The product of two consecutive integers is 110. Use factoring to find these integers.

To find two consecutive integers whose product is 110, we can use algebraic methods involving factoring. Here's a step-by-step solution: ### Step 1: Define the Consecutive Integers Let’s denote the first integer as \( x \). Since the integers are consecutive, the next integer is \( x + 1 \). ### Step 2: Set Up the Equation The product of the two consecutive integers is 110: \[ x \times (x + 1) = 110 \] \[ x(x + 1) = 110 \] ### Step 3: Expand and Rearrange the Equation Expand the left side of the equation: \[ x^2 + x = 110 \] Bring all terms to one side to form a quadratic equation: \[ x^2 + x - 110 = 0 \] ### Step 4: Factor the Quadratic Equation We need to factor the quadratic equation \( x^2 + x - 110 = 0 \). To do this, find two numbers that multiply to **-110** and add up to **1**. - **Factors of -110:** - (11, -10) → \( 11 \times (-10) = -110 \) and \( 11 + (-10) = 1 \) Using these factors, the equation can be factored as: \[ (x + 11)(x - 10) = 0 \] ### Step 5: Solve for \( x \) Set each factor equal to zero and solve for \( x \): \[ x + 11 = 0 \quad \Rightarrow \quad x = -11 \] \[ x - 10 = 0 \quad \Rightarrow \quad x = 10 \] ### Step 6: Determine the Consecutive Integers - **If \( x = -11 \):** - The consecutive integers are \(-11\) and \(-10\). - **Product:** \((-11) \times (-10) = 110\) - **If \( x = 10 \):** - The consecutive integers are \(10\) and \(11\). - **Product:** \(10 \times 11 = 110\) ### **Conclusion** There are two pairs of consecutive integers whose product is 110: - **\(-11\) and \(-10\)** - **\(10\) and \(11\)** Both pairs satisfy the condition \( x(x + 1) = 110 \).