Algebra: Expressions, Indices, Expanding & Factorising

Algebraic notation and simplifying expressions

Algebra uses letters to stand for numbers. A clear notation saves marks, so write expressions tidily.

$3a$ means $3 \times a$; never write $a3$.
$ab$ means $a \times b$; the multiplication sign is hidden.
$a^2$ means $a \times a$ (a squared); $a^3$ means $a \times a \times a$.
$\dfrac{a}{b}$ means $a \div b$.

A term is a single number/letter product such as $5x^2$. Like terms have exactly the same letters and powers, so they can be collected.

Key terms

Expression: letters and numbers joined by $+$, $-$, $\times$, $\div$ (no equals sign), e.g. $3x + 2y$.

Coefficient: the number multiplying a term; in $7x$ the coefficient is $7$.

Like terms: terms with identical letters and powers, e.g. $4x$ and $-9x$, but not $4x$ and $4x^2$.

Worked example

Simplify $5x + 3y - 2x + 7y - 4$.

Collect $x$ terms: $5x - 2x = 3x$. Collect $y$ terms: $3y + 7y = 10y$.

Answer: $3x + 10y - 4$.

Watch out

$x^2$ and $x$ are not like terms. $x^2 + x$ does not simplify to $2x^2$ or $2x$ — leave it as $x^2 + x$.

The laws of indices

An index (plural indices) is a power. The same three laws you use with numbers apply to algebra.

Three special results complete the set:

$a^0 = 1$ (anything to the power zero is $1$).
$a^{-n} = \dfrac{1}{a^n}$ (a negative power means reciprocal).
$a^{1/n} = \sqrt[n]{a}$, so $a^{1/2} = \sqrt{a}$.

Worked example

Simplify $\dfrac{12x^5 y^3}{4x^2 y}$.

Numbers: $12 \div 4 = 3$. Powers of $x$: $x^{5-2} = x^3$. Powers of $y$: $y^{3-1} = y^2$.

Answer: $3x^3 y^2$.

Watch out

The laws of indices only apply when the base is the same. $x^2 \times y^3$ stays as $x^2 y^3$ — you cannot add the powers.

Expanding brackets

To expand (or multiply out) means to remove brackets by multiplying every term inside by the term outside.

Worked example

Expand $3x(2x - 5)$.

$3x \times 2x = 6x^2$ and $3x \times (-5) = -15x$.

Answer: $6x^2 - 15x$.

Take care with signs. A minus outside flips every sign inside:

Worked example

Expand and simplify $4(2x + 3) - 2(3x - 1)$.

$4(2x+3) = 8x + 12$ and $-2(3x-1) = -6x + 2$.

Combine: $8x - 6x + 12 + 2 = 2x + 14$.

Expanding two binomials

To expand $(x + a)(x + b)$, multiply each term in the first bracket by each term in the second — four products in all. The grid (box) method keeps every product organised.

Adding the four cells: $x^2 + 3x + 4x + 12 = x^2 + 7x + 12$.

Many people use FOIL (First, Outer, Inner, Last) for the same four products.

Worked example

Expand $(2x - 3)(x + 5)$.

First $2x \times x = 2x^2$; Outer $2x \times 5 = 10x$; Inner $-3 \times x = -3x$; Last $-3 \times 5 = -15$.

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

Exam tip

Squaring a bracket means two brackets: $(x+5)^2 = (x+5)(x+5) = x^2 + 10x + 25$. A common error is to write $x^2 + 25$ — never do this.

Factorising: common factors

Factorising is the reverse of expanding: write an expression as a product. Always look first for the highest common factor of every term.

Worked example

Factorise $6x^2 + 9x$.

HCF of $6x^2$ and $9x$ is $3x$. Divide each term: $6x^2 \div 3x = 2x$, $9x \div 3x = 3$.

Answer: $3x(2x + 3)$. Check by expanding.

Difference of two squares

Any expression of the form $a^2 - b^2$ factorises to $(a + b)(a - b)$.

Worked example

Factorise $x^2 - 49$.

$49 = 7^2$, so $x^2 - 49 = (x + 7)(x - 7)$.

Factorise $9y^2 - 25$. Here $9y^2 = (3y)^2$ and $25 = 5^2$, giving $(3y + 5)(3y - 5)$.

Watch out

Difference of two squares needs a minus sign between two squares. $x^2 + 49$ does not factorise.

Factorising quadratics $x^2 + bx + c$

Find two numbers that multiply to give $c$ and add to give $b$.

Worked example

Factorise $x^2 + 7x + 12$.

Two numbers multiplying to $12$ and adding to $7$: $3$ and $4$.

Answer: $(x + 3)(x + 4)$.

Factorise $x^2 - 2x - 15$. Need product $-15$, sum $-2$: that is $-5$ and $+3$. Answer: $(x - 5)(x + 3)$.

Factorising quadratics $ax^2 + bx + c$

When $a \neq 1$, use the AC method: find two numbers multiplying to $a \times c$ and adding to $b$, split the middle term, then factorise in pairs.

Worked example

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

$a \times c = 2 \times 3 = 6$. Two numbers multiplying to $6$, adding to $7$: $6$ and $1$.

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

Algebraic fractions

Treat algebraic fractions exactly like number fractions: factorise, then cancel common factors (not common terms).

Worked example

Simplify $\dfrac{x^2 - 9}{x^2 + 7x + 12}$.

Factorise both: top $(x+3)(x-3)$, bottom $(x+3)(x+4)$.

Cancel $(x+3)$: $\dfrac{x - 3}{x + 4}$.

To add or subtract, use a common denominator.

Worked example

Simplify $\dfrac{2}{x} + \dfrac{3}{x + 1}$.

Common denominator $x(x+1)$: $\dfrac{2(x+1) + 3x}{x(x+1)} = \dfrac{2x + 2 + 3x}{x(x+1)} = \dfrac{5x + 2}{x(x+1)}$.

To multiply, multiply tops and bottoms; to divide, multiply by the reciprocal (flip the second fraction).

Worked example

Simplify $\dfrac{3}{x} \div \dfrac{6}{x^2}$.

Flip and multiply: $\dfrac{3}{x} \times \dfrac{x^2}{6} = \dfrac{3x^2}{6x} = \dfrac{x}{2}$.

Changing the subject of a formula

The subject is the single letter on its own. Rearranging uses the same inverse operations as solving an equation — do the same thing to both sides.

Worked example

Make $r$ the subject of $A = \pi r^2$.

Divide by $\pi$: $\dfrac{A}{\pi} = r^2$. Square root: $r = \sqrt{\dfrac{A}{\pi}}$.

Worked example

Make $x$ the subject of $y = \dfrac{3x - 1}{2}$.

Multiply by $2$: $2y = 3x - 1$. Add $1$: $2y + 1 = 3x$. Divide by $3$: $x = \dfrac{2y + 1}{3}$.

When the subject appears twice, collect those terms on one side and factorise it out.

Worked example

Make $x$ the subject of $ax + b = cx + d$.

Gather $x$ terms: $ax - cx = d - b$. Factorise: $x(a - c) = d - b$.

Divide: $x = \dfrac{d - b}{a - c}$.

Exam tip

Whenever the new subject appears more than once, the key move is always the same: get every term containing it on one side, factorise it out as a bracket, then divide by the bracket.

Watch out

When you square-root to find a subject, the context may require only the positive value (e.g. a length or radius), even though $\pm$ is algebraically correct.