Problem 25 - Introductory Algebra for College Students

Question

Factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. $$x^{2}-3 x+6$$

Step-by-Step Solution

Answer

\(x^{2} - 3x + 6\) cannot be factorized using real numbers. But, it can be factorized using complex numbers as \((x - \frac{3 - \sqrt{15}i}{2}) (x - \frac{3 + \sqrt{15}i}{2})\).

Step 1: Try to Factorize the Trinomial

Try to find two numbers that multiply to 6 (the third term) and add to -3 (the second term). Since no two real numbers have these properties, the trinomial cannot be factorized using real numbers.

Step 2: Check With Complex Numbers

The trinomial could be factorizable using complex numbers. To check this, consider the quadratic formula, which says the relationship of the roots \(p\) and \(q\) to the coefficients of the trinomial: \(p, q = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\). Plugging \(a = 1\), \(b = -3\), and \(c = 6\) into the formula yields \(p, q = \frac{3 \pm \sqrt{(-3)^{2}-4(1)(6)}}{2} = \frac{3 \pm \sqrt{9 - 24}}{2} = \frac{3 \pm \sqrt{-15}}{2} = \frac{3 \pm \sqrt{15}i}{2}\).

Step 3: Use the Roots to Factorize

Using the roots, the trinomial may be factorized as: \(x^{2} - 3x + 6 = (x - p)(x - q) = (x - \frac{3 - \sqrt{15}i}{2}) (x - \frac{3 + \sqrt{15}i}{2})\).

Step 4: Check the Factorization

By checking the factorization using FOIL multiplication as follows: First: \(x * x = x^2\), Outside: \(x * -\frac{3 + \sqrt{15}i}{2} = -\frac{3x + x\sqrt{15}i}{2}\), Inside: \(-\frac{3 - \sqrt{15}i}{2} * x = -\frac{3x - x\sqrt{15}i}{2}\), and Last: \(-\frac{3 - \sqrt{15}i}{2} * -\frac{3 + \sqrt{15}i}{2} = 6\). Add up all terms to find: \(x^2 - 3x + 6\), which confirms that the trinomial was correctly factorized.