Chapter 6: Problem 14
Factor each expression. (a) \(3 S^{2}+2 S-8\) (b) \(3 \sec ^{2} \beta+2 \sec \beta-8\)
Short Answer
Expert verified
(a) \((3S - 4)(S + 2)\); (b) \((3\sec\beta - 4)(\sec\beta + 2)\)."
Step by step solution
01
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 \((3S - 4)(S + 2)\).
02
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 \((3\sec\beta - 4)(\sec\beta + 2)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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:
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.
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:
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.