Problem 41 - Introductory Algebra for College Students

Question

Factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. $$a^{2}-18 a b+45 b^{2}$$

Step-by-Step Solution

Answer

The factorization of the trinomial \(a^{2}-18 a b+45 b^{2}\) is \((a - 3b)(a - 15b)\).

Step 1: Identify the terms in the trinomial

The trinomial \(a^{2}-18 a b+45 b^{2}\) has three terms: \(a^{2}\), \(-18ab\), and \(45b^{2}\). The coefficient of \(a^{2}\) is 1, coefficient of \(-18ab\) is -18, and the constant term is \(45b^{2}\).

Step 2: Find two numbers that add to -18 and multiply to 45

We are looking for two numbers that add to -18 and multiply to 45. These numbers are -3 and -15.

Step 3: Rewrite the middle term

The middle term \(-18ab\) of the trinomial can be written as \(-3ab - 15ab\). Thus the trinomial \(a^{2}-18 a b+45 b^{2}\) can be expressed as \(a^{2} - 3ab - 15ab + 45b^{2}\).

Step 4: Factor by grouping

Group the first two terms together and the last two terms together in the expression \(a^{2} - 3ab - 15ab + 45b^{2}\). This results in \((a^{2} - 3ab) - (15ab - 45b^{2})\). Now factor out the greatest common factor from each group. This results in \(a(a - 3b) - 15b(a - 3b)\).

Step 5: Apply the distributive property

Now, factor out the common binomial factor \((a - 3b)\) from the two terms. This results in \((a - 3b)(a - 15b)\).

Step 6: Verify the factorization using the FOIL method

Apply the FOIL method to check the factorization. Multiply the first, outer, inner, and last terms together: First terms: \(a*a = a^{2}\), Outer terms: \(a*-15b = -15ab\), Inner terms: \(-3b*a = -3ab\), Last terms: \(-3b*-15b = 45b^{2}\). Adding these together gives the original trinomial \(a^{2}-18 a b+45 b^{2}\). So \((a - 3b)(a - 15b)\) is the correct factorization.