Factorise x² + 5x + 6 by splitting the middle term

Question

Factorise the expression $x^2 + 5x + 6$ by splitting the middle term.

(NCERT Class 8)

Solution — Step by Step

We need two numbers that:

Multiply to give the constant term: $6$ (product of first and last coefficients: $1 \times 6 = 6$ )
Add to give the coefficient of the middle term: $5$

Let’s find these two numbers.

Factor pairs of 6: $(1, 6)$ and $(2, 3)$ .

Check sums: $1 + 6 = 7$ (not 5), $2 + 3 = 5$ (yes!)

The two numbers are 2 and 3.

Replace $5x$ with $2x + 3x$ :

x^2 + 2x + 3x + 6

Group in pairs:

= x(x + 2) + 3(x + 2)

Take out the common factor $(x + 2)$ :

= \mathbf{(x + 2)(x + 3)}

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

Why This Works

Factorisation reverses expansion. When we multiply $(x + a)(x + b)$ , we get $x^2 + (a+b)x + ab$ . So the middle coefficient is the sum and the constant is the product of $a$ and $b$ . Splitting the middle term essentially finds these values of $a$ and $b$ .

This method works for any quadratic of the form $x^2 + bx + c$ where we can find two integers whose sum is $b$ and product is $c$ .

Alternative Method — Direct Observation

Since we need $(x + ?)(x + ?)$ with the blanks adding to 5 and multiplying to 6, we can mentally test: $2 + 3 = 5$ and $2 \times 3 = 6$ . So the answer is $(x + 2)(x + 3)$ .

For harder quadratics like $6x^2 + 11x + 3$ , the product to consider is $6 \times 3 = 18$ , and you need two numbers adding to 11 with product 18. Those are 9 and 2. Split: $6x^2 + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (3x + 1)(2x + 3)$ .

Common Mistake

Students sometimes pick numbers that add to 6 and multiply to 5 (swapping the conditions). Remember: the two numbers must add to the middle coefficient (5) and multiply to the constant term (6). Getting this backwards leads to wrong factors.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Observation

Common Mistake

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