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Chapter 4: Problem 59

Perform each division. Assume no division by \(0 .\) $$\frac{x^{3}-8}{x-2}$$

Short Answer

Expert verified
The division \(\frac{x^{3}-8}{x-2}\) simplifies to \(x^{2}+2x+4\).

Step by step solution

01

Identify the Form of the Expression

The given expression is \(\frac{x^{3}-8}{x-2}\), which has the form \(x^{3}-y^{3}\) where \(x=x\) and \(y=2\). The difference of cubes formula can be applied here where \(a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})\).
02

Factor the Numerator

Apply the difference of cubes formula on the numerator \(x^{3}-2^{3}\). This will result in \((x-2)(x^{2}+2x+4)\).
03

Cancel Out the Denominator

The denominator \(x-2\) now cancels out with the same factor in the numerator. This leaves \((x^{2}+2x+4)\). So, \(\frac{x^{3}-8}{x-2} = x^{2}+2x+4\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Cubes
The difference of cubes is a specific algebraic pattern where two cube numbers are subtracted. Consider the expression \(a^3 - b^3\). This formula can be factored into \((a - b)(a^2 + ab + b^2)\). Think of it as neatly packing a more complex polynomial expression into factors that are simpler to manage.
The key is identifying the cubic terms in terms of a base number, like \(x^3\) or \(2^3\) in our original exercise.
  • When you encounter a term like \(x^3 - 8\), determine the cube roots: \(x\) and \(2\).
  • Using the difference of cubes formula simplifies factoring by providing a direct path.
This technique is quite powerful as it transforms a hard-to-simplify cubic polynomial into a simpler set of factors.
Factoring Polynomials
Factoring polynomials is like unraveling the puzzle of algebraic expressions to find simpler expressions that multiply back into the original. Let's dive deep into the process using the given problem: \(x^3 - 8\).
Polynomials can often be broken down by identifying recognizable patterns like the difference of cubes, as we've seen.
  • First, see if the polynomial fits a known formula or pattern, like a difference of cubes.
  • Apply this pattern to neatly break apart the expression into two factors as we did with \(x^3 - 8\).
Here, the cubic expression \(x^3 - 8\) was factored into \((x - 2)(x^2 + 2x + 4)\).
This technique makes solving equations and simplifying expressions much more manageable.
Algebraic Expressions
Algebraic expressions are mathematical phrases that include numbers, variables, and operations. In essence, they form the building blocks of many algebraic equations and are crucial in learning math.
Our example \(\frac{x^3-8}{x-2}\) represents an expression where you can see a division setup.
  • Breaking down complex algebraic expressions into simpler parts, as with the numerator \(x^3-8\), is crucial for understanding.
  • Use techniques such as polynomial factoring to transform and simplify.
Simplifying and manipulating these expressions ultimately makes solving many mathematical problems cleaner and quicker.
Mastering the intricacies of these expressions enhances both comprehension and problem-solving skills in algebra.

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Most popular questions from this chapter

Classify each polynomial as a monomial, a binomial, a trinomial, or none of these. $$x^{3}-1$$

Give an example of a polynomial that is \(\ldots\) not a monomial, a binomial, or a trinomial

Graph the function \(f(x)=-x^{2} .\) What do you discover?

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Give an example of a polynomial that is \(\ldots\) of degree 1
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