Algebra and Functions

This lesson covers Algebra and Functions as required by the A-Level Mathematics Pure 1 specification. This topic forms the foundation for almost every other area of pure mathematics. You must be confident with algebraic manipulation, the laws of indices, surds, quadratic functions, simultaneous equations, inequalities, polynomial division, the factor theorem, and graph transformations.

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Laws of Indices

The laws of indices allow you to simplify expressions involving powers. Let a > 0 and m, n be rational numbers:

Rule	Law
Multiplication	aᵐ × aⁿ = aᵐ⁺ⁿ
Division	aᵐ ÷ aⁿ = aᵐ⁻ⁿ
Power of a power	(aᵐ)ⁿ = aᵐⁿ
Zero index	a⁰ = 1
Negative index	a⁻ⁿ = 1/aⁿ
Fractional index	a^(1/n) = ⁿ√a
Combined fractional	a^(m/n) = (ⁿ√a)ᵐ

Example: Simplify 8^(2/3) × 2⁻⁴.

8^(2/3) = (∛8)² = 2² = 4
So 8^(2/3) × 2⁻⁴ = 4 × 1/16 = 1/4

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Surds

A surd is an irrational number left in root form (e.g., √2, √5). Surds are exact values and should not be converted to decimals in exam answers.

Rules for Surds

Rule	Example
√(ab) = √a × √b	√12 = √4 × √3 = 2√3
√(a/b) = √a / √b	√(9/4) = 3/2
(√a)² = a	(√7)² = 7
Like surds combine	3√2 + 5√2 = 8√2

Rationalising the Denominator

Monomial denominator: Multiply top and bottom by the surd.

1/√3 = 1/√3 × √3/√3 = √3/3

Binomial denominator: Multiply by the conjugate.

5/(3 + √2) × (3 − √2)/(3 − √2) = 5(3 − √2)/(9 − 2) = 5(3 − √2)/7

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Quadratic Functions

A quadratic function has the form f(x) = ax² + bx + c where a ≠ 0.

Solving Quadratics

1. Factorising: x² − 5x + 6 = (x − 2)(x − 3) = 0, so x = 2 or x = 3.
2. Completing the square: Write ax² + bx + c in the form a(x + p)² + q.
3. Quadratic formula: x = (−b ± √(b² − 4ac)) / 2a.

The Discriminant

The discriminant Δ = b² − 4ac determines the nature of the roots:

Condition	Nature of Roots
b² − 4ac > 0	Two distinct real roots
b² − 4ac = 0	One repeated real root
b² − 4ac < 0	No real roots

Example: Find the values of k for which 2x² + kx + 8 = 0 has no real roots.

b² − 4ac < 0
k² − 4(2)(8) < 0
k² < 64
−8 < k < 8

Completing the Square

Example: Write 2x² − 12x + 5 in completed-square form.

2x² − 12x + 5 = 2(x² − 6x) + 5 = 2((x − 3)² − 9) + 5 = 2(x − 3)² − 13

The minimum value is −13, occurring when x = 3.

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Polynomial Division and the Factor Theorem

Factor Theorem

If f(a) = 0 then (x − a) is a factor of f(x).

Example: Show that (x − 2) is a factor of f(x) = x³ − 5x² + 8x − 4.

f(2) = 8 − 20 + 16 − 4 = 0 ✓

So (x − 2) is a factor. Dividing: x³ − 5x² + 8x − 4 = (x − 2)(x² − 3x + 2) = (x − 1)(x − 2)².

Remainder Theorem

When f(x) is divided by (x − a), the remainder is f(a).

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Simultaneous Equations

At A-Level, you may need to solve simultaneous equations where one is linear and one is quadratic.

Example: Solve y = x + 1 and x² + y² = 13.

Substitute: x² + (x + 1)² = 13

x² + x² + 2x + 1 = 13
2x² + 2x − 12 = 0
x² + x − 6 = 0
(x + 3)(x − 2) = 0

So x = −3, y = −2 or x = 2, y = 3.

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Inequalities

Quadratic Inequalities

1. Solve the corresponding equation to find the critical values.
2. Sketch the quadratic graph.
3. Read the solution from the graph.

Example: Solve x² − 4x − 5 > 0.

Factorise: (x − 5)(x + 1) > 0. Critical values: x = −1 and x = 5.

Since the parabola opens upwards, the expression is positive when x < −1 or x > 5.

