Problem 14 - Precalculus

Question

Factor each expression. (a) \(3 S^{2}+2 S-8\) (b) \(3 \sec ^{2} \beta+2 \sec \beta-8\)

Step-by-Step Solution

Answer

(a) \((3S - 4)(S + 2)\); (b) \((3\sec\beta - 4)(\sec\beta + 2)\)."

Step 1: Factor the quadratic expression (a)

The expression in (a) is a quadratic, given by:\[ 3S^2 + 2S - 8 \]First, identify the coefficients: \(a = 3\), \(b = 2\), and \(c = -8\). We need factors of \(a \times c = 3 \times -8 = -24\) that add up to \(b = 2\). These numbers are \(6\) and \(-4\).Rewrite the middle term using these factors:\[ 3S^2 + 6S - 4S - 8 \]Then, group the terms:\[ (3S^2 + 6S) + (-4S - 8) \]Factor by grouping:\[ 3S(S + 2) - 4(S + 2) \]Now, factor out the common factor \((S + 2)\):\[ (3S - 4)(S + 2) \]Thus, the completely factored form is \((3S - 4)(S + 2)\).

Step 2: Factor the quadratic expression (b)

For expression (b), the quadratic form is:\[ 3\sec^2\beta + 2\sec\beta - 8 \]Identify the coefficients as \(a = 3\), \(b = 2\), and \(c = -8\). Just like in Part (a), find factors of \(-24\) (\(a \times c\)) that add to \(b = 2\). Again, these factors are 6 and -4.Rewrite the middle term using these factors:\[ 3\sec^2\beta + 6\sec\beta - 4\sec\beta - 8 \]Group the terms:\[ (3\sec^2\beta + 6\sec\beta) + (-4\sec\beta - 8) \]Factor by grouping:\[ 3\sec\beta(\sec\beta + 2) - 4(\sec\beta + 2) \]Factor out the common factor \((\sec\beta + 2)\):\[ (3\sec\beta - 4)(\sec\beta + 2) \]The expression is factorized as \((3\sec\beta - 4)(\sec\beta + 2)\).

Key Concepts

Quadratic Equations

A quadratic equation is any equation that can be written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a\) is not zero. Quadratics feature a squared term, making them second-degree polynomials. They're ubiquitous in various scenarios such as physics, engineering, and everyday life problems like projectile motion.

Solving quadratic equations can be done through several methods:

• Factorization: This involves expressing the quadratic as a product of two binomials. It's efficient when the quadratic factors neatly over the integers.
• Completing the Square: This method rewrites the quadratic in a perfect square form, allowing for direct solution extraction.
• Quadratic Formula: For complex or non-factorable quadratics, the formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) offers a universal solution.

Recognizing which method to use is part of the strategy in handling quadratic equations efficiently.

Algebraic Expressions

Algebraic expressions involve numbers, variables, and arithmetic operations. A simple example is "\(3x + 5\)". They're building blocks in mathematics, used for representing real-world scenarios abstractly.

Key features of algebraic expressions include:

• Constants: These are fixed numbers like \(5\), \(-3\), or \(\pi\).
• Variables: Represented by letters such as \(x\), \(y\), or \(z\), variables stand in for unknown values.
• Operations: Addition, subtraction, multiplication, and division are used to combine constants and variables.
• Coefficients: Numbers in front of the variables, such as \(3\) in \(3x\), multiplying with the variable.

Understanding algebraic expressions lays the foundation for algebra, enabling problem-solving and equation formulation.

Factoring by Grouping

Factoring by grouping is a method often employed to simplify polynomials into more manageable parts. It is especially useful when straightforward factorization isn't apparent.

The process involves the following steps:

• Group Terms: Divide the polynomial into groups that can be factored easily. For instance, with \(3S^2 + 6S - 4S - 8\), group into \((3S^2 + 6S)\) and \((-4S - 8)\).
• Factor Each Group: Factor out the greatest common factor from each group. For example, \(3S(S + 2)\) and \(-4(S + 2)\).
• Combine Like Terms: Notice the common term between groups and factor it out. Here, \((S + 2)\) is the common factor.
• Verify the Expression: Ensure the resulting expression is simpler and correct by expanding it to check your work.

Factoring by grouping not only simplifies polynomials but also reinforces understanding of factorization principles and provides insight into structural polynomial relationships.