Expansion and Factorisation

For $(x + 4)(x + 2)$ the four products are $x^2$, $2x$, $4x$ and $8$. Adding them: $x^2 + 2x + 4x + 8 = x^2 + 6x + 8.$

The two middle terms $2x$ and $4x$ are like terms, so they collect to $6x$.

A few more, watching the signs:

• $(x - 3)(x + 5) = x^2 + 5x - 3x - 15 = x^2 + 2x - 15$.
• $(x + 2)^2 = (x + 2)(x + 2) = x^2 + 4x + 4$.

A grid keeps you sure you have caught all four products. Multiply, fill the grid, then collect the like middle terms.

• Multiplying two brackets gives four products.
• Multiply every term in the first by every term in the second.
• The two middle terms usually collect together.
• A grid keeps track of all four products.

A quadratic of this kind always has three parts: an $x^2$ term, an $x$ term, and a number.

We describe this shape as $x^2 + bx + c$, where:

• $b$ is the coefficient of $x$ — here $b = 6$.
• $c$ is the number term — here $c = 8$.

Why does this matter? Because expanding and factorising are opposites. You have just seen $(x + 4)(x + 2)$ open up into $x^2 + 6x + 8$. Soon you will go the other way — taking a quadratic and finding the two brackets it came from. Spotting the form $x^2 + bx + c$ is the first step of that journey.

• A quadratic expression has an $x^2$ term as its highest power.
• Expanding two brackets always produces a quadratic.
• The form $x^2 + bx + c$ has three parts.
• $b$ is the coefficient of $x$; $c$ is the number term.

Always take out the highest factor. $12x^2 + 8x = 2(6x^2 + 4x)$ is partly factorised — but $4x(3x + 2)$ is fully done.

This skill matters here because not every quadratic factorises into two brackets — some just have a common factor. Always check for one first.

• A common factor divides into every term.
• Find the highest common factor, including any letter.
• Write the HCF outside, divide each term to fill the bracket.
• Always take out the highest factor, not a smaller one.

The method is a number hunt. You need two numbers that:

• add together to make $b$ (the coefficient of $x$), and
• multiply together to make $c$ (the number term).

Example: Factorise $x^2 + 6x + 8$. You need two numbers that add to 6 and multiply to 8. Try the factor pairs of 8: $1 \times 8$ (adds to 9, no) and $2 \times 4$ (adds to 6, yes!). So the two numbers are 2 and 4, and: $x^2 + 6x + 8 = (x + 4)(x + 2).$

The two numbers drop straight into the brackets. Always check by expanding your answer — if it returns to the original quadratic, you are right.

• Find two numbers that add to $b$ and multiply to $c$.
• List the factor pairs of $c$, then test which pair adds to $b$.
• The two numbers go straight inside the two brackets.
• Check by expanding the brackets back out.

Two cases cover almost everything:

• $c$ positive: the two numbers have the same sign, matching the sign of $b$. For $x^2 - 7x + 12$, you need two negatives that multiply to $+12$ and add to $-7$ — that is $-3$ and $-4$, giving $(x - 3)(x - 4)$.
• $c$ negative: the two numbers have opposite signs. For $x^2 + 2x - 15$, you need a positive and a negative multiplying to $-15$ and adding to $+2$ — that is and , giving .

Work calmly: list the factor pairs of $|c|$, then decide the signs from these two rules. Test each candidate by adding — and always expand at the end to be sure.

• If $c$ is positive, both numbers share the sign of $b$.
• If $c$ is negative, the two numbers have opposite signs.
• List factor pairs of $|c|$, then choose the signs.
• Test by adding, and expand at the end to check.

In everyday life, factorising is spotting hidden structure — seeing that a tricky-looking total is really two simple amounts multiplied together.

If quadratic work feels hard later, it is usually a sign slip in the two-number hunt underneath. Come back to this guide whenever you need a refresher — there is no prize for rushing.

• Solving quadratic equations depends on factorising into brackets.
• Quadratic graphs are sketched from the factorised form.
• Algebraic fractions are simplified by factorising first.
• Rectangle areas model the expansion of two brackets.

Step 2

Step 1

1. Step 1

   Multiply every term, taking care with the minus sign.

   $x \times x = x^2,\quad x \times 5 = 5x$

2. Step 2

   Now the products from the $-3$.

   $-3 \times x = -3x,\quad -3 \times 5 = -15$

3. Step 3

   Collect the like middle terms $5x$ and $-3x$.

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