1Cademy - Finding Two Consecutive Integers Whose Product Is 132

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Solving a Quadratic Equation by Factoring

Example

Finding Two Consecutive Integers Whose Product Is 132

Apply the seven-step problem-solving strategy to find two consecutive integers when their product is known, producing a quadratic equation rather than the linear equation that arises from sum problems.

Problem: The product of two consecutive integers is 132. Find the integers.

Read the problem.
Identify what to find: two consecutive integers.
Name the unknowns: Let $n$ = the first integer. Then $n + 1$ = the next consecutive integer.
Translate into an equation: "The first integer times the next integer is 132" becomes:

$n(n + 1) = 132$

Solve the equation. Distribute on the left side:

$n^2 + n = 132$

Subtract 132 from both sides to obtain standard form:

$n^2 + n - 132 = 0$

Factor the trinomial by finding two numbers whose product is $-132$ and whose sum is $1$ . The pair $12$ and $-11$ works ( $12 \cdot (-11) = -132$ and $12 + (-11) = 1$ ):

$(n - 11)(n + 12) = 0$

Apply the Zero Product Property and solve each equation:

$n - 11 = 0 \quad \text{or} \quad n + 12 = 0$

$n = 11 \quad \text{or} \quad n = -12$

Because there are two values of $n$ , two pairs of consecutive integers satisfy the condition:

If $n = 11$ , then $n + 1 = 12$ , giving the pair 11 and 12.
If $n = -12$ , then $n + 1 = -11$ , giving the pair $-12$ and $-11$ .

Check: $11 \cdot 12 = 132$ ✓ and $(-11)(-12) = 132$ ✓.
Answer: The consecutive integers are 11, 12 and $-12$ , $-11$ .

Unlike consecutive-integer problems involving sums — which produce linear equations — problems involving products lead to quadratic equations. Because a quadratic equation can yield two solutions, there may be two valid pairs of consecutive integers. Here, a negative pair also works because multiplying two negative numbers produces a positive product.

Updated 2026-04-21

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