Problem 60 - Algebra A Combined Function

Question

Factor each trinomial completely. See Examples 1 through 7. \(24 y^{2} x+7 y x-5 x\)

Step-by-Step Solution

Answer

The factored form is \(x(3y - 1)(8y + 5)\).

Step 1: Identify Common Factors

First, look for any common factors in each of the terms of the expression. We notice that every term contains the factor \(x\). Extract \(x\) from the trinomial to begin simplifying.\[ 24y^2x + 7yx - 5x = x(24y^2 + 7y - 5) \]

Step 2: Factor the Quadratic Expression

Now, focus on factoring the quadratic expression inside the parentheses: \(24y^2 + 7y - 5\). Look for two numbers whose product is the coefficient of \(y^2\) multiplied by the constant term, and whose sum is the coefficient of \(y\).Calculate: \(24 imes -5 = -120\) and we need numbers that add up to \(7\). These numbers are \(15\) and \(-8\).

Step 3: Rewrite the Quadratic Expression

Rewrite the quadratic \(24y^2 + 7y - 5\) by splitting the middle term using the numbers found: \(15\) and \(-8\).\[ 24y^2 + 15y - 8y - 5 \]

Step 4: Group and Factor by Grouping

Next, group the four terms into two pairs and factor each pair:\[ (24y^2 + 15y) + (-8y - 5) \]Factor out the greatest common factor from each group:\[ 3y(8y + 5) - 1(8y + 5) \]

Step 5: Factor Out the Common Binomial

Notice that both groups contain the common binomial \((8y + 5)\). Factor this binomial out:\[ (3y - 1)(8y + 5) \]

Step 6: Write the Complete Factorization

Finally, combine the factors from Step 1 and Step 5 to write the complete factorization of the original expression:\[ x(3y - 1)(8y + 5) \]

Key Concepts

Quadratic Expression

A quadratic expression is a polynomial of degree two, characterized by its highest power of a variable being squared. Typically, it takes the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). In our exercise, the quadratic expression inside the original trinomial is \(24y^2 + 7y - 5\). This form allows us to apply various factoring techniques to simplify or solve the polynomial.When tackling a quadratic expression, especially in trinomials, it's important to recognize the leading coefficient \(a\), the linear coefficient \(b\), and the constant term \(c\). By understanding the role of each component, students are equipped to apply efficient problem-solving techniques, such as factoring by grouping, which relies heavily on these coefficients.

Common Factors

Common factors are crucial initial steps in simplifying algebraic expressions. They refer to the shared components present in all terms of a polynomial, which can be factored out to simplify calculations. In the trinomial \(24y^2x + 7yx - 5x\), the variable \(x\) is a common factor in each term, allowing us to extract it: \[ 24y^2x + 7yx - 5x = x(24y^2 + 7y - 5) \]Identifying common factors not only simplifies the expression but also sets the stage for further factoring techniques. The process involves looking for the greatest common divisor among the terms' coefficients and any common variables. This step is foundational because it reduces the overall complexity of the problem, making it easier to apply further factorization methods.

Factor by Grouping

Factor by grouping is a method typically used for polynomials with four terms. It involves rearranging and grouping terms so that common factors can be identified and extracted. In the expression \(24y^2 + 15y - 8y - 5\), factor by grouping involves splitting the middle term, resulting in the possibility to group into \((24y^2 + 15y) + (-8y - 5)\).From here, the next step is to factor out common terms from each group:

• Extract \(3y\) from \(24y^2 + 15y\) to get \(3y(8y + 5)\)
• Extract \(-1\) from \(-8y - 5\) to get \(-1(8y + 5)\)

Now, notice both groups share the binomial \((8y + 5)\). The final factorization at this stage leads us to \((3y - 1)(8y + 5)\). This step is crucial in solving higher degree polynomials and is particularly useful in this type of algebraic expression.

Binomial Factorization

Binomial factorization is a technique where a two-term algebraic expression, or binomial, is factored. In our exercise, after factoring the trinomial partially, we found the same binomial factor \((8y + 5)\) in both groups: \(3y(8y + 5) - 1(8y + 5)\).By recognizing this repeated binomial, we can extract it as a common factor:\[ (3y - 1)(8y + 5) \]This reflection involves understanding how binomials can be distributed within larger polynomial equations and effectively simplifies previously complex expressions. When executed correctly, this specific technique not only aids in verifying the equality but also simplifies complex equations into more manageable factors. This skill is particularly beneficial for solving quadratic equations or simplifying polynomial expressions for further analysis.