Problem 281 - Elementary Algebra

Question

Factor completely. \(y^{2}+10 y-39\)

Step-by-Step Solution

Answer

(y - 3)(y + 13)

Step 1: Identify the quadratic equation

The given quadratic equation is: y^{2}+10y-39

Step 2: Identify coefficients

Identify the coefficients in the equation: a = 1, b = 10, c = -39

Step 3: Find two numbers that multiply to ac and add to b

Here, we need to find two numbers that multiply to a*c (1*-39 = -39) and add up to b (10). After testing a few pairs, we find that 13 and -3 work: 13 * (-3) = -39 13 + (-3) = 10

Step 4: Rewrite the middle term using the two numbers

Rewrite the quadratic equation by breaking the middle term into two terms using the numbers found: y^{2} + 13y - 3y - 39

Step 5: Factor by grouping

Group the terms into pairs and factor each pair separately: (y^{2} + 13y) + (-3y - 39) y(y + 13) - 3(y + 13)

Step 6: Factor out the common binomial factor

Factor out the common binomial factor (y + 13): (y - 3)(y + 13)

Key Concepts

Quadratic Equations

Quadratic equations are critical in algebra and many real-life applications. They are polynomial equations of degree two, typically written in the form \(ax^{2} + bx + c = 0\). In this form, \(a\), \(b\), and \(c\) are coefficients where \(a ≠ 0\). The solutions to these equations are the values of \(x\) that make the equation true. These solutions, also known as roots, can be found using various methods:

• Factoring
• Completing the square
• Using the quadratic formula: \(x = \frac{{-b \pm \sqrt{{b^{2} - 4ac}}}}{2a}\)

Understanding how to manipulate and solve quadratic equations is essential, as they form the foundation for more advanced mathematical concepts.

Factoring by Grouping

Factoring by grouping is a valuable technique when dealing with complex polynomials like quadratic equations. This method is particularly useful when simple factoring isn't obvious. Here’s a closer look: First, break down the middle term of the quadratic equation to create a four-term polynomial. Let's consider the given equation: \(y^{2} + 10y - 39\) Identify numbers that multiply to the product of the first and last coefficients (\(a*c\)) and add up to the middle coefficient (\(b\)). For our example, 13 and -3 work because:

• \(13 * (-3) = -39\)
• \(13 + (-3) = 10\)

Next, rewrite the equation by splitting the middle term: \(y^{2} + 13y - 3y - 39\) Group terms into pairs and factor each pair: \((y^{2} + 13y) + (-3y - 39)\) Factor out the greatest common factor (GCF) from each pair:

• \(y(y + 13)\)
• \(-3(y + 13)\)

Finally, we factor out the common binomial factor: \((y - 3)(y + 13)\) This technique simplifies polynomials and finds factors efficiently.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations. They are the building blocks of algebra and are used to describe patterns and solve equations. An expression like \(y^{2} + 10y - 39\) can be factored to simplify and solve it. Here are some key points to remember:

• Terms: Individual parts separated by addition or subtraction (e.g., \(y^{2}\), \(10y\), \(-39\)).
• Coefficients: Numbers multiplying variables (e.g., 10 in \(10y\)).
• Constants: Numbers without variables (e.g., -39).

Factoring expressions helps to simplify them. This makes solving equations more manageable. For instance, factoring \(y^{2} + 10y - 39\) into \((y - 3)(y + 13)\) reveals solutions by setting each factor to zero:

• \((y - 3) = 0\) implies \(y = 3\)
• \((y + 13) = 0\) implies \(y = -13\)

Mastering algebraic expressions and their manipulations is the first step in understanding higher-level math concepts.