Problem 23 - Introductory Algebra for College Students

Question

Factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. $$y^{2}-16 y+48$$

Step-by-Step Solution

Answer

The factored form of the trinomial \(y^{2}-16 y+48\) is \((y-8)(y-6)\)

Step 1: Understand the Trinomial

The trinomial given is \(y^{2}-16y+48\). Here, the coefficient of \(y^{2}\) is 1, the coefficient of y is -16, and the constant term is 48.

Step 2: Factor the Trinomial

To start factoring, look for two numbers that multiply to 48 (the constant term) and add up to -16 (the coefficient of y). The numbers -8 and -6 fit this criterion because \(-8\times -6 = 48\) and \(-8 + -6 = -16\). Hence, the factored form of \(y^{2}-16y+48\) is \((y-8)(y-6)\).

Step 3: Verification Using FOIL

To verify the factored form, use the FOIL method to expand \((y-8)(y-6)\). FOIL stands for First, Outer, Inner, Last. This method multiplies the first terms in each binomial, then the outer terms, then the inner terms, and finally the last terms. The FOIL multiplication for our factored trinomial gives us: First: \(y\times y = y^{2}\), Outer: \(y\times -6 = -6y\), Inner: \(-8\times y = -8y\), Last: \(-8\times -6 = 48\). Add these together to get \(y^{2}-6y-8y +48\). Simplifying gives back the original trinomial \(y^{2}-16y+48\), confirming that the factored form is correct.