Problem 82 - Algebra for College Students

Question

Should help you pull together all of the factoring techniques of this chapter. Factor completely each polynomial, and indicate any that are not factorable using integers. $$x^{3}+125$$

Step-by-Step Solution

Answer

The polynomial is completely factored as \((x + 5)(x^2 - 5x + 25)\).

Step 1: Identify the polynomial form

The given polynomial is \(x^3 + 125\). Notice that this expression is in the form of \(a^3 + b^3\), where \(a = x\) and \(b = 5\).

Step 2: Recall the sum of cubes formula

The sum of cubes formula is \((a^3 + b^3) = (a + b)(a^2 - ab + b^2)\). We'll use this formula to factor the polynomial.

Step 3: Substitute values into the formula

Using the values \(a = x\) and \(b = 5\), substitute into the formula: \((x + 5)(x^2 - 5x + 25)\).

Step 4: Determine if further factoring is possible

The expression \(x^2 - 5x + 25\) cannot be factored further using integers, as its discriminant \((-5)^2 - 4 \cdot 1 \cdot 25\) is negative, indicating no real solutions.

Key Concepts

Sum of Cubes

When we talk about the sum of cubes, we are referring to a specific expression format: \(a^3 + b^3\). This is a special type of polynomial that has a unique factoring formula. The sum of cubes formula is:

• \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\).

Understanding this formula is critical because it allows us to factor expressions of this form into smaller, more manageable components. In our original exercise of factoring \(x^3 + 125\), identifying it as a sum of cubes is the first step to finding its factors. Here, \(a = x\) and \(b = 5\). We used this formula to break down the expression into \((x + 5)(x^2 - 5x + 25)\), which cannot be simplified further using integer factors.

Polynomial Factorization

Polynomial factorization involves breaking down a polynomial into products of simpler polynomials. Our goal is to express the polynomial as a product of its irreducible factors. For the polynomial \(x^3 + 125\), recognizing it as a sum of cubes allowed us to factor it into \((x + 5)(x^2 - 5x + 25)\).
In general, factorization can save us time and effort when solving equations, as it simplifies the problem into smaller parts. Each of these parts can be solved or analyzed individually. Always remember to look for familiar patterns like the sum of cubes or difference of cubes, and other algebraic identities, as these can simplify your work greatly.

Discriminant

The discriminant is a valuable tool when dealing with quadratic equations, particularly when checking if a quadratic can be factored over the integers. For a quadratic equation \(ax^2 + bx + c\), the discriminant is found using the formula \(b^2 - 4ac\).
In our specific polynomial factorization, after using the sum of cubes formula to factor \(x^3 + 125\), we need to analyze whether \(x^2 - 5x + 25\) can be further factored. By calculating the discriminant, \((-5)^2 - 4 \times 1 \times 25\), we find it is negative. A negative discriminant means that there are no real roots, and thus, \(x^2 - 5x + 25\) cannot be factored further using real numbers or integers.

Integers in Factoring

Factoring over the integers means expressing the polynomial as a product of polynomial factors with integer coefficients. This is a common requirement in many algebra exercises because it ensures solutions can be readily understood and verified without needing complex or irrational numbers.
In our exercise with \(x^3 + 125\), after applying the sum of cubes formula, we sought to factor further into polynomials with integer coefficients. However, as the discriminant of \(x^2 - 5x + 25\) indicated no real roots, further integer factoring was not possible. Understanding these limitations is crucial as it guides whether you need to look for other mathematical forms or conclude with the factoring achieved. When dealing with polynomials, always check these criteria to determine possible approaches and limits in factorization.