Where the two relations come from

The zeros of x² − 5x + 6 are 2 and 3. They add to 5. The middle number is −5. Is that luck? Open the brackets with letters and see.

The rule

It is not luck. Opening (x − p)(x − q) gives both relations. Here p and q are the zeros. In ax² + bx + c, a is the number before x². The letter b is the number before x. The c stands alone.

A quadratic with zeros p and q has factors (x − p) and (x − q). Multiply out: x² − qx − px + pq. The two x terms join into −(p + q)x. So (x − p)(x − q) = x² − (p + q)x + pq.

Take a = 1. Write the same quadratic as x² + Bx + C. Here B is the middle number and C is the last. The middle numbers match, so B = −(p + q). The last numbers match, so C = pq.

If a is not 1, divide every term by a first. Dividing does not move a zero, since 0 ÷ a = 0. This gives x² + (b/a)x + c/a. So B = b/a and C = c/a. Then p + q = −b/a and pq = c/a.

If `ax² + bx + c` has zeros, their product is `c/a`.

If the two zeros are equal, count that one value twice.

The polynomial x² − 5 cannot be written as brackets with whole numbers. Its zeros are √5 and −√5. Here a = 1, b = 0 and c = −5. The sum is 0, and −b/a = 0. The product is −5, and c/a = −5.