Problem 87
Question
85–90 ? Factor the expression completely. (This type of expression arises in calculus when using the “product rule.”) $$ 3(2 x-1)^{2}(2)(x+3)^{1 / 2}+(2 x-1)^{3}\left(\frac{1}{2}\right)(x+3)^{-1 / 2} $$
Step-by-Step Solution
Verified Answer
The expression is \((2x-1)^{2}(x+3)^{-1/2}\left(\frac{13x}{2} + \frac{35}{2}\right)\).
1Step 1: Identify Common Factors
Look at each term in the expression. We see that both terms have a common factor of \((2x-1)^{2}(x+3)^{-1/2}\). This is the part of each term that can be factored out.
2Step 2: Factor Out the Common Factor
Take the common factor \((2x-1)^{2}(x+3)^{-1/2}\) outside the expression. This leaves us with:\[(2x-1)^{2}(x+3)^{-1/2}\left[3\cdot2\cdot(x+3) + (2x-1)\cdot\frac{1}{2}\right]\]
3Step 3: Simplify Inside the Brackets
Now focus on simplifying the expression inside the brackets:\[3\cdot2\cdot(x+3) + (2x-1)\cdot\frac{1}{2}\]which simplifies to:\[6(x+3) + \frac{1}{2}(2x-1)\].
4Step 4: Distribute and Combine Like Terms
Distribute and add the new terms inside the brackets:\[6x + 18 + x - \frac{1}{2} = \frac{13x}{2} + \frac{35}{2}. \]
5Step 5: Final Factorization
Combine these steps, the expression is now completely factored:\[(2x-1)^{2}(x+3)^{-1/2}\left( rac{13x}{2} + \frac{35}{2}\right).\]
Key Concepts
Understanding Common FactorsGrasping the Product RuleSteps to SimplificationExploring Factorization in Algebra
Understanding Common Factors
When dealing with algebraic expressions, one of the first steps to simplify an equation is to identify common factors. A common factor is an expression or number that divides each term of a problem evenly. This makes it easier to simplify or factor the equation further.
In the exercise given, both terms in the expression have the common factor \((2x-1)^{2}(x+3)^{-1/2}\). Recognizing this shared part allows us to pull it out of the equation, greatly simplifying the problem. By finding and removing the common factor, you turn a complex expression into something much more manageable.
In the exercise given, both terms in the expression have the common factor \((2x-1)^{2}(x+3)^{-1/2}\). Recognizing this shared part allows us to pull it out of the equation, greatly simplifying the problem. By finding and removing the common factor, you turn a complex expression into something much more manageable.
Grasping the Product Rule
The product rule is a calculus concept, but its application can also be seen in algebra. It refers to the differentiation of a product of two functions. However, in this context, it can help understand how terms combine and how factors work together in expressions.
In our given expression, we have a product of terms, specifically \((2x-1)\) and \((x+3)\) raised to various powers. Recognizing how these terms interact allows us to break down the expression methodically. This makes factoring more straightforward. Understanding the application of the product rule in integrating terms helps simplify checking if they can be treated as a unified product in line with factorization.
In our given expression, we have a product of terms, specifically \((2x-1)\) and \((x+3)\) raised to various powers. Recognizing how these terms interact allows us to break down the expression methodically. This makes factoring more straightforward. Understanding the application of the product rule in integrating terms helps simplify checking if they can be treated as a unified product in line with factorization.
Steps to Simplification
The simplification process involves clarifying an expression to its simplest form. When you have complex expressions, breaking them down by pulling out common factors is key.
In the exercise, once the common factor \((2x-1)^{2}(x+3)^{-1/2}\) was removed, you are left with an easier part to handle. Inside the brackets, compute and simplify: \(3\cdot2\cdot(x+3)\) becomes \(6(x+3)\) while \((2x-1)\cdot\frac{1}{2}\) becomes \(\frac{1}{2}(2x-1)\).
Adding these results gives a more streamlined equation: \(6x + 18 + x - \frac{1}{2}\).
Through simplification, you'll get: \(\frac{13x}{2} + \frac{35}{2}\), making the equation much easier to handle.
In the exercise, once the common factor \((2x-1)^{2}(x+3)^{-1/2}\) was removed, you are left with an easier part to handle. Inside the brackets, compute and simplify: \(3\cdot2\cdot(x+3)\) becomes \(6(x+3)\) while \((2x-1)\cdot\frac{1}{2}\) becomes \(\frac{1}{2}(2x-1)\).
Adding these results gives a more streamlined equation: \(6x + 18 + x - \frac{1}{2}\).
Through simplification, you'll get: \(\frac{13x}{2} + \frac{35}{2}\), making the equation much easier to handle.
Exploring Factorization in Algebra
Factorization refers to breaking down an expression into simpler components (or "factors") that multiply together to give the original expression. It is a cornerstone of algebra, allowing complex problems to be tackled by segmenting them into smaller parts.
In the given exercise, factorization involves expressing the entire equation as a product of its factors: \((2x-1)^{2}(x+3)^{-1/2}\) and \(\frac{13x}{2} + \frac{35}{2}\).
Such breakdown leverages algebraic rules and properties efficiently, streamlining problem-solving processes. Factorization is crucial not only for simplifying expressions but also for solving equations and performing calculus operations effectively. By mastering this concept, you'll be able to handle increasingly challenging algebraic expressions with confidence.
In the given exercise, factorization involves expressing the entire equation as a product of its factors: \((2x-1)^{2}(x+3)^{-1/2}\) and \(\frac{13x}{2} + \frac{35}{2}\).
Such breakdown leverages algebraic rules and properties efficiently, streamlining problem-solving processes. Factorization is crucial not only for simplifying expressions but also for solving equations and performing calculus operations effectively. By mastering this concept, you'll be able to handle increasingly challenging algebraic expressions with confidence.
Other exercises in this chapter
Problem 86
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