Algebra

Info: Board Coverage AQA Paper 1 & 2 | Edexcel Paper 1 & 2 | OCR Paper 1, 2, 3 | WJEC Unit 1 & 2

Like terms have the same variable part. They can be combined by addition and subtraction.

Worked Example. Simplify $4x^2 - 3x + 7x^2 + 5x - 2$.

$(4 + 7)x^2 + (-3 + 5)x - 2 = 11x^2 + 2x - 2$

Worked Example. Expand and simplify $3(2x - 5) - 2(x + 4)$.

$6x - 15 - 2x - 8 = 4x - 23$

Single bracket: $a(b + c) = ab + ac$

Double brackets (FOIL): $(a + b)(c + d) = ac + ad + bc + bd$

Worked Example. Expand and simplify $(2x + 3)(x - 5)$.

$2x^2 - 10x + 3x - 15 = 2x^2 - 7x - 15$

Worked Example. Expand $(x + 4)^2$.

$(x + 4)(x + 4) = x^2 + 4x + 4x + 16 = x^2 + 8x + 16$

Worked Example (Higher Tier). Expand and simplify $(2x - 1)(3x + 4)(x - 2)$.

First expand two brackets:

$(2x - 1)(3x + 4) = 6x^2 + 8x - 3x - 4 = 6x^2 + 5x - 4$

Now multiply by the third:

$(6x^2 + 5x - 4)(x - 2) = 6x^3 - 12x^2 + 5x^2 - 10x - 4x + 8 = 6x^3 - 7x^2 - 14x + 8$

Worked Example (Higher Tier). Expand and simplify $(x + 2)(x - 3)(2x - 1)$.

First: $(x + 2)(x - 3) = x^2 - 3x + 2x - 6 = x^2 - x - 6$.

Then: $(x^2 - x - 6)(2x - 1) = 2x^3 - x^2 - 2x^2 + x - 12x + 6 = 2x^3 - 3x^2 - 11x + 6$.

Common factor: $ab + ac = a(b + c)$

Difference of two squares: $a^2 - b^2 = (a - b)(a + b)$

Quadratic trinomial: $x^2 + bx + c = (x + p)(x + q)$ where $p + q = b$ and $pq = c$.

Worked Example. Factorise $x^2 - 9$.

$x^2 - 9 = (x - 3)(x + 3)$

Worked Example. Factorise $x^2 + 5x + 6$.

We need $p + q = 5$ and $pq = 6$Giving $p = 2$ and $q = 3$:

$x^2 + 5x + 6 = (x + 2)(x + 3)$

Worked Example. Factorise $6x^2 + 17x + 12$.

We need $pq = 6 \times 12 = 72$ and $p + q = 17$. The values are $p = 8$ and $q = 9$.

$6x^2 + 8x + 9x + 12 = 2x(3x + 4) + 3(3x + 4) = (2x + 3)(3x + 4)$

Worked Example (Higher Tier). Factorise $x^3 - x$.

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

Worked Example (Higher Tier). Factorise $4x^2 - 25y^2$.

$4x^2 - 25y^2 = (2x - 5y)(2x + 5y)$

Worked Example (Higher Tier). Factorise $x^2 + 2x - 8$.

We need two numbers with product $-8$ and sum $2$: these are $4$ and $-2$.

$x^2 + 2x - 8 = (x + 4)(x - 2)$

Worked Example (Higher Tier). Factorise $12x^2 - 27$.

$12x^2 - 27 = 3(4x^2 - 9) = 3(2x - 3)(2x + 3)$

Worked Example (Higher Tier). Factorise $x^4 - 16$.

This is a difference of two squares twice:

$x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$

Note that $x^2 + 4$ cannot be factorised further over the reals since $x^2 + 4 = 0$ has no real Solutions.

Worked Example. Simplify $\frac{x^2 - 9}{x^2 + 5x + 6}$.

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

Worked Example. Simplify $\frac{x^2 - 4x + 4}{x^2 - 4}$.

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

Worked Example (Higher Tier). Simplify $\frac{2x^2 - 8}{x^2 + 4x + 4}$.