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Graph Transformations

Transformation	Effect
y = f(x) + a	Translate up by a units
y = f(x) − a	Translate down by a units
y = f(x + a)	Translate left by a units
y = f(x − a)	Translate right by a units
y = af(x)	Vertical stretch, scale factor a
y = f(ax)	Horizontal stretch, scale factor 1/a
y = −f(x)	Reflection in the x-axis
y = f(−x)	Reflection in the y-axis

Example: Describe the transformation from y = x² to y = (x − 3)² + 2.

Translation by the vector (3, 2) — that is, 3 units to the right and 2 units up.

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Summary

• Use the laws of indices to simplify expressions with powers.
• Leave answers in surd form and rationalise denominators where required.
• Solve quadratics by factorising, completing the square, or the quadratic formula.
• The discriminant tells you the number and type of roots.
• Use the factor theorem and polynomial division to factorise cubics.
• Solve simultaneous equations by substitution when one equation is nonlinear.
• Solve quadratic inequalities by sketching the graph.
• Know all graph transformations of y = f(x).

Exam Tip: Graph transformation questions often carry several marks. Always state the transformation type (translation, stretch, reflection) and give specific details (direction, scale factor, vector). When completing the square with a ≠ 1, always factor out a first. Show all working when rationalising denominators — marks are awarded for method.

A-Level Deep Dive: Algebra and Functions

Spec mapping

AQA 7357 Pure section B — Algebra and Functions. This subject content covers indices and surds, quadratic functions and the discriminant, simultaneous equations, linear and quadratic inequalities, polynomial manipulation, the factor and remainder theorems, algebraic fractions and partial fractions, the modulus function and its graphs, and the full family of graph transformations $y = f(x) + a$y=f(x)+a, $y = f(x + a)$y=f(x+a), $y = af(x)$y=af(x), $y = f(ax)$y=f(ax) and combinations. Although it is the second strand of the Pure content, Algebra and Functions is synoptic with every other Pure topic: trigonometric identities reduce to quadratics in $\sin x$sinx; logarithmic equations reduce to polynomials in $\ln x$lnx; differentiation of $x^n$xn relies on rational-index fluency; integration by partial fractions is an algebra-and-functions question dressed in calculus clothing. The AQA formula booklet does not list index laws, the discriminant, or the factor theorem — these must be memorised.

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Worked example with full mark scheme

Question (8 marks):

(a) Express $\dfrac{(3 - \sqrt{2})^2}{2 + \sqrt{2}}$2+2​(3−2​)2​ in the form $a + b\sqrt{2}$a+b2​, where $a$a and $b$b are rational. (5)

(b) Hence solve $3^{2x - 1} \cdot 9^{x + 1} = \dfrac{1}{27}$32x−1⋅9x+1=271​, giving $x$x as an exact rational. (3)

Solution with mark scheme:

(a) Step 1 — expand the numerator.

$(3 - \sqrt{2})^2 = 9 - 6\sqrt{2} + 2 = 11 - 6\sqrt{2}$(3−2​)2=9−62​+2=11−62​

M1 — correct expansion of the squared bracket. The middle term $-6\sqrt{2}$−62​ comes from $2 \cdot 3 \cdot (-\sqrt{2})$2⋅3⋅(−2​). A frequent slip is writing $(3 - \sqrt{2})^2 = 9 + 2 = 11$(3−2​)2=9+2=11, which discards the cross term and forfeits both M1s.

A1 — correct simplified numerator $11 - 6\sqrt{2}$11−62​.

Step 2 — rationalise the denominator.

Multiply numerator and denominator by the conjugate $(2 - \sqrt{2})$(2−2​):

$\dfrac{(11 - 6\sqrt{2})(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})}$(2+2​)(2−2​)(11−62​)(2−2​)​

M1 — multiplying by the correct conjugate. Multiplying by $(2 + \sqrt{2})$(2+2​) (the same factor) produces an irrational denominator and earns nothing.

Step 3 — expand the new numerator and denominator.

$(11 - 6\sqrt{2})(2 - \sqrt{2}) = 22 - 11\sqrt{2} - 12\sqrt{2} + 6 \cdot 2 = 22 - 23\sqrt{2} + 12 = 34 - 23\sqrt{2}$(11−62​)(2−2​)=22−112​−122​+6⋅2=22−232​+12=34−232​

The denominator: $(2 + \sqrt{2})(2 - \sqrt{2}) = 4 - 2 = 2$(2+2​)(2−2​)=4−2=2.

M1 — correct expansion of both numerator and denominator, simplifying $\sqrt{2}\cdot\sqrt{2} = 2$2​⋅2​=2 at the point it appears.

Step 4 — combine.

