Problem 175 - Elementary Algebra

Question

Factor. \(6 y^{2}+y-15\)

Step-by-Step Solution

Answer

The factored form of \(6y^2 + y - 15\) is \((2y - 3)(3y + 5)\).

Step 1 - Identify coefficients

In the quadratic expression \(6y^2 + y - 15\), identify the coefficients: \(a = 6\), \(b = 1\), and \(c = -15\).

Step 2 - Multiply coefficients

Calculate the product of \(a\) and \(c\): \(6 \times -15 = -90\).

Step 3 - Find two numbers that multiply to \(-90\) and add to \(1\)

Determine two numbers that multiply to \(-90\) and add up to \(1\). These numbers are \(10\) and \(-9\) because \(10 \times -9 = -90\) and \(10 + (-9) = 1\).

Step 4 - Rewrite the middle term

Express the middle term \(y\) as the sum \(10y - 9y\). The quadratic becomes \(6y^2 + 10y - 9y - 15\).

Step 5 - Factor by grouping

Group the terms: \((6y^2 + 10y) + (-9y - 15)\). Factor out the greatest common factor (GCF) from each group: \(2y(3y + 5) - 3(3y + 5)\).

Step 6 - Factor out the common binomial

Factor out the common binomial \((3y + 5)\): \((2y - 3)(3y + 5)\).

Key Concepts

factoring by grouping

When factoring quadratic expressions, one common technique is called 'factoring by grouping.' This method is especially useful when you have a quadratic equation where the leading coefficient, the number in front of the squared term, is not 1.

Let's break it down using a simple structure:

• Step 1: Identify the coefficients.
• Step 2: Multiply the first (leading) coefficient and the constant term.
• Step 3: Find two numbers that multiply to the product and add to the middle coefficient.
• Step 4: Rewrite the quadratic expression using these two new numbers in place of the middle term.
• Step 5: Group the first two terms and the last two terms separately.
• Step 6: Factor out the greatest common factor (GCF) from each group.
• Step 7: Factor out the common binomial from the two groups, if able.

For example, given the expression 6y^2 + y - 15, we first group the terms into pairs: (6y^2 + 10y) + (-9y - 15). Factoring out the GCF from each pair, we get 2y(3y + 5) - 3(3y + 5). Finally, we factor out the common binomial, which results in (2y - 3)(3y + 5). By using factoring by grouping, you can simplify many quadratic expressions efficiently.

polynomial coefficients

In the context of a quadratic expression, polynomial coefficients play a crucial role. These coefficients are the numbers that multiply the variables. In a standard quadratic equation of the form ax^2 + bx + c, the coefficients are:

• a: Coefficient of x^2
• b: Coefficient of x
• c: Constant term

For instance, in the expression 6y^2 + y - 15, the coefficients are:

• a = 6
• b = 1
• c = -15

Understanding these coefficients helps in applying methods like factoring by grouping.

When you multiply the first and last coefficients (a and c), you get a product used to break down the middle term (b). In our example, multiplying 6 and -15 gives -90, and we look for two numbers that multiply to -90 and add up to 1. Decoding the coefficients lays the groundwork for successful factoring.

quadratic equation

A quadratic equation is a polynomial equation of degree 2, generally in the form ax^2 + bx + c = 0. It is called 'quadratic' because 'quad' means square, and the variable is raised to the second power.

Key features of a quadratic equation include:

• The leading term involves the variable squared.
• It can be factored into two binomials.
• It typically has two solutions, which can be real or complex numbers.
• When plotted on a graph, it forms a parabola.

Factoring quadratic expressions helps in solving quadratic equations because it breaks the equation into simpler parts that make finding the roots (solutions) easier. In our solved example, 6y^2 + y - 15 is factored into (2y - 3)(3y + 5), which can then be set to zero to find the solutions as 2y - 3 = 0 and 3y + 5 = 0.

Knowing how to handle quadratic equations through factoring allows for easier problem-solving and a better understanding of how these equations behave in algebra.