Factoring trinomials
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Factor x^2 + 7x + 12.
Common mix-up
In this example, a student wrote x^2 + 7x + 12 = (x + 2)(x + 6).
The student picked 2 and 6 because 2 * 6 = 12, but forgot that the pair must also add to the middle coefficient.
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See why this works
Which pair of numbers factors x^2 + 7x + 12?
A trinomial like x^2 + 7x + 12 hides a pair of numbers. Find the pair, and the factored form falls into place.
For x^2 + bx + c, find two numbers whose product is c and whose sum is b.
Factoring a trinomial starts with a pair of numbers, not with x. Look at x^2 + 7x + 12. The constant 12 must come from multiplying two numbers, and the middle coefficient 7 must come from adding the same two numbers. So the numbers you need satisfy two conditions at once: product 12, sum 7.
List the whole-number factor pairs of 12: 1 and 12, 2 and 6, 3 and 4. Their sums are 13, 8, and 7. Only the pair 3 and 4 has a sum of 7, so that pair is the one hiding in the trinomial. This is the moment where picking 2 and 6 fails: the product works, but the sum does not.
Each factor of the trinomial gets one of the numbers: x^2 + 7x + 12 = (x + 3)(x + 4). Multiply back with FOIL to see the pieces: x^2 plus 4x plus 3x plus 12. The middle terms add to 7x, and the constant is 12, so the expansion matches the original trinomial exactly.
Signs change the search, not the strategy. For x^2 - 5x + 6, the product is still positive, so both numbers share a sign. Since the sum is negative, both are negative, and the pair is -2 and -3. One search method handles every case: find the factor pair of the constant, then check the sum.
Keep this idea
For x^2 + bx + c, find two numbers whose product is c and whose sum is b, then multiply back with FOIL to verify.
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