1.1 Quadratic Factorisation

1. Standard Form

A quadratic equation is typically written in the form $a x^{2} + b x + c = 0$ , where $a, b, c$ are constants and $a \neq 0$ .

a x^{2} + b x + c = 0

2. Methods of Factorisation

Common Factor:
Always look for a common factor first (e.g., $2 x^{2} + 4 x = 2 x (x + 2)$ ).
Difference of Two Squares:
$a^{2} - b^{2} = (a - b) (a + b)$ .
Trinomial Factorisation:
Finding two numbers that multiply to $a c$ and add to $b$ .

3. Splitting the Middle Term (The ac Method)

This method is essential for expressions where $a \neq 1$ .

Step 1:
Identify $a, b,$ and $c$ in the expression $a x^{2} + b x + c$ .
Step 2:
Find Two Numbers: Find two integers, say $p$ and $q$ , such that:
$p \times q = a c$ (Product)
$p + q = b$ (Sum)
Step 3:
Split the Middle Term: Write $b x$ as $p x + q x$ .
Step 4:
Factorise by Grouping: Factorise the first two terms and the last two terms separately.

Worked Examples: Factorisation

1. Factorise $x^{2} - 5 x + 6$ .

Find numbers that multiply to 6 and add to -5.

These are -2 and -3.

Result: $(x - 2) (x - 3)$ .

2. Factorise $2 x^{2} + 7 x + 3$ .

$a c = 2 \times 3 = 6$ . Need numbers that multiply to 6 and add to 7 (6 and 1).

Split: $2 x^{2} + 6 x + x + 3$ .

Group: $2 x (x + 3) + 1 (x + 3)$ .

Result: $(2 x + 1) (x + 3)$ .

3. Factorise $3 x^{2} - 10 x + 8$ .

$a c = 24$ . Numbers adding to -10 (-6 and -4).

Split: $3 x^{2} - 6 x - 4 x + 8$ .

Group: $3 x (x - 2) - 4 (x - 2)$ .

Result: $(3 x - 4) (x - 2)$ .

4. Factorise $6 x^{2} - 7 x - 3$ .

$a c = - 18$ . Numbers adding to -7 (-9 and 2).

Split: $6 x^{2} - 9 x + 2 x - 3$ .

Group: $3 x (2 x - 3) + 1 (2 x - 3)$ .

Result: $(3 x + 1) (2 x - 3)$ .

5. Factorise $x^{2} - 9$ (Difference of Squares).

Recognize as $x^{2} - 3^{2}$ .

Apply $a^{2} - b^{2} = (a - b) (a + b)$ .

Result: $(x - 3) (x + 3)$ .

6. Factorise $4 x^{2} - 20 x + 25$ .

$a c = 100$ . Numbers adding to -20 (-10 and -10).

Split: $4 x^{2} - 10 x - 10 x + 25$ .

Group: $2 x (2 x - 5) - 5 (2 x - 5)$ .

Result: $(2 x - 5)^{2}$ .

4. The Quadratic Formula

When a quadratic expression cannot be easily factorised by inspection or splitting the middle term, the Quadratic Formula provides a direct way to find the roots. This formula is derived from the method of completing the square.

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

Where $a x^{2} + b x + c = 0$

AIdentification

The values of $a, b, a n d c$ must be identified from the standard form. Be extremely careful with negative signs.

BSimplification

Evaluate the discriminant first, then substitute and simplify the square root (often results in surds).

Worked Examples: Quadratic Formula

1. Solve $x^{2} - 4 x + 1 = 0$ using the formula.

$a = 1, b = - 4, c = 1$ .

$x = \frac{4 \pm \sqrt{(- 4)^{2} - 4 (1) (1)}}{2 (1)}$ .

$x = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2 \sqrt{3}}{2}$ .

Result: $x = 2 \pm \sqrt{3}$ .

2. Solve $3 x^{2} + 8 x + 2 = 0$ .

$a = 3, b = 8, c = 2$ .

$x = \frac{- 8 \pm \sqrt{64 - 24}}{6} = \frac{- 8 \pm \sqrt{40}}{6}$ .

Result: $x = \frac{- 8 \pm 2 \sqrt{10}}{6} = \frac{- 4 \pm \sqrt{10}}{3}$ .

3. Solve $x^{2} + x - 5 = 0$ .

$a = 1, b = 1, c = - 5$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 (1) (- 5)}}{2} = \frac{- 1 \pm \sqrt{21}}{2}$ .

Result: $x \approx 1.79, x \approx - 2.79$ .

4. Solve $2 x^{2} - 3 x - 1 = 0$ .

$a = 2, b = - 3, c = - 1$ .

$x = \frac{3 \pm \sqrt{9 - 4 (2) (- 1)}}{4} = \frac{3 \pm \sqrt{17}}{4}$ .

Result: $x \approx 1.78, x \approx - 0.28$ .

5. Solve $5 x^{2} + 2 x - 1 = 0$ .

$a = 5, b = 2, c = - 1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 (5) (- 1)}}{10} = \frac{- 2 \pm \sqrt{24}}{10}$ .

Result: $x = \frac{- 2 \pm 2 \sqrt{6}}{10} = \frac{- 1 \pm \sqrt{6}}{5}$ .

6. Solve $x^{2} - 6 x + 7 = 0$ .

$a = 1, b = - 6, c = 7$ .

$x = \frac{6 \pm \sqrt{36 - 28}}{2} = \frac{6 \pm \sqrt{8}}{2}$ .

Result: $x = \frac{6 \pm 2 \sqrt{2}}{2} = 3 \pm \sqrt{2}$ .

5. Completing the Square

Completing the square is a powerful algebraic technique used to solve quadratic equations, find the maximum or minimum values of quadratic functions, and describe the transformation of graphs.

a x^{2} + b x + c = a (x + p)^{2} + q

The "Vertex Form" of a Quadratic

A Process for $a = 1$

1. Divide the coefficient of $x$ by 2 (call it $p$ ).
2. Calculate $p^{2}$ and write as $(x + p)^{2} - p^{2} + c$ .

B Process for $a \neq 1$

1. Factor out $a$ from the terms containing $x$ .
2. Complete the square inside the bracket, then expand.

Worked Examples: Completing the Square

1. Express $x^{2} - 6 x + 13$ in the form $(x + a)^{2} + b$ .

$x^{2} - 6 x + 13$

$= x^{2} - 6 x + (- 3)^{2} - (- 3)^{2} + 13$

$= (x - 3)^{2} - 9 + 13$

$= (x - 3)^{2} + 4$

$∴ a = - 3, b = 4$

2. Express $4 x - x^{2}$ in the form $a - (x + b)^{2}$ .

Factor out $- 1$ : $- (x^{2} - 4 x)$ .

Inside bracket: half of $- 4$ is $- 2$ . Square of $- 2$ is 4.

Complete square inside: $- [(x - 2)^{2} - 4]$ .

Distribute negative: $- (x - 2)^{2} + 4$ .

Final Form: $4 - (x - 2)^{2}$ .

3. Express $2 x^{2} + 12 x + 5$ in the form $a (x + b)^{2} + c$ .

Factor out 2 from $x$ terms: $2 (x^{2} + 6 x) + 5$ .

Inside bracket: half of 6 is 3. Square of 3 is 9.

Complete square: $2 [(x + 3)^{2} - 9] + 5$ .

Distribute 2: $2 (x + 3)^{2} - 18 + 5$ .

Final Form: $2 (x + 3)^{2} - 13$ .

4. Solve $x^{2} + 8 x - 2 = 0$ by completing the square.

Half of 8 is 4. Square of 4 is 16.

Complete square: $(x + 4)^{2} - 16 - 2 = 0$ .

Simplify: $(x + 4)^{2} = 18$ .

Square root: $x + 4 = \pm \sqrt{18} = \pm 3 \sqrt{2}$ .

Results: $x = - 4 \pm 3 \sqrt{2}$ .

5. Find the turning point for $y = 3 x^{2} - 6 x + 10$ .

Factor out 3: $3 (x^{2} - 2 x) + 10$ .

Half of $- 2$ is $- 1$ . Square of $- 1$ is 1.

Complete square: $3 [(x - 1)^{2} - 1] + 10$ .

Distribute 3: $3 (x - 1)^{2} - 3 + 10$ .

Turning point: $(1, 7)$ .

6. Express $4 x^{2} - 4 x + 5$ in the form $a (x + b)^{2} + c$ .

Factor out 4: $4 (x^{2} - x) + 5$ .

Inside bracket: half of $- 1$ is $- 0.5$ . Square of $- 0.5$ is 0.25.

Complete square: $4 [(x - 0.5)^{2} - 0.25] + 5$ .

Distribute 4: $4 (x - 0.5)^{2} - 1 + 5$ .

Final Form: $4 (x - 0.5)^{2} + 4$ .

6. Graphing Method

Sketching a quadratic graph involves identifying the shape, the intercepts, and the turning point. This provides a visual representation of the function's behavior.

A Shape & Intercepts

Shape: Determined by $a$ . $a > 0$ is a minimum (U), $a < 0$ is a maximum (n).
Intercepts: $y$ -intercept at $(0, c)$ . $x$ -intercepts by solving $y = 0$ .

B Key Features

The Turning Point (vertex) can be found at $x = - b / 2 a$ or from the vertex form $a (x + p)^{2} + q$ .

Worked Examples: Graphing Method

1. Sketch $y = x^{2} - 4 x + 3$

x	-1	0	1	2	3	4	5
y	8	3	0	-1	0	3	8

Vertex

(2, - 1)

x-Intercepts

1, 3

2. Sketch $y = - x^{2} + 2 x + 8$

x	-3	-2	-1	0	1	2	3	4	5
y	-7	0	5	8	9	8	5	0	-7

Vertex

(1, 9)

x-Intercepts

- 2, 4

3. Sketch $y = x^{2} - 6 x + 9$

x	0	1	2	3	4	5	6
y	9	4	1	0	1	4	9

Vertex

(3, 0)

x-Intercepts

3

(Repeated)

4. Sketch $y = 2 x^{2} + 4 x + 5$

x	-3	-2	-1	0	1
y	11	5	3	5	11

Vertex

(- 1, 3)

x-InterceptsNo real roots

5. Sketch $y = (x - 2)^{2} - 1$

x	0	1	2	3	4
y	3	0	-1	0	3

Vertex

(2, - 1)

x-Intercepts

1, 3

6. Sketch $y = 4 x - x^{2}$

x	-1	0	1	2	3	4	5
y	-5	0	3	4	3	0	-5

Vertex

(2, 4)

x-Intercepts

0, 4

7. Difference of Two Squares (DOTS)

The Difference of Two Squares is a special case of factorisation that occurs when a perfect square is subtracted from another perfect square.

a^{2} - b^{2} = (a - b) (a + b)

The DOTS Identity

PIdentifying the Pattern

Ensure there are only two terms, both are perfect squares, and they are subtracted.

SSolution Steps

Take the square root of both terms. One bracket gets the sum (+), and the other gets the difference (-).

Worked Examples: Difference of Two Squares

1. Factorise $9 x^{2} - 16$ .

Identify terms as squares: $(3 x)^{2} - (4)^{2}$ .

Apply pattern $a^{2} - b^{2} = (a - b) (a + b)$ .

Result: $(3 x - 4) (3 x + 4)$ .

2. Factorise $25 a^{2} - 49 b^{2}$ .

Identify terms: $(5 a)^{2} - (7 b)^{2}$ .

Apply DOTS rule.

Result: $(5 a - 7 b) (5 a + 7 b)$ .

3. Factorise $2 x^{2} - 50$ .

Factor out common factor 2: $2 (x^{2} - 25)$ .

Inside bracket: $x^{2} - 5^{2}$ .

Result: $2 (x - 5) (x + 5)$ .

4. Factorise $\frac{1}{4} x^{2} - 1$ .

Identify terms: $(\frac{1}{2} x)^{2} - (1)^{2}$ .

Apply DOTS rule.

Result: $(\frac{1}{2} x - 1) (\frac{1}{2} x + 1)$ .

5. Factorise $(x + 3)^{2} - 4$ .

Identify terms: $(x + 3)^{2} - (2)^{2}$ .

Apply DOTS: $[(x + 3) - 2] [(x + 3) + 2]$ .

Simplify: $(x + 1) (x + 5)$ .

6. Factorise $100 - 49 x^{2}$ .

Identify terms: $10^{2} - (7 x)^{2}$ .

Apply DOTS rule.

Result: $(10 - 7 x) (10 + 7 x)$ .

8. Mixed Practice Questions

Test your understanding with these mixed problems involving algebraic fractions, indices (hidden quadratics), and geometric applications.

1. Solve $x - \frac{6}{x - 5} = 0$ .

Solution

Multiply by $(x - 5)$ : $x (x - 5) - 6 = 0$ .

Expand: $x^{2} - 5 x - 6 = 0$ .

Factorise: $(x - 6) (x + 1) = 0$ .

Result: $x = 6, x = - 1$ .

2. Solve $\frac{5 x + 1}{4} - \frac{2 x - 1}{2} = x^{2}$ .

Solution

Multiply by 4: $(5 x + 1) - 2 (2 x - 1) = 4 x^{2}$ .

Simplify: $5 x + 1 - 4 x + 2 = 4 x^{2} \Rightarrow x + 3 = 4 x^{2}$ .

Rearrange: $4 x^{2} - x - 3 = 0$ .

Factorise: $(4 x + 3) (x - 1) = 0$ .

Result: $x = 1, x = - 0.75$ .

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

Solution

Combine LHS: $\frac{3 x + 1}{x (x + 1)} = 2$ .

Cross multiply: $3 x + 1 = 2 x (x + 1) = 2 x^{2} + 2 x$ .

Rearrange: $2 x^{2} - x - 1 = 0$ .

Factorise: $(2 x + 1) (x - 1) = 0$ .

Result: $x = 1, x = - 0.5$ .

4. Solve $\frac{x^{2} + x - 6}{x^{2} + 5} = 0$ .

Solution

Numerator must be zero: $x^{2} + x - 6 = 0$ .

Factorise: $(x + 3) (x - 2) = 0$ .

Result: $x = - 3, x = 2$ .

5. Solve $2^{x^{2} - 4 x + 6} = 8$ .

Solution

Write 8 as base 2: $2^{3}$ .

Equate indices: $x^{2} - 4 x + 6 = 3$ .

$x^{2} - 4 x + 3 = 0$ .

Factorise: $(x - 3) (x - 1) = 0$ .

Result: $x = 3, x = 1$ .

6. Solve $(x^{2} - 7 x + 11)^{8} = 1$ .

Solution

Case 1: Base is 1

$x^{2} - 7 x + 11 = 1 \Rightarrow x^{2} - 7 x + 10 = 0$

$(x - 5) (x - 2) = 0 \Rightarrow x = 2, 5$ .

Case 2: Base is -1 (power even)

$x^{2} - 7 x + 11 = - 1 \Rightarrow x^{2} - 7 x + 12 = 0$

$(x - 3) (x - 4) = 0 \Rightarrow x = 3, 4$ .

Case 3: Power is 0

(Not applicable here as power is constant 8).

Result: $x = 2, 3, 4, 5$ .

7. Right-angled triangle has sides $x, x + 1$ and hypotenuse $x + 2$ . Find $x$ .

Solution

Pythagoras: $x^{2} + (x + 1)^{2} = (x + 2)^{2}$ .

Expand: $x^{2} + x^{2} + 2 x + 1 = x^{2} + 4 x + 4$ .

Simplify: $x^{2} - 2 x - 3 = 0$ .

Factorise: $(x - 3) (x + 1) = 0$ .

Since length > 0, $x = 3$ . Sides are 3, 4, 5.

8. Rectangle has area 20. Sides are $x$ and $x - 1$ . Find $x$ .

Solution

Area = Length × Width: $x (x - 1) = 20$ .

$x^{2} - x - 20 = 0$ .

Factorise: $(x - 5) (x + 4) = 0$ .

Since length > 0, $x = 5$ .

9. Solve $x^{4} - 5 x^{2} + 4 = 0$ (Hidden Quadratic).

Solution

Let $u = x^{2}$ .

Equation becomes: $u^{2} - 5 u + 4 = 0$ .

Factorise: $(u - 4) (u - 1) = 0$ .

$u = 4 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$ .

$u = 1 \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1$ .

Result: $x = \pm 1, \pm 2$ .

10. Solve $3^{2 x} - 12 (3^{x}) + 27 = 0$ .

Solution

Let $u = 3^{x}$ .

Equation becomes: $u^{2} - 12 u + 27 = 0$ .

Factorise: $(u - 9) (u - 3) = 0$ .

$u = 9 \Rightarrow 3^{x} = 9 \Rightarrow x = 2$ .

$u = 3 \Rightarrow 3^{x} = 3 \Rightarrow x = 1$ .

Result: $x = 1, x = 2$ .

1.1 Quadratic Factorisation

1. Standard Form

2. Methods of Factorisation

3. Splitting the Middle Term (The ac Method)

Worked Examples: Factorisation

4. The Quadratic Formula

AIdentification

BSimplification

Worked Examples: Quadratic Formula

5. Completing the Square

A Process for a=1

B Process for a≠1

Worked Examples: Completing the Square

6. Graphing Method

A Shape & Intercepts

B Key Features

Worked Examples: Graphing Method

7. Difference of Two Squares (DOTS)

PIdentifying the Pattern

SSolution Steps

Worked Examples: Difference of Two Squares

8. Mixed Practice Questions

A Process for $a = 1$

B Process for $a \neq 1$