Problem 63 - Introductory Algebra for College Students

Question

Factor completely. $$15 x y^{2}+45 x y-60 x$$

Step-by-Step Solution

Answer

The factorized form of the given expression is \(15x(y + 4)(y - 1)\).

Step 1: Identify the common factor

The first thing to do is to identify if there is any common factor in all the terms of the given expression \(15xy^2\), \(45xy\), and \(-60x\). From these terms, we can see that \(15x\) can be factored out from all of them.

Step 2: Factor out the common factor

Now we factor out the common factor \(15x\) from each term. So, by doing this, the expression becomes \(15x(y^2 + 3y - 4)\).

Step 3: Factor the resulting expression further

The remaining quadratic inside the parenthesis - \(y^2 + 3y - 4\) - can be factored further. This gives us \(y^2 + 3y - 4 = (y + 4)(y - 1)\), using the factors of \(a*c = -4\) (from the quadratic \(y^2 + 3y - 4\) in the form \(ax^2 + bx + c\)). Thus, the initially given expression can be fully factored into \(15x(y + 4)(y - 1)\).

Key Concepts

Common Factor

When you have a polynomial expression like \(15xy^2 + 45xy - 60x\), finding the **common factor** is the first essential step in the factoring process. A common factor is a number or variable that is present in all terms of an expression. In simpler terms, it's something that each part of the expression "shares." To identify it, you assess each term separately.

• For the expression given, each term—\(15xy^2\), \(45xy\), and \(-60x\)—includes a \(15x\).

This means that you can pull this \(15x\) out front to simplify the expression. This step helps in reducing the clutter of the expression and makes further factoring easier. By factoring out \(15x\), the given expression becomes \(15x(y^2 + 3y - 4)\). Recognizing and removing the common factor not only simplifies the calculation but sets you up perfectly for the next steps in the factoring process.

Quadratic Expressions

Once you have factored out the common factor from a polynomial expression, you may find yourself dealing with a quadratic expression. In this case, after removing \(15x\) from the original expression, the result was a simpler expression inside the parentheses: \(y^2 + 3y - 4\).

• A quadratic expression has the standard form \(ax^2 + bx + c\), which contains a variable raised to the second power.

The goal is to identify a way to express it as a product of two binomials. This step aids in determining the basic roots or solutions of the equation if it's set to zero. Quadratics are a key focus in algebra because they appear often in problems involving areas, trajectories, and many real-world applications. Proper handling of quadratics requires patience and practice, and understanding how to break them down will greatly improve your problem-solving skills in mathematics.

Factoring Techniques

After identifying and factoring out all possible common factors, and recognizing any quadratic expressions, it's important to apply the appropriate factoring techniques. For the example expression, the quadratic \(y^2 + 3y - 4\) utilizes a specific technique that looks for two numbers that multiply to \(-4\) (the product of \(a\) and \(c\) in the quadratic) and add up to \(3\) (the middle coefficient, \(b\)).

• These numbers are \(4\) and \(-1\).
• Thus, the quadratic expression factors as \((y + 4)(y - 1)\).

Factoring techniques vary based on the expressions you are dealing with. While the method above is specific to quadratics, other techniques may involve decomposing, grouping, or using special identities like difference of squares. Practicing with different techniques will help in catching tricks and shortcuts, ultimately making the factoring process quicker and more intuitive.