Algebraic Expressions

Board Coverage

Board	Paper	Notes
AQA	Paper 1	Surds, indices, polynomials, factor theorem
Edexcel	P1, P2	Same core; P2 includes harder factorisation
OCR (A)	Paper 1	Similar coverage
CIE (9709)	P1	Surds, quadratics, factor theorem

---

1. Surds

Definition. A surd is an irrational number expressible as the root of a rational number — that Is, an expression of the form $\sqrt{n}$ where $n \in \mathbb{Q}^+$ and $\sqrt{n} \notin \mathbb{Q}$.

The most familiar surds are \sqrt{2}$$\sqrt{3}$$\sqrt{5}And so on. The set of surds is a Subset of the algebraic numbers, and they arise whenever we solve equations of degree 2 or Higher.

1.1 Laws of Surds

For positive real numbers $a$ and $b$:

$$\begin{aligned} \sqrt{a}\sqrt{b} &= \sqrt{ab} \\ \frac◆LB◆\sqrt{a}◆RB◆◆LB◆\sqrt{b}◆RB◆ &= \sqrt◆LB◆\frac{a}{b}◆RB◆ \\ (\sqrt{a})^n &= a^{n/2} \end{aligned}$$

These follow directly from the laws of indices (which we prove in Section 2), since $\sqrt{a} = a^{1/2}$.

Rationalising the denominator. If a fraction has a surd in the denominator, we multiply Numerator and denominator by the surd (or its conjugate) to eliminate it.

For a denominator of the form $\sqrt{a}$Multiply by $\frac◆LB◆\sqrt{a}◆RB◆◆LB◆\sqrt{a}◆RB◆$:

$$\begin{aligned} \frac◆LB◆1◆RB◆◆LB◆\sqrt{a}◆RB◆ &= \frac◆LB◆1◆RB◆◆LB◆\sqrt{a}◆RB◆ \cdot \frac◆LB◆\sqrt{a}◆RB◆◆LB◆\sqrt{a}◆RB◆ = \frac◆LB◆\sqrt{a}◆RB◆◆LB◆a◆RB◆ \end{aligned}$$

For a denominator of the form $a + b\sqrt{c}$Multiply by the conjugate $a - b\sqrt{c}$:

$$\begin{aligned} \frac◆LB◆1◆RB◆◆LB◆a + b\sqrt{c}◆RB◆ &= \frac◆LB◆a - b\sqrt{c}◆RB◆◆LB◆(a + b\sqrt{c})(a - b\sqrt{c})◆RB◆ = \frac◆LB◆a - b\sqrt{c}◆RB◆◆LB◆a^2 - b^2 c◆RB◆ \end{aligned}$$

:::caution A common error is to “rationalise” by splitting the fraction: $\frac◆LB◆a + b\sqrt{c}◆RB◆◆LB◆d◆RB◆ \neq \frac{a}{d} + b\sqrt◆LB◆\frac{c}{d}◆RB◆$ . Always distribute the Denominator correctly. :::

1.2 Irrationality of $\sqrt{2}$

Theorem. $\sqrt{2}$ is irrational.

Proof. We proceed by contradiction. Assume $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{p}{q}$ Where p, q \in \mathbb{Z}^+$$q \neq 0And $\gcd(p, q) = 1$ (i.e., the fraction is in lowest Terms).

$$\begin{aligned} \sqrt{2} &= \frac{p}{q} \\ 2 &= \frac{p^2}{q^2} \\ P^2 &= 2q^2 \end{aligned}$$

Since $p^2 = 2q^2$We have that $p^2$ is even. A standard result (proved below) tells us that if $p^2$ is even, then $p$ is even. So $p = 2k$ for some integer $k$.

Substituting: $(2k)^2 = 2q^2$So $4k^2 = 2q^2$Hence $q^2 = 2k^2$.

By the same argument, $q^2$ is even, so $q$ is even. But this contradicts $\gcd(p, q) = 1$ since Both $p$ and $q$ are divisible by 2. Therefore our assumption was false, and $\sqrt{2}$ is Irrational. $\blacksquare$

Lemma. If $p^2$ is even, then $p$ is even.

