Question

Expressions that occur in calculus are given. Factor each expression. Express your answer so that only positive exponents occur. $$6 x^{1 / 2}\left(x^{2}+x\right)-8 x^{3 / 2}-8 x^{1 / 2} \quad x \geq 0$$

Short answer

The factored expression is \(2x^{1/2}(3x^2 - x - 4)\).

Step-by-step solution

- Identify Common Factors

Identify the common factors among the terms in the expression. Here, the common factor is \(2x^{1/2}\). Factoring out \(2x^{1/2}\) from each term will simplify the expression.

- Factor Out the Common Factor

Extract \(2x^{1/2}\) from each term and rewrite the expression: \[ 6x^{1/2}(x^2 + x) - 8x^{3/2} - 8x^{1/2} \rightarrow 2x^{1/2} \big[ 3(x^2 + x) - 4x - 4 \big] \]

- Simplify Inside the Bracket

Distribute and combine like terms inside the brackets: \[ 2x^{1/2} \big[ 3x^2 + 3x - 4x - 4 \big] \] Then combine like terms: \[ 2x^{1/2} \big[ 3x^2 - x - 4 \big] \]

- Final Simplified Form

The expression is now factored and simplified with only positive exponents: \[ 2x^{1/2}(3x^2 - x - 4) \]

Key concepts

algebra

Algebra is a branch of mathematics that uses symbols and letters to represent numbers and quantities in formulas and equations. Understanding algebra is crucial for simplifying and solving mathematical expressions. In our worked exercise, we use algebraic techniques to factor and simplify the given expression by identifying common factors and combining like terms. The factorization process involves:

• **Identifying Common Factors:** Recognize common elements to factor out from each term. Here, the common factor is identified as \(2x^{1/2}\).
• **Extracting the Common Factor:** Factoring \(2x^{1/2}\) out simplifies the expression immensely, transforming it to \(2x^{1/2} \big[ 3(x^2 + x) - 4x - 4 \big]\).
• **Combining Like Terms:** Inside the brackets, we then distribute and add/subtract similar terms to get a simplified result, leading to the final form where the expression is reduced to \(2x^{1/2}(3x^2 - x - 4)\).

Algebra helps in breaking down complex expressions into simpler, more manageable parts, making it easier to handle more intricate calculations.

calculus preparation

Mastering algebraic techniques such as factoring and simplifying expressions is essential for preparing for calculus. Calculus involves the study of change and motion through derivatives and integrals.

**Importance of Factoring**: In calculus, recognizing common factors and simplifying expressions aids in:
• **Differentiation**: Simplified expressions make finding derivatives straightforward and less error-prone.
• **Integration**: Simplified forms are easier to integrate, ensure more accurate calculations.

In our example, by factoring the given expression \[ 6 x^{1 / 2}(x^{2}+x)-8 x^{3 / 2}-8 x^{1 / 2} \rightarrow 2x^{1/2} \big[ 3(x^2 + x) - 4x - 4 \big] \],we reduce complexity. This reduction is crucial for both differentiation and integration processes in calculus.

Developing these algebraic skills builds a good foundation for the advanced topics encountered in calculus.

positive exponents

Working with positive exponents is fundamental in algebra and calculus, as they align with standard mathematical conventions, making expressions easier to interpret and manipulate. Positive exponents indicate straightforward growth or increase in the respective variable's power.

In our exercise, we ensure all exponents remain positive:
\[ 6 x^{1 / 2} (x^2 + x)-8 x^{3 / 2}-8 x^{1 / 2} \rightarrow 2x^{1/2} (3x^2 - x - 4) \]

**Why Focus on Positive Exponents?**: Positive exponents:
• Simplify computations and interpretations, removing negative exponent complications.
• Ensure consistency in results when applying algebraic rules.
• Facilitate the direct application of calculus techniques like differentiation and integration, which often rely on positive exponent forms.

By adhering to positive exponents, we maintain clarity and accuracy in our mathematical problem-solving.