Factorising , multiplication in reverse

If multiplication expands $(x+2)(x+3) = x^2 + 5x + 6$, factorising does the reverse: given $x^2 + 5x + 6$, produce $(x+2)(x+3)$. Why bother? Because a factored expression reveals its zeros, its common pieces, and its structure , information that an expanded expression hides.

Concept

Step 1: Take out the highest common factor (HCF).

Look at all the terms. Find the greatest factor (numerical and/or variable) that divides every one of them. Pull it out.

Example: $12x^3 - 18x^2 + 6x$. The HCF is $6x$. So $= 6x(2x^2 - 3x + 1)$.

Step 2: Recognise an identity.

If your expression is a perfect square, a difference of squares, or a sum/difference of cubes, factorise using the matching identity:

• $a^2 + 2ab + b^2 = (a+b)^2$.
• $a^2 - 2ab + b^2 = (a-b)^2$.
• $a^2 - b^2 = (a+b)(a-b)$.
• $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$.
• $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.

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

Example: $x^2 + 6x + 9 = (x+3)^2$.

Step 3: Factor by grouping.

When an expression has four terms, you can sometimes group them in pairs to find a common bracket.

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

Step 4: Splitting the middle term (for $x^2 + bx + c$).

To factor a quadratic like $x^2 + 5x + 6$:

1. Find two numbers that multiply to $c = 6$ and add to $b = 5$.
2. Try $2$ and $3$: $2 \times 3 = 6$, $2 + 3 = 5$. Match.
3. Write $x^2 + 5x + 6 = x^2 + 2x + 3x + 6 = x(x+2) + 3(x+2) = (x+2)(x+3)$.

If the coefficient of $x^2$ is not $1$ , say $2x^2 + 7x + 3$:

1. Multiply the leading coefficient and constant: $2 \times 3 = 6$.
2. Find two numbers that multiply to $6$ and add to $7$: $6$ and $1$.
3. Split: $2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (x+3)(2x+1)$.

Always check by expanding the factored form. If it matches the original, you are correct.

Worked examples

Example 1. Factorise $15x^3 y - 20x^2 y^2 + 25 x y^3$.

• HCF $= 5xy$.
• $= 5xy(3x^2 - 4xy + 5y^2)$. (Inner expression doesn't factor further.)

Example 2. Factorise $16a^2 - 49b^2$.

• Difference of squares: $(4a)^2 - (7b)^2 = (4a + 7b)(4a - 7b)$.

Example 3. Factorise $x^2 - 7x + 12$.

• Need two numbers with product $12$ and sum $-7$: $-3$ and $-4$.
• $= x^2 - 3x - 4x + 12 = x(x-3) - 4(x-3) = (x-3)(x-4)$.

Example 4. Factorise $ax + ay + bx + by$.

• Group: $a(x+y) + b(x+y) = (a+b)(x+y)$.

Try it yourself

1. Factorise $4x^2 - 16$.
2. Factorise $x^2 + 10x + 25$.
3. Factorise $x^2 - 13x + 42$.
4. Factorise $9 - a^2$.
5. Factorise $2x^2 + 5x + 2$ by splitting the middle term.
6. Factorise $3a^2 + 6ab - 4a - 8b$ by grouping.
7. Factorise $a^3 - 27$.
8. Factorise $x^2 - 4x - 12$.

Activity / Insight

Reverse FOIL race. With a friend, take turns reading out a quadratic in the form $x^2 + bx + c$ where $b, c$ are small integers. The other has to factorise it in under $30$ seconds. After ten rounds, switch roles. You will quickly internalise the "two numbers with product $c$, sum $b$" recipe , the foundation of all algebraic factoring you will do for years.