Algebraic manipulation and factorisation

Algebraic fluency means changing the form of an expression without changing its value. It is the working language of A-level Mathematics: calculus, trigonometry, coordinate geometry and mechanics all assume that you can simplify, expand and factorise accurately.

The central question is always:

What operation is valid here, and will the new expression be equivalent to the old one?

For example,

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

for every value of $x$. The two expressions look different but are equivalent.

Prerequisites

You should already be able to:

• use negative numbers and the order of operations;
• multiply and divide integers and simple fractions;
• recognise powers such as $x^2$ and $x^3$;
• substitute a number into a simple expression.

If numerical fractions or powers are a barrier, review exact arithmetic and indices, roots and surds.

Terms, coefficients and factors

In

$4x^2-3x+7,$

the terms are $4x^2$, $-3x$ and $7$. The coefficient of $x^2$ is $4$, the coefficient of $x$ is $-3$, and $7$ is the constant term.

Terms are separated by addition or subtraction. Factors are multiplied. Thus $3x$ is one term with factors $3$ and $x$, while $3+x$ contains two terms.

This distinction controls what you may combine:

$3x+5x=8x,$

but $3x+5$ cannot be simplified because the terms are unlike. Similarly,

$2x^2+7x^2=9x^2,$

but $2x^2+7x$ cannot be combined. Powers are part of the identity of a term.

Worked example: collect like terms

Simplify

$5a-3+2a^2-8a+7-a^2.$

Group terms with exactly the same variable part:

$$\begin{aligned} 5a-3+2a^2-8a+7-a^2 &=(2a^2-a^2)+(5a-8a)+(-3+7)\\ &=a^2-3a+4. \end{aligned}$$

The subtraction signs belong to the terms following them. In particular, the term is $-a^2$, not $a^2$.

Expanding brackets

Expanding uses the distributive law:

$a(b+c)=ab+ac.$

Every term outside a bracket multiplies every relevant term inside it.

A single bracket

Expand and simplify $4(2x-3)-5(x+1)$.

$$\begin{aligned} 4(2x-3)-5(x+1) &=8x-12-5x-5\\ &=3x-17. \end{aligned}$$

The second bracket is multiplied by $-5$, so both signs change. Writing the multiplier as $-5$ rather than treating the minus separately prevents a common error.

Two brackets

To expand $(x+3)(x-5)$, multiply each term in the first bracket by each term in the second:

$$\begin{aligned} (x+3)(x-5) &=x^2-5x+3x-15\\ &=x^2-2x-15. \end{aligned}$$

A useful general identity is

$(x+a)(x+b)=x^2+(a+b)x+ab.$

The coefficient of $x$ is the sum of $a$ and $b$; the constant is their product.

Worked example: non-unit coefficients

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

$$\begin{aligned} (3x-2)(2x+5) &=6x^2+15x-4x-10\\ &=6x^2+11x-10. \end{aligned}$$

There are four products before like terms are collected. Keeping this intermediate line makes sign errors easier to find.

Three brackets

Expand two brackets first, simplify, then multiply by the third. For example,

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

Do not try to multiply all three mentally. A controlled sequence is usually faster because it avoids repair work.

Algebraic identities

An identity is true for every permitted value of the variable. These three identities recur throughout A-level work:

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

The middle term in a squared bracket is essential. For instance,

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

not $x^2+16$. Squaring means multiplying the bracket by itself.

Worked example: use an identity efficiently

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

Take $a=2x$ and $b=3$:

$$\begin{aligned} (2x-3)^2 &=(2x)^2-2(2x)(3)+3^2\\ &=4x^2-12x+9. \end{aligned}$$

Factorising by taking out a common factor

Factorising reverses expansion. It rewrites a sum or difference as a product.

For $12x^3-18x^2$, the greatest common numerical factor is $6$, and both terms contain $x^2$:

$12x^3-18x^2=6x^2(2x-3).$

Check by expanding:

$6x^2(2x-3)=12x^3-18x^2.$

Taking out the greatest common factor usually gives the most useful form, but any valid common factor produces an equivalent expression.

Worked example: a negative common factor

Factorise $-4x^2+12x$.

One valid answer is

$-4x(x-3).$

Taking out $-4x$ leaves a positive leading term inside the bracket, which is often easier to use later. Expanding confirms the signs:

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

Factorising monic quadratics

A monic quadratic has coefficient $1$ on $x^2$. To factorise

$x^2+bx+c,$

find two numbers whose sum is $b$ and whose product is $c$.

Worked example: both signs positive

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

We need two numbers with sum $7$ and product $12$. They are $3$ and $4$, so

$x^2+7x+12=(x+3)(x+4).$

Worked example: mixed signs

Factorise $x^2-2x-15$.

The product is negative, so the two numbers have opposite signs. We need sum $-2$ and product $-15$. The numbers are $3$ and $-5$:

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

The sign of the sum decides which number has the larger magnitude.

Factorising non-monic quadratics

For $ax^2+bx+c$ with $a\ne1$, use splitting the middle term. Find two numbers with product $ac$ and sum $b$, split $bx$, then factorise by grouping.

