Factorise x² + 7x + 12 by splitting the middle term

Question

Factorise $x^2 + 7x + 12$ by splitting the middle term.

Solution — Step by Step

We have $x^2 + 7x + 12$ . In the method of splitting the middle term, we split the coefficient of $x$ (which is 7) into two numbers $p$ and $q$ such that:

$p + q = 7$ (sum = coefficient of $x$ )
$p \times q = 12$ (product = constant term × coefficient of $x^2 = 12 \times 1 = 12$ )

We need two numbers that add to 7 and multiply to 12. Let’s think systematically:

Pairs with product 12: $(1,12)$ , $(2,6)$ , $(3,4)$ , $(4,3)$ , $(6,2)$ , $(12,1)$

Check which pair has sum 7:

$3 + 4 = 7$ ✓

So $p = 3$ and $q = 4$ .

Replace $7x$ with $3x + 4x$ :

x^2 + 7x + 12 = x^2 + 3x + 4x + 12

Now group in pairs:

= (x^2 + 3x) + (4x + 12)

Factor out common terms from each group:

= x(x + 3) + 4(x + 3)

$(x + 3)$ is common in both terms:

= (x + 3)(x + 4)

\boxed{x^2 + 7x + 12 = (x + 3)(x + 4)}

Verification: $(x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$ ✓

Why This Works

Every quadratic $ax^2 + bx + c$ with rational roots can be factorised this way. The condition $p + q = b$ and $pq = ac$ comes from the fact that when we expand $(x + p)(x + q)$ , we get $x^2 + (p+q)x + pq$ . So the “middle term” is $(p+q)$ and the “last term” is $pq$ .

By reversing this: given $b$ and $c$ , find $p$ and $q$ with $p+q = b$ and $pq = c$ (when $a = 1$ ). This is always possible if the discriminant $b^2 - 4ac \geq 0$ .

Alternative Method — Quadratic Formula

The roots of $x^2 + 7x + 12 = 0$ are:

x = \frac{-7 \pm \sqrt{49 - 48}}{2} = \frac{-7 \pm 1}{2}

x = -3 \text{ or } x = -4

So the factors are $(x+3)(x+4)$ .

The splitting method is faster for simple cases; the formula is more reliable for complex coefficients.

Common Mistake

Finding pairs that sum to the constant term and multiply to the middle coefficient — the reverse of what’s needed. The correct rule: sum = middle coefficient (7), product = last term × leading coefficient (12 × 1 = 12). Students sometimes do sum = 12 and product = 7, which gives non-integer solutions and leads nowhere.

For $ax^2 + bx + c$ where $a \neq 1$ : the product condition changes to $pq = a \times c$ (not just $c$ ). Then after splitting and grouping, you’ll need to factor out $\frac{1}{a}$ or handle the $a$ within the grouping. Always multiply $a \times c$ first.

Question

Solution — Step by Step

Why This Works

Alternative Method — Quadratic Formula

Common Mistake

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