Problem 18 - Introductory Algebra for College Students

Question

Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$4 x^{3}-100 x$$

Step-by-Step Solution

Answer

The completely factored form of the polynomial \(4x^3-100x\) is \(4x(x-5)(x+5)\).

Step 1: Identify Any Common Factors

First step would be to identify if there are any common factors between the components of the polynomial. In this case, our polynomial \(4x^3-100x\) has a common factor of \(4x\).

Step 2: Factor Out the Common Factor

The second step is to factor out the common factor. Factoring out \(4x\) from \(4x^3-100x\) would give us \(4x(x^2-25)\).

Step 3: Continue Factoring if Possible

Observe that \(x^2-25\) is a binomial that continues to factor given two perfect square terms. This binomial represents the difference of squares which can be factored as \(a^2-b^2 = (a-b)(a+b)\). In this instance, factoring \(x^2-25\) yields \(x-5)(x+5)\).

Step 4: Incorporate All Factors into Final Answer

Lastly, we incorporate all factors that we have identified into our final answer. The last step gives us the fully factored form \(4x(x-5)(x+5)\).

Key Concepts

Common Factors

To start factoring a polynomial, it's important to look for any common factors that might be present. A common factor is a term that can be evenly divided into each component of the polynomial. Identifying these factors can simplify the expression and make further factoring easier.

In the given polynomial, \(4x^3 - 100x\), notice that both terms have a factor of \(x\). Additionally, the coefficients 4 and 100 share a common multiple, which is 4. So, \(4x\) is the common factor here.

Once the common factor is identified, we'll factor it out of each term. This changes \(4x^3 - 100x\) into \(4x(x^2 - 25)\). Removing the common factor simplifies the polynomial and sets it up for further factoring methods.

Difference of Squares

The concept of a difference of squares is a reliable and straightforward technique for factoring binomials. A difference of squares occurs when you have two squared terms separated by a subtraction sign. Mathematically, it takes the form \(a^2 - b^2\) and factors into \((a-b)(a+b)\).

In our polynomial, after factoring out the common factor, we are left with the term \(x^2 - 25\). This is a classic example of a difference of squares. Here, we recognize that both \(x^2\) and 25 are perfect squares, \(x^2\) equates to \(x \cdot x\) and 25 equates to \(5 \cdot 5\).

To factor this binomial using the difference of squares method, simply apply the formula: this results in \((x-5)(x+5)\). It's a simple step that can significantly break down complex expressions into manageable pieces.

Binomials

Binomials are expressions composed of just two terms. They are quite significant in algebra due to their prevalence in various mathematical problems, including those that involve factoring.

The binomial \(x^2 - 25\), as derived from our original expression \(4x^3 - 100x\), is a prime example. Once we've simplified a polynomial to just a binomial form, several techniques—like the difference of squares—can be applied to further explore its properties and simplify it.

In working with binomials, recognizing special patterns such as perfect squares or sum and difference of cubes helps to proceed with factoring effectively. Always be attentive to these patterns since they can lead to quicker and more efficient solutions.