Problem 31 - Introductory Algebra for College Students

Question

Use the method of your choice to factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. $$15 y^{2}-y-2$$

Step-by-Step Solution

Answer

The factors of the trinomial \(15y^{2} - y - 2\) are \((3y - 1)(5y + 2)\).

Step 1: Identify the coefficients

Identify the coefficients of the quadratic, linear, and constant terms. In \(15y^{2} - y -2\), the coefficients are a = 15, b = -1, and c = -2 respectively.

Step 2: Find the numbers that meet the factoring requirements

Find two numbers m and n that multiply to give ac (a*c = -30 here) and add up to give b (b = -1 here). The numbers that fulfil these conditions are -5 and 6 because -5*6 = -30 and -5+6 = 1.

Step 3: Rewrite the trinomial

Rewrite the trinomial by breaking up the middle term using the two numbers found. So, \(15y^{2} - y - 2\) becomes \(15y^{2} - 5y + 6y - 2\).

Step 4: Factor by grouping

Group the terms and factor by grouping. This gives: \(5y(3y - 1) + 2(3y - 1)\).

Step 5: Extract common factor

Now, since both terms in Step 4 have a common factor of \(3y - 1\), this can be further factored to \((3y - 1)(5y + 2)\)

Step 6: Verify the factorization

The factors obtained in Step 5 can be multiplied using the FOIL method to check if the original trinomial is obtained. In this case, when you multiply \((3y - 1)\) with \((5y + 2)\) using the FOIL method, you will get back the original trinomial, verifying the factorization.

Key Concepts

Quadratic Equations

Quadratic equations are a cornerstone in algebra, and they commonly appear in the form \(ax^2 + bx + c = 0\). In our specific example, we are dealing with a trinomial, which is a type of quadratic expression with three terms. The goal is often to solve for the variable \(y\) by factoring the expression or finding its roots.

For this exercise, the expression given is \(15y^2 - y - 2\). Here, \(a\) is 15, \(b\) is -1, and \(c\) is -2. Understanding how to manipulate these terms is critical in the process of factoring, and ultimately, solving the equation. Quadratic equations can be solved through various methods, including factoring, completing the square, or using the quadratic formula, but factoring often provides a straightforward solution when applicable.

FOIL Method

The FOIL method is a handy tool for multiplying two binomials. FOIL stands for First, Outer, Inner, Last, which are the four steps to consider when multiplying binomials.

In our context, when verifying the factorization \((3y - 1)(5y + 2)\), each term should be multiplied systematically:

• First: Multiply the first terms from each binomial: \(3y \times 5y = 15y^2\).
• Outer: Multiply the outer terms: \(3y \times 2 = 6y\).
• Inner: Multiply the inner terms: \(-1 \times 5y = -5y\).
• Last: Multiply the last terms: \(-1 \times 2 = -2\).

Then, you would combine these results to get back the original trinomial: \(15y^2 - y - 2\). The FOIL method is an effective way to ensure your factorization was performed correctly.

Factor by Grouping

Factoring by grouping involves rearranging and grouping a polynomial to make factoring easier. It works well for quadratic trinomials like our example \(15y^2 - y - 2\). Once the middle term has been split into two parts that satisfy both multiplication and addition conditions, the expression can be grouped for easier factoring.

In our steps, after rewriting the equation as \(15y^2 - 5y + 6y - 2\), grouping is applied as follows:

• Group the terms: \((15y^2 - 5y) + (6y - 2)\).
• Factor out a common factor from each group: \(5y(3y - 1) + 2(3y - 1)\).

Observe that \((3y - 1)\) is a common factor, allowing further simplification to \((3y - 1)(5y + 2)\). This technique of factor by grouping effectively simplifies a complex trinomial into an easily solvable equation.

Algebraic Expressions

Algebraic expressions form the basis of solving equations, allowing us to manipulate numbers and variables together. An algebraic expression like \(15y^2 - y - 2\) comprises coefficients, variables, and constants, working together to create complex relationships which can be simplified or factored.

Understanding the components and the structure of these expressions is key to solving any algebra problem.
Key parts include:

• Variable: \(y\), which can have different values.
• Coefficients: Numbers like 15 and -1, which multiply the variable.
• Constant: A standalone number such as -2.

Factorization requires a strong grasp of these elements to break down the expression into simpler, manageable parts. Learning to work with algebraic expressions is a vital skill for tackling more advanced algebraic problems and understanding the relationships represented within equations.