added at the end of each year.

(a) Write a recurrence relationship for the amount $A_n$ after $n$ years.   (b) Find the amount after $3$ years.   (c) Explain why the balance cannot settle down to a fixed value.

▶ Show solution

(a) Each year the balance is multiplied by $1.04$ and then $500$ is added:

$A_{n+1} = 1.04\,A_n + 500$, with $A_0 = 2000$.

(b) $A_1 = 1.04(2000) + 500 = 2580$

$A_2 = 1.04(2580) + 500 = 3183.20$

$A_3 = 1.04(3183.20) + 500 = 3810.53$ (to the nearest penny)

(c) A fixed value $A$ would need $A = 1.04A + 500$, giving $-0.04A = 500$ and so $A = -12\,500$.

That fixed point is negative, so it can never be reached from a positive starting balance. Because the multiplier $1.04$ is greater than $1$, the sequence moves away from the fixed point and grows without limit.

Algebra (AL1–AL11) · OCR FSMQ Additional Maths · Created with MathJax