Proof. The contrapositive: if $p$ is odd, then $p^2$ is odd. If $p = 2k + 1$Then $p^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$Which is odd. $\blacksquare$

Intuition. This proof exploits the structure of divisibility: the number 2 has a unique prime Factorisation, and squaring preserves parity. The contradiction arises because $2$ “forces” factors Of 2 into both $p$ and $q$Making the fraction reducible.

---

2. Indices (Exponent Laws)

Definition. For $a > 0$ and $n \in \mathbb{Z}^+$We define $a^n = \underbrace◆LB◆a \times a \times \cdots \times a◆RB◆_{n \mathrm{ times}}$. We extend this Definition as follows:

$$\begin{aligned} A^0 &= 1 \quad \mathrm{for } a \neq 0 \\ A^{-n} &= \frac{1}{a^n} \quad \mathrm{for } a \neq 0 \\ A^{1/n} &= \sqrt[n]{a} \quad \mathrm{(the positive root for } a > 0) \end{aligned}$$

2.1 Laws of Indices

For $a, b > 0$ and $m, n \in \mathbb{R}$:

$$\begin{aligned} A^m \cdot a^n &= a^{m+n} \\ \frac{a^m}{a^n} &= a^{m-n} \\ (a^m)^n &= a^{mn} \\ (ab)^n &= a^n b^n \\ \left(\frac{a}{b}\right)^n &= \frac{a^n}{b^n} \end{aligned}$$

Proof of $a^m \cdot a^n = a^{m+n}$ for positive integer exponents.

$$\begin{aligned} A^m \cdot a^n &= \underbrace◆LB◆a \cdot a \cdots a◆RB◆_{m} \cdot \underbrace◆LB◆a \cdot a \cdots a◆RB◆_{n} \\ &= \underbrace◆LB◆a \cdot a \cdots a◆RB◆_{m + n} = a^{m+n} \end{aligned}$$

The extension to rational and real exponents requires more sophisticated machinery (continuity Arguments and the exponential function), which we develop in Exponentials and Logarithms.

Intuition. Exponentiation is repeated multiplication, so multiplying two powers of the same base Adds the counts. Just as $3 \times 5$ counts $3 + 5$ groups of size 1, $a^3 \cdot a^5$ counts $3 + 5 = 8$ factors of $a$.

---

3. Algebraic Manipulation

3.1 Expanding Brackets

The distributive law is the foundation: $a(b + c) = ab + ac$.

For two brackets: $(a + b)(c + d) = ac + ad + bc + bd$.

Proof. By applying the distributive law twice:

$$\begin{aligned} (a + b)(c + d) &= a(c + d) + b(c + d) \\ &= ac + ad + bc + bd \quad \blacksquare \end{aligned}$$

3.2 Factorisation

Factorisation is the reverse of expansion. The key techniques are:

• Common factors: $ab + ac = a(b + c)$
• Difference of two squares: $a^2 - b^2 = (a - b)(a + b)$
• Trinomial (quadratic): $ax^2 + bx + c = (px + q)(rx + s)$ where pr = a$$ps + qr = b $qs = c$
• Grouping: $ab + ac + db + dc = a(b + c) + d(b + c) = (a + d)(b + c)$

Theorem (Difference of Two Squares). $a^2 - b^2 = (a - b)(a + b)$.

Proof. Direct expansion:

$$\begin{aligned} (a - b)(a + b) &= a^2 + ab - ab - b^2 = a^2 - b^2 \quad \blacksquare \end{aligned}$$

3.3 Factorising Cubics

A cubic $ax^3 + bx^2 + cx + d$ can be factorised by finding one root $\alpha$ (using the Factor Theorem), then dividing by $(x - \alpha)$ to obtain a quadratic, which can then be factorised.

---

4. Polynomial Division

4.1 Long Division

Polynomial long division mirrors integer long division exactly. We demonstrate with an example, then State the general theorem.

