Problem 22 - Introductory Algebra for College Students

Question

Factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. $$x^{2}+4 x+5$$

Step-by-Step Solution

Answer

The trinomial \(x^{2}+4 x+5\) is prime in the integers, but it can be factored in the complex numbers as \((x-(-2+i))(x-(-2-i))\).

Step 1: Identifying the relevant terms

Identifying the relevant terms of trinomial \(x^{2}+4 x+5\). Here, a equals to 1 (coefficient of \(x^{2}\)), b equals to 4 (coefficient of x), and c equals to 5 (the constant term).

Step 2: Factoring the Trinomial

Start factoring the trinomial. Look for two numbers that multiply to \(a*c=(1*5)=5\) and add up to b=4. However, there are no such numbers, meaning the quadratic trinomial cannot be factored by traditional factoring. The trinomial is prime in the integers but not in the complex numbers.

Step 3: The Complex Factorization

A prime trinomial in the integers can still often be factored in the complex numbers. Use the quadratic formula \(x = [-b \pm sqrt(b^{2}-4ac)]/(2a)\). Substitute a=1, b=4, c=5. Calculating, we get two complex roots \(x = -2+ i\) and \(x = -2- i\). Hence, trinomial in the complex numbers can be factored as \(x^{2}+4 x+5=(x-(-2+i))(x-(-2-i))\).

Step 4: Verifying with FOIL Multiplication

To verify the factorization, FOIL (First, Outer, Inner, Last) multiplication is applied to distribute the binomials. First, multiply the first terms of each binomial. Then, multiply the outer terms, followed by the inner terms. Last, multiply the last terms of each binomial. Add up all the products to verify if it equals to the original trinomial.