Problem 56 - Algebra A Combined Function

Question

Factor each trinomial completely. See Examples 1 through 7. \(6 x^{3}-31 x^{2}+5 x\)

Step-by-Step Solution

Answer

The completely factored form is \(x(6x - 1)(x - 5)\).

Step 1: Find the Greatest Common Factor (GCF)

The first step in factoring a trinomial is to find the greatest common factor (GCF) of all terms. In the trinomial \(6x^3 - 31x^2 + 5x\), the GCF of all the terms is \(x\). So, we factor out \(x\) from each term: \(x(6x^2 - 31x + 5)\).

Step 2: Factor the Quadratic

Now, we need to factor the quadratic expression \(6x^2 - 31x + 5\). We look for two numbers that multiply to the product of the coefficient of \(x^2\) term (6) and the constant term (5), which is 30, and add up to the coefficient of the \(x\) term (-31). These numbers are -30 and -1.

Step 3: Rewrite the Middle Term

Using the numbers -30 and -1, rewrite the middle term \(-31x\) as \(-30x - x\). This gives us \(6x^2 - 30x - x + 5\).

Step 4: Group and Factor

Group the terms into two pairs: \((6x^2 - 30x)\) and \((-x + 5)\). Factor each group separately: \(6x(x - 5) - 1(x - 5)\).

Step 5: Factor by Grouping

Since both terms have a common factor of \((x - 5)\), factor it out: \((6x - 1)(x - 5)\).

Step 6: Write the Complete Factorization

Combine the factored terms with the GCF \(x\) from Step 1 to write the complete factorization: \(x(6x - 1)(x - 5)\).

Key Concepts

Greatest Common Factor

Factoring polynomials often begins with identifying the Greatest Common Factor (GCF). The GCF is the largest factor shared by all terms in a polynomial. For instance, in the polynomial

• \(6x^3 - 31x^2 + 5x\), each term contains the variable \(x\), making \(x\) the GCF. By factoring out \(x\), you simplify the expression to \(x(6x^2 - 31x + 5)\).

Think of the GCF as the biggest number or variable part you can divide all terms by evenly. It's like finding a common theme among elements. Start by checking the coefficients (numerical parts) and the variables. Look for what is similar across all terms.
Once identified, factoring out the GCF can significantly simplify expressions and make further factoring steps easier.

Quadratic Expression

A quadratic expression is a polynomial of the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. In this exercise:

• The expression you need to factor is \(6x^2 - 31x + 5\).
  

Quadratics are fascinating because they often form parabolas when graphed. Factoring them allows you to solve equations by finding the roots or zeroes.
To factor a quadratic, you'll look for two numbers that multiply to the product of \(a\) (the coefficient of \(x^2\)) and \(c\) (the constant term), and add up to \(b\) (the coefficient of \(x\)). In this case, those two numbers are -30 and -1, multiplying to 30 and adding to -31.
Recognizing these patterns in quadratic expressions is key to mastering factorization.

Factoring by Grouping

Factoring by grouping involves creating groups from a polynomial's terms that can be factored easily. Once terms are grouped and factored, you can factor out any common expressions from these groups. Let’s break it down:

• After rewriting \(-31x\) as \(-30x - x\), the expression becomes \(6x^2 - 30x - x + 5\).
  
• Group terms: \( (6x^2 - 30x) \) and \( (-x + 5) \).
  
• Factor each group separately: From the first pair, \(6x(x - 5)\), and from the second, \(-1(x - 5)\).
  

Now, you'll notice a shared factor: \((x - 5)\), allowing you to finally factor it as \((6x - 1)(x - 5)\).
Factoring by grouping requires a little creativity and practice to see which terms to pair, but with practice, it becomes an intuitive part of polynomial factoring. This technique is efficient for expressions where standard factoring techniques are challenging.