Example. Divide $x^3 + 2x^2 - 5x + 6$ by $x - 1$.

$$\begin{aligned} X^3 + 2x^2 - 5x + 6 &= (x - 1)(x^2) + \mathrm{remainder of } x^3 - x^2 \\ &= (x - 1)(x^2) + 3x^2 - 5x + 6 \\ &= (x - 1)(x^2 + 3x) + \mathrm{remainder of } 3x^2 - 3x \\ &= (x - 1)(x^2 + 3x) - 2x + 6 \\ &= (x - 1)(x^2 + 3x - 2) + \mathrm{remainder of } -2x + 2 \\ &= (x - 1)(x^2 + 3x - 2) - 2(x - 1) \\ &= (x - 1)(x^2 + 3x - 2) + 4 \end{aligned}$$

So the quotient is $x^2 + 3x - 2$ with remainder $4$.

More systematically, the division algorithm tells us:

Theorem (Polynomial Division Algorithm). For polynomials $f(x)$ and $g(x)$ with $g(x) \neq 0$ There exist unique polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that:

$f(x) = g(x) \cdot q(x) + r(x)$

Where $\deg(r) < \deg(g)$ or $r(x) = 0$.

Intuition. This is exactly analogous to integer division: $47 = 5 \times 9 + 2$Where $0 \leq 2 < 5$. In polynomials, the “size” ordering is replaced by degree, and the remainder must Have smaller degree than the divisor.

4.2 Why Polynomial Division Mirrors Integer Long Division

The structural analogy is deep. Both are instances of a Euclidean domain — an algebraic structure Where we can perform division with remainder. In $\mathbb{Z}$The “degree” is the absolute value; In $\mathbb{R}[x]$The degree is the polynomial degree. The algorithm is the same: at each step, Eliminate the leading term.

---

5. The Factor Theorem and Remainder Theorem

5.1 The Remainder Theorem

Theorem (Remainder Theorem). If a polynomial $f(x)$ is divided by $(x - a)$The remainder is $f(a)$.

Proof. By the division algorithm, $f(x) = (x - a) \cdot q(x) + r$Where $r$ is a constant (since $\deg(r) < \deg(x - a) = 1$).

Substituting $x = a$:

$$\begin{aligned} F(a) &= (a - a) \cdot q(a) + r = 0 \cdot q(a) + r = r \quad \blacksquare \end{aligned}$$

Intuition. When you plug in $x = a$The $(x - a)$ factor vanishes, leaving only the remainder. The remainder is the “leftover” that doesn’t contain the factor $(x - a)$.

5.2 The Factor Theorem

Theorem (Factor Theorem). $(x - a)$ is a factor of $f(x)$ if and only if $f(a) = 0$.

Proof.

($\Rightarrow$) If $(x - a)$ is a factor, then $f(x) = (x - a) \cdot q(x)$. Setting $x = a$: $f(a) = (a - a) \cdot q(a) = 0$.

($\Leftarrow$) If $f(a) = 0$Then by the Remainder Theorem, the remainder upon division by $(x - a)$ is $f(a) = 0$. So $f(x) = (x - a) \cdot q(x)$Meaning $(x - a)$ is a factor. $\blacksquare$

Intuition. The Factor Theorem connects algebra (polynomials) to geometry (roots on the $x$-axis). A root $x = a$ means the graph crosses the $x$-axis at $a$Which means $(x - a)$ must divide the Polynomial.

Example

Show that $(x - 2)$ is a factor of $f(x) = x^3 - 3x^2 + 4$.

By the Factor Theorem, we check $f(2)$:

$$\begin{aligned} F(2) &= 2^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0 \end{aligned}$$

Since $f(2) = 0$, $(x - 2)$ is a factor. To find the remaining factor, we perform polynomial Division:

$$\begin{aligned} X^3 - 3x^2 + 4 &= (x - 2)(x^2 - x - 2) \\ &= (x - 2)(x - 2)(x + 1) \\ &= (x - 2)^2(x + 1) \end{aligned}$$

---

6. Simplifying Algebraic Fractions

6.1 Cancellation

$\frac{ab + ac}{ad + ae} = \frac{a(b + c)}{a(d + e)} = \frac{b + c}{d + e}$

:::caution You can only cancel factors, not terms. $\frac{x + 3}{x}$ is NOT equal to $3$. The $x$ In the denominator is a factor of the whole expression, but the $x$ in the numerator is only one Term. :::

6.2 Addition and Subtraction

Find a common denominator, then combine:

$$\begin{aligned} \frac{1}{x + 1} + \frac{2}{x - 3} &= \frac{(x - 3) + 2(x + 1)}{(x + 1)(x - 3)} \\ &= \frac{x - 3 + 2x + 2}{(x + 1)(x - 3)} \\ &= \frac{3x - 1}{x^2 - 2x - 3} \end{aligned}$$

---

7. Problem Set

Problem 1. Simplify $\frac◆LB◆\sqrt{3} + 1◆RB◆◆LB◆\sqrt{3} - 1◆RB◆$.

Solution

$$ \begin{aligned} \frac◆LB◆\sqrt{3} + 1◆RB◆◆LB◆\sqrt{3} - 1◆RB◆ &= \frac◆LB◆(\sqrt{3} + 1)(\sqrt{3} + 1)◆RB◆◆LB◆(\sqrt{3} - 1)(\sqrt{3} + 1)◆RB◆ \\ &= \frac◆LB◆3 + 2\sqrt{3} + 1◆RB◆◆LB◆3 - 1◆RB◆ \\ &= \frac◆LB◆4 + 2\sqrt{3}◆RB◆◆LB◆2◆RB◆ = 2 + \sqrt{3} \end{aligned} $$

If you get this wrong, revise: [Rationalising denominators](#1-surds)

---

Problem 2. Solve $(2^{x})^3 \cdot 2^{4-x} = 128$.

Solution

$$ \begin{aligned} 2^{3x} \cdot 2^{4 - x} &= 128 \\ 2^{3x + 4 - x} &= 2^7 \\ 2^{2x + 4} &= 2^7 \\ 2x + 4 &= 7 \\ X &= \frac{3}{2} \end{aligned} $$

If you get this wrong, revise: [Laws of indices](#2-indices-exponent-laws)

---

Problem 3. Factorise fully $x^3 - 6x^2 + 11x - 6$.

Solution

Try $f(1) = 1 - 6 + 11 - 6 = 0$. So $(x - 1)$ is a factor.

Divide $x^3 - 6x^2 + 11x - 6$ by $(x - 1)$:

$$\begin{aligned} X^3 - 6x^2 + 11x - 6 &= (x - 1)(x^2 - 5x + 6) \\ &= (x - 1)(x - 2)(x - 3) \end{aligned}$$

If you get this wrong, revise: [Factor theorem](#52-the-factor-theorem)

---

Problem 4. When $f(x) = 2x^3 + ax^2 + bx - 12$ is divided by $(x - 1)$ the remainder is $-6$ And $(x + 2)$ is a factor. Find $a$ and $b$.

Solution

By the Remainder Theorem: $f(1) = -6$: $$\begin{aligned} 2(1)^3 + a(1)^2 + b(1) - 12 &= -6 \\ 2 + a + b - 12 &= -6 \\ A + b &= 4 \quad \mathrm{--- (1)} \end{aligned}$$

By the Factor Theorem: $f(-2) = 0$:

$$\begin{aligned} 2(-2)^3 + a(-2)^2 + b(-2) - 12 &= 0 \\ -16 + 4a - 2b - 12 &= 0 \\ 4a - 2b &= 28 \\ 2a - b &= 14 \quad \mathrm{--- (2)} \end{aligned}$$

Adding (1) and (2): $3a = 18$So $a = 6$. Then $b = 4 - 6 = -2$.

If you get this wrong, revise: [Remainder theorem](#51-the-remainder-theorem)

---

Problem 5. Simplify $\left(\frac{8x^6}{27y^3}\right)^{-2/3}$.

Solution

$$ \begin{aligned} \left(\frac{8x^6}{27y^3}\right)^{-2/3} &= \left(\frac{27y^3}{8x^6}\right)^{2/3} \\ &= \frac◆LB◆(27)^{2/3} \cdot (y^3)^{2/3}◆RB◆◆LB◆(8)^{2/3} \cdot (x^6)^{2/3}◆RB◆ \\ &= \frac{9y^2}{4x^4} \end{aligned} $$

Since $27^{1/3} = 3$ and $8^{1/3} = 2$.