$\frac{2(x^2 - 4)}{(x + 2)^2} = \frac{2(x - 2)(x + 2)}{(x + 2)^2} = \frac{2(x - 2)}{x + 2}$

::warning When cancelling factors in algebraic fractions, always factorise first. Never cancel Individual terms across addition or subtraction.

A linear equation has the general form $ax + b = c$ where the highest power of $x$ is 1.

Worked Example. Solve $3(2x - 1) = 4x + 7$.

$6x - 3 = 4x + 7$ $2x = 10$ $x = 5$

Worked Example. Solve $\frac{2x + 1}{3} = \frac{x - 2}{4} + 1$.

Multiply through by 12:

$4(2x + 1) = 3(x - 2) + 12$ $8x + 4 = 3x - 6 + 12$ $8x + 4 = 3x + 6$ $5x = 2$ $x = \frac{2}{5}$

Worked Example (Higher Tier). Solve $\frac{3x - 1}{4} - \frac{x + 2}{3} = \frac{x - 1}{6}$.

Multiply through by 12 (the LCM of 4, 3, and 6):

$3(3x - 1) - 4(x + 2) = 2(x - 1)$ $9x - 3 - 4x - 8 = 2x - 2$ $5x - 11 = 2x - 2$ $3x = 9$ $x = 3$

A quadratic equation has the form $ax^2 + bx + c = 0$ with $a \neq 0$.

Method 1: Factorisation. If $ax^2 + bx + c$ factorises, set each factor equal to zero.

Worked Example. Solve $x^2 - 5x + 6 = 0$.

$(x - 2)(x - 3) = 0$ x = 2 \mathrm{ or x = 3

Method 2: The quadratic formula. For $ax^2 + bx + c = 0$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Theorem. The discriminant $\Delta = b^2 - 4ac$ determines the nature of the roots:

Proof sketch. The formula gives $x = \frac{-b \pm \sqrt{\Delta}}{2a}$. If $\Delta \gt 0$The Square root is a positive real, giving two distinct values. If $\Delta = 0$Both roots equal $\frac{-b}{2a}$. If $\Delta \lt 0$The square root is not real, so no real roots exist.

Worked Example. Solve $2x^2 + 3x - 5 = 0$ using the formula.

$x = \frac{-3 \pm \sqrt{9 + 40}}{4} = \frac{-3 \pm \sqrt{49}}{4} = \frac{-3 \pm 7}{4}$

x = 1 \mathrm{ or x = -\frac{5}{2}

Worked Example (Higher Tier). Find the values of $k$ for which $x^2 + 4x + k = 0$ has equal Roots.

Equal roots means $\Delta = 0$:

$16 - 4k = 0 \implies k = 4$

Method 3: Completing the square.

$ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}$

Worked Example. Write $x^2 + 6x + 2$ in completed square form and solve $x^2 + 6x + 2 = 0$.

$x^2 + 6x + 2 = (x + 3)^2 - 9 + 2 = (x + 3)^2 - 7$

$(x + 3)^2 = 7 \implies x + 3 = \pm\sqrt{7} \implies x = -3 \pm \sqrt{7}$

Worked Example (Higher Tier). Find the minimum value of $f(x) = 2x^2 - 12x + 5$ and the value of $x$ at which it occurs.

$f(x) = 2(x^2 - 6x) + 5 = 2\left((x - 3)^2 - 9\right) + 5 = 2(x - 3)^2 - 18 + 5 = 2(x - 3)^2 - 13$

Since $(x - 3)^2 \geq 0$The minimum value is $-13$ when $x = 3$.

Worked Example (Higher Tier). Solve $3x^2 - 5x - 2 = 0$.

$\Delta = 25 + 24 = 49$.

$x = \frac{5 \pm 7}{6}$

x = 2 \mathrm{ or x = -\frac{1}{3}

Linear-linear systems: Solve by elimination or substitution.

Worked Example. Solve the simultaneous equations:

$\begin{cases} 3x + 2y = 12 \\ 5x - y = 7 \end{cases}$

Multiply the second equation by 2: $10x - 2y = 14$.

Add to the first: $13x = 26$So $x = 2$.

Substitute back: $10 - y = 7$So $y = 3$.

Linear-quadratic systems: Substitute the linear equation into the quadratic.

Worked Example. Solve:

$\begin{cases} y = x^2 - 3x + 2 \\ y = 2x - 2 \end{cases}$

Substituting: $x^2 - 3x + 2 = 2x - 2$

$x^2 - 5x + 4 = 0$ $(x - 1)(x - 4) = 0$ x = 1 \mathrm{ or x = 4

When $x = 1$: $y = 0$. When $x = 4$: $y = 6$.

Solutions: $(1, 0)$ and $(4, 6)$.

Worked Example (Higher Tier). Solve:

$\begin{cases} x + 2y = 5 \\ x^2 + y^2 = 10 \end{cases}$

From the first equation: $x = 5 - 2y$.

Substituting: $(5 - 2y)^2 + y^2 = 10$

$25 - 20y + 4y^2 + y^2 = 10$ $5y^2 - 20y + 15 = 0$ $y^2 - 4y + 3 = 0$ $(y - 1)(y - 3) = 0$

$y = 1$: $x = 3$. $y = 3$: $x = -1$.

Solutions: $(3, 1)$ and $(-1, 3)$.

Worked Example (Higher Tier). Solve:

$\begin{cases} 2x + y = 8 \\ x^2 - y^2 = 12 \end{cases}$

From the first equation: $y = 8 - 2x$.

Substituting: $x^2 - (8 - 2x)^2 = 12$

$x^2 - (64 - 32x + 4x^2) = 12$

$x^2 - 64 + 32x - 4x^2 = 12$

$-3x^2 + 32x - 76 = 0$

$3x^2 - 32x + 76 = 0$

$\Delta = 1024 - 912 = 112 = 16 \times 7$.

$x = \frac{32 \pm 4\sqrt{7}}{6} = \frac{16 \pm 2\sqrt{7}}{3}$

Solving linear inequalities follows the same rules as equations, with one critical difference.

:::caution When multiplying or dividing both sides of an inequality by a negative number, you must Reverse the inequality sign. :::

Worked Example. Solve $3 - 2x \gt 7$.

$-2x \gt 4$ $x \lt -2$

Worked Example. Solve $\frac{5 - x}{2} \geq 3$.

$5 - x \geq 6$ $-x \geq 1$ $x \leq -1$

Worked Example. Solve $x^2 - 5x + 6 \leq 0$.

Factorise: $(x - 2)(x - 3) \leq 0$.

The expression changes sign at $x = 2$ and $x = 3$. We need the region where it is negative or zero:

$2 \leq x \leq 3$

Method for quadratic inequalities:

1. Factorise the quadratic (or find roots).
2. Sketch the sign of the expression in each region.
3. Select the region(s) satisfying the inequality.

Worked Example (Higher Tier). Solve $2x^2 + 3x - 5 \lt 0$.

Roots: $x = 1$ and $x = -\frac{5}{2}$.

Since the coefficient of $x^2$ is positive, the parabola opens upward. The expression is negative Between the roots:

$-\frac{5}{2} \lt x \lt 1$

Worked Example (Higher Tier). Solve $-x^2 + 4x + 5 \geq 0$.

Multiply by $-1$ (reversing the inequality): $x^2 - 4x - 5 \leq 0$.

Factorise: $(x - 5)(x + 1) \leq 0$.

$-1 \leq x \leq 5$

• Open circle at an endpoint: the value is not included ($\lt$ or $\gt$).
• Closed circle at an endpoint: the value is included ($\leq$ or $\geq$).

Worked Example. Solve $-3 \lt 2x + 1 \leq 7$.

Subtract 1 from all parts: $-4 \lt 2x \leq 6$.

Divide by 2: $-2 \lt x \leq 3$.

The solution $x \gt 3$ can be written as $\{x \in \mathbb{R} : x \gt 3\}$ or using interval notation $(3, \infty)$.

An arithmetic sequence has a constant common difference $d$. The $n$-th term is:

$u_n = a + (n - 1)d$

Where $a$ is the first term.

The sum of the first $n$ terms is:

$S_n = \frac{n}{2}(2a + (n - 1)d) = \frac{n}{2}(a + l)$

Where $l$ is the last term.

Proof of the sum formula. Write the sum forwards and backwards:

$S_n = a + (a + d) + (a + 2d) + \cdots + (l - d) + l$ $S_n = l + (l - d) + (l - 2d) + \cdots + (a + d) + a$

Adding: $2S_n = n(a + l)$So $S_n = \frac{n(a + l)}{2}$. $\blacksquare$

Worked Example. Find the 20th term and the sum of the first 20 terms of $3, 7, 11, 15, \ldots$.

Here $a = 3$ and $d = 4$.

$u_{20} = 3 + 19 \times 4 = 79$

$S_{20} = \frac{20}{2}(2 \times 3 + 19 \times 4) = 10 \times 82 = 820$

Worked Example (Higher Tier). An arithmetic sequence has first term 5 and common difference 3. Find the value of $n$ for which $u_n = 137$.

$5 + (n - 1) \times 3 = 137$ $3(n - 1) = 132$ $n - 1 = 44$ $n = 45$

Worked Example (Higher Tier). An arithmetic sequence has $u_3 = 14$ and $u_7 = 26$. Find $a$ and $d$.

a + 2d = 14 \quad \mathrm{and \quad a + 6d = 26

Subtracting: $4d = 12$So $d = 3$. Then $a = 14 - 6 = 8$.

The sequence is $8, 11, 14, 17, \ldots$

A geometric sequence has a constant common ratio $r$. The $n$-th term is:

$u_n = ar^{n-1}$

Worked Example. Find the 8th term of $2, 6, 18, 54, \ldots$.

Here $a = 2$ and $r = 3$.

$u_8 = 2 \times 3^7 = 2 \times 2187 = 4374$

Worked Example (Higher Tier). Find the sum of the first 10 terms of $1, -2, 4, -8, \ldots$.

a = 1$$r = -2$$n = 10.

$S_{10} = \frac{1((-2)^{10} - 1)}{-2 - 1} = \frac{1024 - 1}{-3} = \frac{1023}{-3} = -341$

A quadratic sequence has second differences that are constant.

Worked Example. Find the $n$-th term of $2, 6, 12, 20, 30, \ldots$.

First differences: $4, 6, 8, 10, \ldots$

Second differences: $2, 2, 2, \ldots$

Since the second difference is 2, the coefficient of $n^2$ is $2/2 = 1$. So $u_n = n^2 + bn + c$.

When $n = 1$: $1 + b + c = 2 \implies b + c = 1$

When $n = 2$: $4 + 2b + c = 6 \implies 2b + c = 2$

Subtracting: $b = 1$So $c = 0$.

$u_n = n^2 + n$

Worked Example (Higher Tier). Find the $n$-th term of $1, 0, -3, -8, -15, \ldots$.

First differences: $-1, -3, -5, -7, \ldots$

Second differences: $-2, -2, -2, \ldots$

The coefficient of $n^2$ is $-2/2 = -1$. So $u_n = -n^2 + bn + c$.

When $n = 1$: $-1 + b + c = 1 \implies b + c = 2$

When $n = 2$: $-4 + 2b + c = 0 \implies 2b + c = 4$

Subtracting: $b = 2$So $c = 0$.

$u_n = -n^2 + 2n = n(2 - n)$

Each term is the sum of the two preceding terms: $u_n = u_{n-1} + u_{n-2}$.

Example: $1, 1, 2, 3, 5, 8, 13, 21, \ldots$

Proposition. Every third Fibonacci number is even.

Proof. Consider the Fibonacci sequence modulo 2. The sequence of parities is: $1, 1, 0, 1, 1, 0, 1, 1, 0, \ldots$ (period 3). Therefore $F_n$ is even if and only if $3 \mid n$. $\blacksquare$

Proposition. $\gcd(F_m, F_n) = F_{\gcd(m, n)}$.

This beautiful identity connects the Fibonacci sequence to the greatest common divisor. It explains Why consecutive Fibonacci numbers are coprime: $\gcd(F_n, F_{n+1}) = F_1 = 1$.

Adjust the parameters in the graph above to explore the relationships between variables.

The equation of a straight line is $y = mx + c$ where $m$ is the gradient and $c$ is the $y$-intercept.

Gradient formula: For two points $(x_1, y_1)$ and $(x_2, y_2)$:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Parallel lines have the same gradient. Perpendicular lines have gradients whose product is $-1$: $m_1 m_2 = -1$.

Proof of the perpendicular gradient property. If two lines with gradients $m_1$ and $m_2$ are Perpendicular, then the angle between them is $90^{\circ}$. The tangent of the angle between two Lines is $\frac{m_2 - m_1}{1 + m_1 m_2}$. Setting this to be undefined (as $\tan 90^{\circ}$ is Undefined), the denominator $1 + m_1 m_2 = 0$Giving $m_1 m_2 = -1$. $\blacksquare$

Worked Example. Find the equation of the line through $(2, 5)$ perpendicular to $y = 3x - 1$.

The gradient of the given line is 3. The perpendicular gradient is $-\frac{1}{3}$.

$y - 5 = -\frac{1}{3}(x - 2)$ $y = -\frac{1}{3}x + \frac{2}{3} + 5 = -\frac{1}{3}x + \frac{17}{3}$

Worked Example (Higher Tier). Find the equation of the perpendicular bisector of the line Segment joining $A(1, 3)$ and $B(7, -1)$.

Midpoint: $M = \left(\frac{8}{2}, \frac{2}{2}\right) = (4, 1)$.

Gradient of AB: $m = \frac{-1 - 3}{7 - 1} = \frac{-4}{6} = -\frac{2}{3}$.

Perpendicular gradient: $\frac{3}{2}$.

Equation: $y - 1 = \frac{3}{2}(x - 4)$I.e. $2y - 2 = 3x - 12$Or $3x - 2y = 10$.

The graph of $y = ax^2 + bx + c$ is a parabola. If $a \gt 0$ it opens upward; if $a \lt 0$ it Opens downward.

The turning point is at $x = -\frac{b}{2a}$.

The line of symmetry is $x = -\frac{b}{2a}$.

Worked Example. Sketch the graph of $y = x^2 - 4x + 3$.

Factorise: $y = (x - 1)(x - 3)$So the roots are $x = 1$ and $x = 3$.

$y$-intercept: $(0, 3)$.

Vertex: $x = -\frac{-4}{2} = 2$, $y = 4 - 8 + 3 = -1$. Vertex at $(2, -1)$.

The parabola opens upward with minimum at $(2, -1)$Crossing the $x$-axis at $x = 1$ and $x = 3$.

Worked Example (Higher Tier). Sketch the graph of $y = -2x^2 + 8x - 6$.

Factorise: $-2(x^2 - 4x + 3) = -2(x - 1)(x - 3)$. Roots at $x = 1$ and $x = 3$.

$y$-intercept: $(0, -6)$.

Vertex: $x = -\frac{8}{-4} = 2$, $y = -8 + 16 - 6 = 2$. Vertex at $(2, 2)$.

The parabola opens downward with maximum at $(2, 2)$.

Worked Example (Higher Tier). The graph of $y = f(x)$ passes through $(2, 5)$. State the Coordinates of the corresponding point on the graph of $y = f(x - 3) + 1$.

Translation right by 3, up by 1: $(2 + 3, 5 + 1) = (5, 6)$.

Worked Example (Higher Tier). Given $f(x) = x^2$Sketch $y = -f(2x) + 3$.

$y = -(2x)^2 + 3 = -4x^2 + 3$. This is a parabola opening downward with vertex at $(0, 3)$Narrower Than $y = x^2$ by a factor of 2 in the $x$-direction.

Worked Example. Simplify $\frac{x^2 - 9}{x^2 + 5x + 6}$.

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

Find a common denominator, then combine.

Worked Example. Simplify $\frac{3}{x + 1} + \frac{2}{x - 2}$.

$\frac{3(x - 2) + 2(x + 1)}{(x + 1)(x - 2)} = \frac{3x - 6 + 2x + 2}{(x + 1)(x - 2)} = \frac{5x - 4}{(x + 1)(x - 2)}$

Worked Example (Higher Tier). Simplify $\frac{2x}{x^2 - 1} - \frac{1}{x + 1}$.

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

Worked Example. Solve $\frac{x + 1}{3} + \frac{x - 2}{4} = 5$.

Multiply through by 12 (the LCM of 3 and 4):

$4(x + 1) + 3(x - 2) = 60$ $4x + 4 + 3x - 6 = 60$ $7x - 2 = 60$ $7x = 62$ $x = \frac{62}{7}$

Worked Example (Higher Tier). Solve $\frac{2x + 1}{x - 3} = \frac{x + 4}{x + 1}$.

Cross-multiply: $(2x + 1)(x + 1) = (x + 4)(x - 3)$.

$2x^2 + 3x + 1 = x^2 + x - 12$ $x^2 + 2x + 13 = 0$

$\Delta = 4 - 52 = -48 \lt 0$. No real solutions.

List all possible cases and verify each one.

Example. Prove that every integer squared leaves remainder 0, 1, or 4 when divided by 5.

Every integer is of the form $5k + r$ where $r \in \{0, 1, 2, 3, 4\}$.

$(5k)^2 = 25k^2$: remainder 0. $(5k + 1)^2 = 25k^2 + 10k + 1$: remainder 1. $(5k + 2)^2 = 25k^2 + 20k + 4$: remainder 4. $(5k + 3)^2 = 25k^2 + 30k + 9$: remainder 4. $(5k + 4)^2 = 25k^2 + 40k + 16$: remainder 1.

All cases give remainder 0, 1, or 4. $\blacksquare$

Start from known facts and use logical steps to reach a conclusion.

Example. Prove that the sum of two consecutive odd numbers is divisible by 4.

Let the consecutive odd numbers be $2n + 1$ and $2n + 3$ for integer $n \geq 0$.

Sum $= (2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1)$.

Since $n + 1$ is an integer, the sum is a multiple of 4. $\blacksquare$

Example (Higher Tier). Prove that the product of any three consecutive integers is divisible By 6.

Let the three consecutive integers be $n$, $n + 1$And $n + 2$.

Among any three consecutive integers, one is divisible by 2 (even) and one is divisible by 3. Since 2 and 3 are coprime, their product 6 divides the product of the three integers. $\blacksquare$

Example (Higher Tier). Prove that the sum of the squares of any two consecutive integers, minus 1, is divisible by 8.

Let the consecutive integers be $n$ and $n + 1$.

$n^2 + (n+1)^2 - 1 = n^2 + n^2 + 2n + 1 - 1 = 2n^2 + 2n = 2n(n + 1)$.

Since $n$ and $n + 1$ are consecutive, one is even, so $n(n+1)$ is even, meaning $n(n+1) = 2k$ for Some integer $k$.

Therefore $2n(n+1) = 4k$Which is divisible by 4 but not necessarily by 8. The original claim that This is always divisible by 8 is false. For example, when $n = 1$: $1 + 4 - 1 = 4$Which is not Divisible by 8.

::warning Not every claim about numbers is true. When asked to prove something, first check whether The statement is actually correct with a small example.

To disprove a statement, find a single example where it fails.

Example. “All prime numbers are odd.” Counterexample: 2 is prime and even.

Example. “If $n^2$ is divisible by 4, then $n$ is divisible by 4.” Counterexample: $n = 6$: $n^2 = 36$ is divisible by 4, but 6 is not.

Example (Higher Tier). “The sum of two irrational numbers is irrational.” Counterexample: $\sqrt{2} + (-\sqrt{2}) = 0$Which is rational.

Example. Given $f(x) = 2x - 1$ and $g(x) = x^2 + 3$Prove that $fg(-2) = gf(1)$.

$f(-2) = -5$, $g(-5) = 25 + 3 = 28$.

$f(1) = 1$, $g(1) = 1 + 3 = 4$.

$fg(-2) = g(f(-2)) = 28 \neq 4 = gf(1)$.

So $fg(-2) \neq gf(1)$ — the statement is false. This illustrates that $f \circ g \ne G \circ f$.

• Forgetting to reverse the inequality sign when multiplying or dividing by a negative.
• Incorrectly expanding $(a + b)^2$. It is $a^2 + 2ab + b^2$Not $a^2 + b^2$.
• Squaring brackets incorrectly in completing the square. $(x + 3)^2 = x^2 + 6x + 9$Not $x^2 + 9$.
• Confusing the turning point formula. The $x$-coordinate is $-\frac{b}{2a}$Not $\frac{b}{2a}$.
• Dropping solutions when solving quadratics. Always check both values from $\pm$ in the formula.
• Assuming all sequences are arithmetic. Always check the first differences.
• Cancelling terms instead of factors in algebraic fractions. Factorise first.
• Mistaking graph transformations. $y = f(x + 2)$ shifts LEFT by 2, not right.
• Forgetting domain restrictions when simplifying algebraic fractions (e.g., $x \neq 1$).
• Losing a negative sign when expanding a bracket preceded by a minus sign. Expand $-(x - 3)$ as $-x + 3$Not $-x - 3$.
• Incorrectly identifying the common difference of a geometric sequence as addition rather than multiplication. Arithmetic sequences add $d$; geometric sequences multiply by $r$.

1. Expand and simplify $(3x - 2)(2x + 5) - (x + 1)^2$.

2. Solve the simultaneous equations $2x + 3y = 13$ and $4x - y = 5$.

3. Solve $x^2 - 7x + 10 = 0$ by factorisation, and verify using the quadratic formula.

4. Write $2x^2 - 12x + 5$ in completed square form. Hence state the minimum value and where it occurs.

5. Solve the inequality $x^2 - 4x - 5 \gt 0$.

6. The $n$-th term of a sequence is $n^2 + 3n$. Find the first 5 terms and the 50th term.

7. Find the equation of the line through $(-1, 4)$ that is parallel to $2y = 6x + 5$.

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

9. Prove that the product of three consecutive integers is always divisible by 6.

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

11. Find the equation of the line through $(3, -1)$ perpendicular to $2x + 5y = 10$.

12. Express $\frac{x^3 - x}{x^2 - 1}$ as a simplified algebraic fraction, stating any restriction on $x$.

13. Prove that the difference between the squares of any two consecutive integers is always odd.

14. Solve $x^2 - 6x + 2 \leq 0$Giving your answer in exact form.

15. The first three terms of a geometric sequence are $k, k + 3, k + 12$. Find the value of $k$ and the common ratio.

16. Find the $n$-th term of the sequence $5, 10, 17, 26, 37, \ldots$.

17. Disprove by counterexample: “The square root of any irrational number is irrational.”

18. Find the values of $k$ for which $kx^2 - 6x + 4 = 0$ has two distinct real roots.

19. Simplify $\frac{x^3 - 8}{x^2 - 4}$Stating any restriction on $x$.

20. Prove that for any positive integer $n$The number $n^3 - n$ is divisible by 6.

Example 1:

A typical exam question on Algebra requires you to apply your knowledge to an unfamiliar context. Read the question carefully, identify the key concept being tested, and structure your answer using the appropriate terminology.

Example 2:

Multi-step problems in Algebra often combine two or more concepts. Break the problem down: identify what you need to find, recall the relevant formula or principle, substitute values, and state your answer with correct units or formatting.

This topic covers the mathematical techniques and concepts related to algebra, including key theorems, methods, and problem-solving approaches.

Key concepts include:

• quadratic equations and the discriminant
• simultaneous equations
• polynomial division and the factor theorem
• partial fractions
• binomial expansion

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.