Question

Factor each polynomial completely. If the polynomial cannot be factored, say it is prime. $$ 3 x^{2}-12 x+15 $$

Short answer

The factored form is \(3(x^2 - 4x + 5)\).

Step-by-step solution

Identify the Greatest Common Factor (GCF)

Find the greatest common factor of all the terms in the polynomial. For the polynomial \(3x^2 - 12x + 15\), the GCF of 3, -12, and 15 is 3.

Factor Out the GCF

Factor out the GCF from each term in the polynomial. This gives: \[3(x^2 - 4x + 5)\].

Check for Further Factoring

Examine the quadratic expression \(x^2 - 4x + 5\) to see if it can be factored further. In this case, it cannot be factored into real number factors, so the expression inside the parentheses is a prime polynomial.

Write the Final Factored Form

The polynomial is already fully factored and cannot be simplified further. So, the final factored form is: \[3(x^2 - 4x + 5)\].

Key concepts

Greatest Common Factor

Let's dive into the concept of the Greatest Common Factor (GCF). Think of the GCF as the largest number that can evenly divide all terms in a polynomial. To find it, identify the common factors in each coefficient. In our example polynomial \(3x^2 - 12x + 15\), the coefficients are 3, -12, and 15. The GCF of these numbers is 3, because 3 is the highest number that divides them all without leaving a remainder. Here's how you find it:

• List the factors of each coefficient. For example, the factors of 3 are 1 and 3.
• Compare the lists to find the highest common number. For our coefficients, 1 and 3 are common, but 3 is the greatest.

Once you identify the GCF, you factor it out from each term in the polynomial, simplifying it step by step.

Quadratic Expressions

Moving on to quadratic expressions, these are polynomials of the form \(ax^2 + bx + c\). Quadratics are widely used in algebra, making it crucial to understand how to factor them. When the leading coefficient (the coefficient of \(x^2\)) is not 1, factoring becomes slightly more challenging. In our case, after factoring out the GCF, we get the quadratic expression \(x^2 - 4x + 5\).
Here's a quick guide to factoring quadratics:

• Determine whether the quadratic can be factored further by testing for factor pairs that multiply to \(ac\) (the product of the leading coefficient and the constant term) and add up to \(b\). In simpler quadratics, these pairs give us our factors.
• Check if the quadratic is a perfect square trinomial or fits any special factoring cases, which can simplify your work.

In the given example, the expression \(x^2 - 4x + 5\) does not factor into real numbers because no pairs of numbers multiply to 5 and add up to -4, which indicates it's a prime quadratic within real numbers.

Prime Polynomials

Lastly, let's explore prime polynomials. Just like prime numbers, prime polynomials cannot be factored further over a given number set. For our purposes, we'll stick to real numbers. If no factor pairs exist that satisfy the necessary conditions (multiplying to \(ac\) and summing to \(b\)), the polynomial is prime. After factoring out the GCF in our example, we end up with \(3(x^2 - 4x + 5)\).
Why is \(x^2 - 4x + 5\) prime?

• Factor pairs for \(ac\): We look for pairs that multiply to 5 and add to -4. However, no such pairs exist using real numbers.
• Square roots: The discriminant (\(b^2 - 4ac\)) check can confirm primeness. Here, \((-4)^2 - 4(1)(5) = 16 - 20 = -4\) is not a perfect square, indicating no real roots.

Therefore, it cannot be factored further using real numbers and stands as a prime polynomial.