$\dfrac{34 - 23\sqrt{2}}{2} = 17 - \dfrac{23}{2}\sqrt{2}$234−232​​=17−223​2​

A1 — final form $a = 17$a=17, $b = -\tfrac{23}{2}$b=−223​, presented in the requested form $a + b\sqrt{2}$a+b2​.

(b) Step 1 — convert all terms to base 3.

$9 = 3^2$9=32, so $9^{x + 1} = 3^{2(x + 1)} = 3^{2x + 2}$9x+1=32(x+1)=32x+2. Also $\dfrac{1}{27} = 3^{-3}$271​=3−3.

M1 — converting all bases to a common base.

Step 2 — combine indices on the LHS.

$3^{2x - 1} \cdot 3^{2x + 2} = 3^{(2x - 1) + (2x + 2)} = 3^{4x + 1}$32x−1⋅32x+2=3(2x−1)+(2x+2)=34x+1

M1 — correct application of the multiplication law of indices.

Step 3 — equate indices.

$4x + 1 = -3 \implies 4x = -4 \implies x = -1$4x+1=−3⟹4x=−4⟹x=−1

A1 — exact rational answer.

Total: 8 marks (M4 A4, split as shown).

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): Given that $y = 6x^{1/2} + 10x^{-1/2}$y=6x1/2+10x−1/2 for $x > 0$x>0:

(a) Show that $y$y can be written in the form $\dfrac{6x + 10}{\sqrt{x}}$x​6x+10​. (2)

(b) Hence find $\dfrac{dy}{dx}$dxdy​, simplifying your answer. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — rewriting $6x^{1/2}$6x1/2 as $\dfrac{6x^{1/2} \cdot x^{1/2}}{x^{1/2}} = \dfrac{6x}{\sqrt{x}}$x1/26x1/2⋅x1/2​=x​6x​ using $x^{1/2}\cdot x^{1/2} = x$x1/2⋅x1/2=x.
• A1 (AO1.1b) — combining over the common denominator: $\dfrac{6x + 10}{\sqrt{x}}$x​6x+10​.

(b)

• M1 (AO1.1b) — applying the power rule to $6x^{1/2}$6x1/2: $\dfrac{d}{dx}(6x^{1/2}) = 3x^{-1/2}$dxd​(6x1/2)=3x−1/2.
• M1 (AO1.1b) — applying the power rule to $10x^{-1/2}$10x−1/2: $\dfrac{d}{dx}(10x^{-1/2}) = -5x^{-3/2}$dxd​(10x−1/2)=−5x−3/2.
• A1 (AO1.1b) — correct sum: $\dfrac{dy}{dx} = 3x^{-1/2} - 5x^{-3/2}$dxdy​=3x−1/2−5x−3/2.
• A1 (AO2.5) — simplified form $\dfrac{3}{\sqrt{x}} - \dfrac{5}{x\sqrt{x}}$x​3​−xx​5​ or factorised $\dfrac{1}{x^{3/2}}(3x - 5)$x3/21​(3x−5).

Total: 6 marks split AO1 = 5, AO2 = 1. This is an AO1-dominated question — AQA uses algebra-and-functions questions primarily to test procedural fluency, with AO2 marks reserved for the final elegant simplification.

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Synoptic links

1. Trigonometric identities — equations such as $2\sin^2 x - \sin x - 1 = 0$2sin2x−sinx−1=0 are quadratics in $\sin x$sinx. The substitution $u = \sin x$u=sinx converts a trigonometry problem into a Section B factorisation.
2. Exponentials and logarithms — $e^{2x} - 5e^x + 6 = 0$e2x−5ex+6=0 becomes $u^2 - 5u + 6 = 0$u2−5u+6=0 via $u = e^x$u=ex. Logarithmic equations $(\ln x)^2 - 3\ln x + 2 = 0$(lnx)2−3lnx+2=0 are quadratics in $\ln x$lnx.
3. Differentiation of polynomials — the power rule $\dfrac{d}{dx}x^n = nx^{n - 1}$dxd​xn=nxn−1 requires confidence with rational and negative indices, which is core Section B content.
4. Integration by partial fractions — splitting $\dfrac{1}{(x - 1)(x + 2)}$(x−1)(x+2)1​ into partial fractions before integrating is an algebra-and-functions manipulation, not a calculus one. The calculus is trivial once the algebra is done.
5. Numerical methods (Year 2) — locating roots of $f(x) = 0$f(x)=0 by sign change and the Newton-Raphson iteration both rely on polynomial manipulation, factor-theorem reasoning and graph transformations from Section B.

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Mark-scheme literacy

Algebra-and-functions questions on 7357 split AO marks heavily toward AO1, with a meaningful AO2 reasoning component on graph-transformation and modulus parts: