# The Algebra Toolkit

> Expanding, factorising, index laws, surds and solving equations. The skills every later chapter uses.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/algebra-toolkit/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The carpenter's toolbox
Mia's uncle builds decks. Before he starts a job he checks his toolbox: hammer, saw, tape measure, level. He does not think about them while he works. He just reaches for the right one.

Algebra is the toolbox for all of Year 11 and 12 maths. Calculus, logs and probability all fall apart if you cannot expand a bracket or solve an equation without thinking. This chapter checks your toolbox so the later chapters feel easy.

**In short:** Algebra skills are the tools every later topic uses: expanding brackets, factorising, the index laws and solving equations. Expand by multiplying every term inside by the term outside, factorise by taking out common factors first, and solve by doing the same thing to both sides.

## Starting point: Years 9 and 10

### Like terms and substitution

**Like terms** have exactly the same letters and powers. You can add them: $3x + 5x = 8x$, but $3x + 5x^2$ cannot be joined.

To **substitute**, swap each letter for its number, using brackets. If $x = -2$, then $3x^2 = 3(-2)^2 = 3 \times 4 = 12$.

### Expanding brackets

Multiply everything inside by the term outside.

$$3(2x - 5) = 6x - 15$$

For two brackets, multiply every term in the first by every term in the second.

$$(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15$$

### Key idea: Three special expansions
$$(a + b)^2 = a^2 + 2ab + b^2$$
$$(a - b)^2 = a^2 - 2ab + b^2$$
$$(a + b)(a - b) = a^2 - b^2$$

### Factorising

Factorising is expanding backwards. Always try these in order:

1. **Common factor:** $6x^2 + 9x = 3x(2x + 3)$
2. **Difference of two squares:** $x^2 - 49 = (x + 7)(x - 7)$
3. **Quadratic trinomial:** find two numbers that multiply to the constant and add to the middle number. For $x^2 + 7x + 12$, the numbers 3 and 4 work, so it is $(x + 3)(x + 4)$.
4. **Grouping (four terms):** pair the terms and take out a common factor from each pair. The same bracket appears twice:
$$x^3 + 2x^2 - 3x - 6 = x^2(x + 2) - 3(x + 2) = (x^2 - 3)(x + 2)$$

### Worked example: Factorise with a leading number
Factorise $2x^2 + 7x + 3$.

Multiply $2 \times 3 = 6$. Find two numbers that multiply to 6 and add to 7: they are 6 and 1.

Split the middle: $2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$.

### Index laws

$$a^m \times a^n = a^{m+n} \qquad \frac{a^m}{a^n} = a^{m-n} \qquad (a^m)^n = a^{mn}$$

$$a^0 = 1 \qquad a^{-n} = \frac{1}{a^n} \qquad a^{\frac{1}{n}} = \sqrt[n]{a}$$

So $8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4$.

### Calculator tip: Brackets for negatives and fractional powers
Type $(-3)^2$ with brackets: `( - 3 ) x²` gives 9. Without brackets, `- 3 x²` gives $-9$, because the calculator squares 3 first and then makes it negative.

For a fractional power such as $8^{\frac{2}{3}}$, use the power key (`xʸ` or `^`) and put the whole power in brackets: `8 ^ ( 2 ÷ 3 )` gives 4.

### Surds

A **surd** is a root that is not a whole number, like $\sqrt{2}$. Simplify by finding a square factor:

$$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$$

To **rationalise** a denominator, multiply top and bottom by the surd:

$$\frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$$

### Solving equations

Do the same thing to both sides until the letter is alone.

$$\frac{2x - 1}{3} = 5 \;\Rightarrow\; 2x - 1 = 15 \;\Rightarrow\; 2x = 16 \;\Rightarrow\; x = 8$$

**Simultaneous equations** have two unknowns. Use elimination or substitution.

### Worked example: Elimination
Solve $x + y = 10$ and $x - y = 4$.

Add the equations: $2x = 14$, so $x = 7$. Then $7 + y = 10$, so $y = 3$.

### Watch out: Squaring a bracket
$(x + 3)^2$ is **not** $x^2 + 9$. It is $x^2 + 6x + 9$. Write it as $(x + 3)(x + 3)$ if you are unsure.

### Exam tip: Check by substituting
After solving, put your answer back into the original equation. It takes ten seconds and catches most mistakes.

### Calculator tip: Check a solution with memory
Type your answer, then press `STO` and a letter such as `X` to store it. Now type the left side of the original equation using `X`, then the right side. If both give the same number, your solution is right. If your calculator has a `CALC` key, it does the same job: it asks for the value of $x$ and works out the expression.

**Mind map**

```text
Algebra toolkit
  Expand
    Every term times every term
    Perfect squares
  Factorise
    Common factor first
    Difference of squares
    Trinomials
  Indices
    Add, subtract, multiply powers
    Negative means reciprocal
    Fraction means root
  Surds
    Find square factors
    Rationalise denominators
  Solve
    Same to both sides
    Simultaneous: eliminate
```

### Year 11: Rationalising with a conjugate
For a denominator like $\sqrt{5} + 2$, multiply top and bottom by its **conjugate** $\sqrt{5} - 2$. The difference of two squares removes the surd:

$$\frac{1}{\sqrt{5} + 2} \times \frac{\sqrt{5} - 2}{\sqrt{5} - 2} = \frac{\sqrt{5} - 2}{5 - 4} = \sqrt{5} - 2$$

### Year 11: Changing the subject and inequalities
To make $r$ the subject of $V = \pi r^2 h$: divide by $\pi h$, then take the square root, so $r = \sqrt{\dfrac{V}{\pi h}}$.

Solve inequalities like equations, with one extra rule: **when you multiply or divide by a negative number, flip the sign.**

$$-3x > 12 \;\Rightarrow\; x < -4$$

Absolute value $|x|$ is the distance from zero. $|x - 2| < 5$ means $x$ is less than 5 away from 2, so $-3 < x < 7$.

**Absolute value equations.** $|A| = k$ means $A = k$ or $A = -k$. So $|2x - 5| = 3$ gives $2x - 5 = 3$ or $2x - 5 = -3$, that is $x = 4$ or $x = 1$.

If there is an $x$ on the other side too, solve both cases and **check each answer**, because one may not work. For $|x - 1| = 2x + 4$: the case $x - 1 = 2x + 4$ gives $x = -5$, but then the right side is $-6$, and an absolute value is never negative, so reject it. The case $x - 1 = -(2x + 4)$ gives $x = -1$, which checks: $|-2| = 2$.

For inequalities: $|A| < k$ means $-k < A < k$, and $|A| \geq k$ means $A \leq -k$ or $A \geq k$.

Also, $\sqrt{a^2} = |a|$, not just $a$. For example $\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = |x - 3|$.

### Year 11: More factorising and equations with powers
**Sums and differences of cubes:**
$$a^3 + b^3 = (a + b)(a^2 - ab + b^2) \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$
For example $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ and $27y^3 + 1 = (3y + 1)(9y^2 - 3y + 1)$.

**Equations with the unknown in the power.** Write both sides as powers of the same base, then make the powers equal:
$$4^{x + 2} = 8^x \;\Rightarrow\; 2^{2x + 4} = 2^{3x} \;\Rightarrow\; 2x + 4 = 3x \;\Rightarrow\; x = 4$$
With different bases such as $\dfrac{25^x \times 3^{2x}}{15^{x}}$, break every base into primes ($25 = 5^2$, $15 = 3 \times 5$) and collect the powers of each prime.

**Quadratics in disguise.** In $9^x - 4 \times 3^x + 3 = 0$, notice $9^x = (3^x)^2$. Let $u = 3^x$: then $u^2 - 4u + 3 = 0$, so $u = 1$ or $u = 3$. That gives $3^x = 1$ or $3^x = 3$, so $x = 0$ or $x = 1$.

**A squaring trick.** If $x + \dfrac{1}{x} = 3$, square both sides: $x^2 + 2 + \dfrac{1}{x^2} = 9$, so $x^2 + \dfrac{1}{x^2} = 7$.

### Year 12: Algebraic fractions in calculus
Year 12 calculus often needs fractions joined or split. Use a common denominator:

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

And rewrite roots and fractions as powers before differentiating or integrating: $\dfrac{3}{\sqrt{x}} = 3x^{-\frac{1}{2}}$.

## Quick summary

- Expand: multiply every term inside by every term outside.
- Factorise: common factor, then difference of squares, then trinomial.
- Index laws: add powers when multiplying, subtract when dividing, multiply for a power of a power.
- $a^{-n} = \dfrac{1}{a^n}$ and $a^{\frac{1}{n}} = \sqrt[n]{a}$.
- Simplify surds with square factors. Rationalise to clear roots from the bottom.
- Solve by doing the same thing to both sides. Check by substituting back.

## Quiz

Try these questions from the chapter quiz.

1. Expand $(x + 4)(x - 2)$.
   - A. $x^2 - 8$
   - B. $x^2 + 2x - 8$
   - C. $x^2 + 6x - 8$
   - D. $x^2 - 2x - 8$
   - Answer: B. $x^2 + 2x - 8$. $x^2 - 2x + 4x - 8 = x^2 + 2x - 8$.

2. Expand $(x - 5)^2$.
   - A. $x^2 - 25$
   - B. $x^2 + 25$
   - C. $x^2 - 10x + 25$
   - D. $x^2 - 5x + 25$
   - Answer: C. $x^2 - 10x + 25$. $(a - b)^2 = a^2 - 2ab + b^2$, so $x^2 - 10x + 25$.

3. Factorise $x^2 - 36$.
   - A. $(x - 6)(x + 6)$
   - B. $(x - 6)^2$
   - C. $(x - 18)(x + 2)$
   - D. $x(x - 36)$
   - Answer: A. $(x - 6)(x + 6)$. Difference of two squares: $a^2 - b^2 = (a - b)(a + b)$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/algebra-toolkit/#quiz