If you get this wrong, revise: [Laws of indices](#2-indices-exponent-laws)

---

Problem 6. Express $\frac{2x + 1}{x^2 + x - 6}$ in partial fractions.

Solution

First factorise: $x^2 + x - 6 = (x + 3)(x - 2)$. $$\begin{aligned} \frac{2x + 1}{(x + 3)(x - 2)} &= \frac{A}{x + 3} + \frac{B}{x - 2} \\ 2x + 1 &= A(x - 2) + B(x + 3) \end{aligned}$$

Setting $x = -3$: $-6 + 1 = A(-5)$So $A = 1$.

Setting $x = 2$: $4 + 1 = B(5)$So $B = 1$.

$\frac{2x + 1}{(x + 3)(x - 2)} = \frac{1}{x + 3} + \frac{1}{x - 2}$

If you get this wrong, revise: [Algebraic fractions](#6-simplifying-algebraic-fractions)

---

Problem 7. Given $f(x) = x^4 - x^3 - 7x^2 + x + 6$Show that $(x - 1)$, $(x + 1)$And $(x - 3)$ are all factors, and hence factorise $f(x)$ completely.

Solution

$f(1) = 1 - 1 - 7 + 1 + 6 = 0$ ✓

$f(-1) = 1 + 1 - 7 - 1 + 6 = 0$ ✓

$f(3) = 81 - 27 - 63 + 3 + 6 = 0$ ✓

Since $f$ is degree 4 with three known linear factors:

$$\begin{aligned} F(x) &= (x - 1)(x + 1)(x - 3)(px + q) \end{aligned}$$

The product $(x-1)(x+1)(x-3) = (x^2-1)(x-3) = x^3 - 3x^2 - x + 3$.

Comparing leading coefficients: $p = 1$.

Comparing constant terms: $(-1)(1)(-3)(q) = 6$So $3q = 6$, $q = 2$.

$f(x) = (x - 1)(x + 1)(x - 3)(x + 2)$

If you get this wrong, revise: [Factor theorem](#52-the-factor-theorem)

---

Problem 8. Prove that $\sqrt{3}$ is irrational.

Solution

Assume $\sqrt{3} = \frac{p}{q}$ in lowest terms, $p, q \in \mathbb{Z}^+$, $\gcd(p, q) = 1$. $$\begin{aligned} 3 &= \frac{p^2}{q^2} \\ P^2 &= 3q^2 \end{aligned}$$

So $3 \mid p^2$Which means $3 \mid p$ (since 3 is prime). Write $p = 3k$:

$$\begin{aligned} 9k^2 &= 3q^2 \\ Q^2 &= 3k^2 \end{aligned}$$

So $3 \mid q^2$Hence $3 \mid q$. But $\gcd(p, q) = 1$ and both are divisible by 3 — contradiction. $\blacksquare$

If you get this wrong, revise: [Irrationality of surds](#12-irrationality-of-sqrt2)

---

Problem 9. Divide $3x^3 + 5x^2 - 10x + 8$ by $x + 2$. State the quotient and remainder.

Solution

Using long division:

$3x^3 \div x = 3x^2$. Multiply: $3x^2(x + 2) = 3x^3 + 6x^2$. Subtract: $-x^2 - 10x$.

$-x^2 \div x = -x$. Multiply: $-x(x + 2) = -x^2 - 2x$. Subtract: $-8x + 8$.

$-8x \div x = -8$. Multiply: $-8(x + 2) = -8x - 16$. Subtract: $24$.

Quotient: $3x^2 - x - 8$Remainder: $24$.

Verification: $(x + 2)(3x^2 - x - 8) + 24 = 3x^3 - x^2 - 8x + 6x^2 - 2x - 16 + 24 = 3x^3 + 5x^2 - 10x + 8$ ✓

If you get this wrong, revise: [Polynomial division](#4-polynomial-division)

---

Problem 10. Solve $\frac{3}{x + 1} - \frac{1}{x - 2} = 1$.

