How to factor a quadratic by splitting the middle term
Factoring a quadratic rewrites an expression as a product of brackets. For ax² + bx + c, a useful method is to split bx into two terms that let you factor by grouping. The pair of numbers must satisfy two conditions at once: their product is ac and their sum is b.
Put the expression in order and look for a common factor
Write the terms in descending powers: x², then x, then the constant. In 6x² + 11x + 3, a = 6, b = 11 and c = 3. Include the signs when recording them. In 2x² − 7x + 3, b is −7.
Before splitting the middle term, check whether every term has a common factor. For 2x² + 10x + 12, take out 2 to get 2(x² + 5x + 6). You can then work on the simpler quadratic inside the bracket. Keep the outside factor in the final answer.
If the expression has no x term, record b = 0. For x² − 16, the pair will need a sum of zero. Do not add an x term that was never present.
Start with a quadratic whose leading coefficient is one
Consider x² + 5x + 6. Here ac = 1 × 6 = 6 and b = 5. The positive factor pairs of 6 are 1 and 6, or 2 and 3. Only 2 and 3 add to 5.
Split 5x into 2x + 3x: x² + 2x + 3x + 6. Group the first two terms and the last two terms: x(x + 2) + 3(x + 2). Each group now contains the same bracket, x + 2.
Factor out that shared bracket to get (x + 2)(x + 3). Expanding checks the answer: x² + 3x + 2x + 6 = x² + 5x + 6. If you removed a common factor of 2 earlier, the full answer to 2x² + 10x + 12 is 2(x + 2)(x + 3).
Use ac when the coefficient of x² is not one
For 6x² + 11x + 3, multiply a and c: 6 × 3 = 18. You need two numbers with product 18 and sum 11. The positive pairs are 1 and 18, 2 and 9, or 3 and 6. Choose 2 and 9.
Write 6x² + 9x + 2x + 3. The first group has a common factor of 3x and the second has a common factor of 1, giving 3x(2x + 3) + 1(2x + 3). The shared bracket is 2x + 3, so the answer is (3x + 1)(2x + 3).
Using 3 as the target product would miss the leading coefficient. The pair used to split 11x is based on ac, while the numbers in the final brackets come from the grouped factors. They are not automatically the same pair.
You can split the terms in the other order: 6x² + 2x + 9x + 3 = 2x(3x + 1) + 3(3x + 1). This gives (2x + 3)(3x + 1), the same product with the brackets reversed.
Choose signs from both the product and the sum
If ac is positive, the two splitting numbers have the same sign. A positive sum means both are positive; a negative sum means both are negative. If ac is negative, their signs differ, and the number with the larger absolute value determines the sign of their sum.
For 2x² − 7x + 3, ac = 6 and b = −7. Choose −6 and −1. Splitting gives 2x² − 6x − x + 3. Factor the groups as 2x(x − 3) − 1(x − 3), then combine them into (2x − 1)(x − 3).
Notice the negative factor in the second group: −x + 3 = −1(x − 3). Taking out positive 1 instead would give a different bracket and prevent the groups from matching. Expand your final answer to check all three coefficients, including the sign of the middle term.
For x² − 16, the required pair is 4 and −4. Splitting gives x² + 4x − 4x − 16 = x(x + 4) − 4(x + 4) = (x − 4)(x + 4). The introduced x terms cancel, preserving the original expression.
Know when an integer pair is unavailable
List the integer factor pairs of ac systematically. For x² + x − 1, the integer pairs with product −1 are 1 and −1, whose sum is zero. Neither order gives the required sum of 1. This expression cannot be written as a product of linear factors with integer coefficients.
An unsuccessful integer-pair search does not mean an equation has no solutions. It means this particular factoring method has not produced integer brackets. If your task is to solve the equation, another method such as completing the square or the quadratic formula may be appropriate.
Keep factoring an expression separate from solving an equation. Writing (x + 2)(x + 3) is a factorisation. If the question gives x² + 5x + 6 = 0, the factored equation is (x + 2)(x + 3) = 0, so x = −2 or x = −3. Do not set brackets equal to zero when the original task gives no equation.
Keep a working record you can check
Use this template: original expression; common factor removed; a, b and c including signs; product ac; pair whose product is ac and sum is b; four-term split; factors of each group; shared bracket; final product; expansion check.
Try 3x² + 10x + 8 before checking the answer. Here ac = 24 and the pair is 6 and 4. The split is 3x² + 6x + 4x + 8 = 3x(x + 2) + 4(x + 2), giving (3x + 4)(x + 2). Expanding gives 3x² + 6x + 4x + 8, which collects to the original expression.
If the two grouped brackets do not match, check the split and the sign of the factor taken from each group. If they match but expansion gives the wrong expression, inspect the outside factors and any common factor removed at the beginning.