Worked example

Factorise $6x^2+11x-10$.

Here $ac=6(-10)=-60$. We need two numbers with product $-60$ and sum $11$: $15$ and $-4$.

$$\begin{aligned} 6x^2+11x-10 &=6x^2+15x-4x-10\\ &=3x(2x+5)-2(2x+5)\\ &=(3x-2)(2x+5). \end{aligned}$$

The repeated bracket $(2x+5)$ becomes the common factor. Expanding the answer returns the original quadratic.

If no suitable integer pair exists, the quadratic may not factorise over the integers. Do not invent factors. Later methods such as the quadratic formula may be required.

Difference of two squares

The identity

$a^2-b^2=(a-b)(a+b)$

applies when two square terms are subtracted. There is no middle term.

For example,

$9x^2-25=(3x)^2-5^2=(3x-5)(3x+5).$

Look for a common factor first:

$$\begin{aligned} 8x^2-72 &=8(x^2-9)\\ &=8(x-3)(x+3). \end{aligned}$$

By contrast, $a^2+b^2$ is a sum of squares and does not factorise into real linear factors using this identity.

Choosing the useful form

Expanding and factorising are not competing goals. The question determines the useful form.

• Expanded form makes coefficients and like terms visible.
• Factorised form makes common factors and zeros visible.
• A squared-bracket form makes symmetry and turning points visible.

For instance,

$(x-2)(x+5)=x^2+3x-10.$

The left side immediately shows that the product is zero when $x=2$ or $x=-5$. The right side makes the coefficients easy to compare. Fluency includes moving deliberately between these forms.

Reliable checks

Use at least one of these checks after substantial manipulation:

1. Reverse the operation. Expand a factorisation or factorise an expansion.
2. Substitute a simple value. Equivalent expressions must agree for every permitted value.
3. Check structure. Multiplying two linear expressions should produce a quadratic, not a cubic.

Suppose someone claims

$(2x-1)(x+4)=2x^2+7x-4.$

At $x=1$, the left side is $(1)(5)=5$ and the right side is $2+7-4=5$. This is encouraging but not a proof, because one matching value could be accidental. Full expansion proves the identity:

$(2x-1)(x+4)=2x^2+8x-x-4=2x^2+7x-4.$

Common misconceptions

• $3x+4$ cannot become $7x$. Only like terms combine.
• $x^2+x^2=2x^2$, whereas $x^2\times x^2=x^4$. Addition and multiplication use different rules.
• $-(x-3)=-x+3$. A negative multiplier changes every sign in the bracket.
• $(a+b)^2\ne a^2+b^2$. The correct expansion includes $2ab$.
• $x^2-9=(x-3)(x+3)$, not $(x-3)^2$.
• Factorising changes form, not value. An answer must expand exactly to the original expression.
• Cancelling is division by a common factor, not deletion of matching terms across addition. This is developed in algebraic fractions.

Self-check

Try these without a calculator. Expand your factorised answers to check them.

1. Collect terms

Simplify

$4p^2-7p+3-6p^2+2p-8.$

Answer

$-2p^2-5p-5.$

2. Expand

Expand and simplify

$3(2x-5)-(x-4).$

Answer

$$\begin{aligned} 3(2x-5)-(x-4) &=6x-15-x+4\\ &=5x-11. \end{aligned}$$

3. Expand two brackets

Expand $(2y+3)(y-4)$.

Answer

$2y^2-8y+3y-12=2y^2-5y-12.$

4. Take out a common factor

Factorise fully

$15a^3b-10a^2b^2.$

Answer

The greatest common factor is $5a^2b$:

$15a^3b-10a^2b^2=5a^2b(3a-2b).$

5. Factorise a monic quadratic

Factorise $x^2+x-20$.

Answer

The numbers $5$ and $-4$ have product $-20$ and sum $1$:

$x^2+x-20=(x+5)(x-4).$

6. Factorise a non-monic quadratic

Factorise $3x^2-10x-8$.

Answer

Since $ac=-24$, use $-12$ and $2$:

$$\begin{aligned} 3x^2-10x-8 &=3x^2-12x+2x-8\\ &=3x(x-4)+2(x-4)\\ &=(3x+2)(x-4). \end{aligned}$$

7. Recognise a difference of squares

Factorise fully

$5x^2-80.$

Answer

Take out the common factor before using the identity:

$5x^2-80=5(x^2-16)=5(x-4)(x+4).$

8. Diagnose an error

A student writes $(x-6)^2=x^2+36$. Explain the error and correct it.

Answer

The bracket must be multiplied by itself, producing two middle products:

$$\begin{aligned} (x-6)^2 &=(x-6)(x-6)\\ &=x^2-6x-6x+36\\ &=x^2-12x+36. \end{aligned}$$

Where to go next

Once these manipulations are secure:

• use factors to find roots in quadratic equations;
• practise maintaining equivalence in linear equations and rearranging formulae;
• learn when factors may be cancelled in algebraic fractions;
• extend these skills to A-level quadratics and polynomials.

Aim for accuracy before speed. When every line is an equivalent expression and every factorisation is checked by expansion, speed develops without fragile shortcuts.