Solution

$$ \begin{aligned} \frac{3(x - 2) - (x + 1)}{(x + 1)(x - 2)} &= 1 \\ \frac{3x - 6 - x - 1}{x^2 - x - 2} &= 1 \\ \frac{2x - 7}{x^2 - x - 2} &= 1 \\ 2x - 7 &= x^2 - x - 2 \\ X^2 - 3x + 5 &= 0 \end{aligned} $$

Discriminant: $\Delta = 9 - 20 = -11 < 0$. No real solutions.

If you get this wrong, revise: [Algebraic fractions](#6-simplifying-algebraic-fractions) and [Quadratics](./02-quadratics.md)

---

Problem 11. Given that $x^3 + ax^2 + bx + c$ is exactly divisible by $(x - 1)^2$ and leaves Remainder $12$ when divided by $(x + 2)$Find $a$, $b$And $c$.

Solution

Since $(x - 1)^2$ is a factor, both $f(1) = 0$ and $f'(1) = 0$. $$\begin{aligned} F(1) &= 1 + a + b + c = 0 \quad \mathrm{--- (1)} \\ F'(x) &= 3x^2 + 2ax + b \\ F'(1) &= 3 + 2a + b = 0 \quad \mathrm{--- (2)} \end{aligned}$$

Also $f(-2) = 12$:

$$\begin{aligned} -8 + 4a - 2b + c &= 12 \\ 4a - 2b + c &= 20 \quad \mathrm{--- (3)} \end{aligned}$$

From (2): $b = -3 - 2a$.

From (1): $c = -1 - a - b = -1 - a + 3 + 2a = 2 + a$.

Substituting into (3): $4a - 2(-3 - 2a) + (2 + a) = 20$

$4a + 6 + 4a + 2 + a = 20 \implies 9a + 8 = 20 \implies a = \frac{12}{9} = \frac{4}{3}$

$b = -3 - \frac{8}{3} = -\frac{17}{3}, \quad c = 2 + \frac{4}{3} = \frac{10}{3}$

If you get this wrong, revise: [Remainder and Factor theorems](#5-the-factor-theorem-and-remainder-theorem)

---

Problem 12. Simplify $\frac{x^2 - 9}{x^2 + 5x + 6} \div \frac{x^2 - 4x + 3}{x^2 - x - 6}$.

Solution

Factorise all quadratics: $$\begin{aligned} \frac{(x - 3)(x + 3)}{(x + 2)(x + 3)} \div \frac{(x - 1)(x - 3)}{(x - 3)(x + 2)} &= \frac{(x - 3)(x + 3)}{(x + 2)(x + 3)} \times \frac{(x - 3)(x + 2)}{(x - 1)(x - 3)} \\ &= \frac{(x - 3)^2(x + 3)(x + 2)}{(x + 2)(x + 3)(x - 1)(x - 3)} \\ &= \frac{x - 3}{x - 1} \end{aligned}$$

For $x \neq -3, -2, 1, 3$.

If you get this wrong, revise: [Factorisation](#32-factorisation) and [Algebraic fractions](#6-simplifying-algebraic-fractions)

---

Problem 13. Find the value of $k$ such that $x^2 + kx + (k + 3)$ is a perfect square.

Solution

A perfect square has discriminant $\Delta = 0$: $$\begin{aligned} K^2 - 4(k + 3) &= 0 \\ K^2 - 4k - 12 &= 0 \\ (k - 6)(k + 2) &= 0 \end{aligned}$$

$k = 6$: gives $x^2 + 6x + 9 = (x + 3)^2$ ✓

$k = -2$: gives $x^2 - 2x + 1 = (x - 1)^2$ ✓

If you get this wrong, revise: [Quadratics](./02-quadratics.md)

---

:::tip Diagnostic Test Ready to test your understanding of Algebraic Expressions? The contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Algebraic Expressions with other pure mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

Common Pitfalls

1. Rounding too early in multi-step calculations — carry full precision through and round only the final answer.

2. Forgetting the $+c$ constant of integration in indefinite integrals, or misusing boundary conditions in definite integrals.

3. Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’ — the former requires using previous work.

4. Losing marks by not showing sufficient working — always write out each step, especially in proof questions.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

:::