Problem 24 - 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. $$7 x^{4}-7$$

Step-by-Step Solution

Answer

The factorized form of the polynomial \(7x^{4}-7\) is \(7(x-1)(x+1)(x^2+1)\)

Step 1: Identifying a common factor

Here we can see that both terms have the same coefficient of 7. We can draw this common coefficient out. That simplifies the task. So the polynomial will be \(7(x^{4}-1)\)

Step 2: Applying the difference of squares

The polynomial can now be expressed as the difference of two squares, as \(x^4\) is a perfect square and so is 1. Using the formula \(a^2 - b^2 = (a-b)(a+b)\), we express the polynomial as a two separate binomials. Our polynomial now looks like this: \(7((x^2 - 1)(x^2 + 1))\)

Step 3: Further factorizing

We can see that \(x^2 - 1\) is again a case of difference of squares which can be factorized as \((x-1)(x+1)\) and \(x^2 + 1\) is not factorizable. Hence the further simplified version of the given polynomial is: \(7(x-1)(x+1)(x^2+1)\)

Step 4: Checking the factorization

One can confirm the correctness of the factorization by multiplying the factors and comparing with the original polynomial. In this case \(7(x-1)(x+1)(x^2+1)\) = \(7x^4 - 7\) which is same as the given polynomial, hence the factorization is correct.

Key Concepts

Common Factor Extraction

When factoring algebraic expressions, the first step to simplify the task is to look for a common factor that each term in the polynomial shares. In our textbook problem, the expression starts with two terms: \(7x^4\) and -7. Both terms have the coefficient 7, which serves as a common factor. We can extract this common factor by factoring it out, writing the polynomial as \(7(x^4 - 1)\).

Extracting a common factor makes the polynomial simpler and often reveals a recognizable pattern that can be further factored using other techniques. It's important to always look for common factors first, as this step paves the way for more complicated factorization methods that follow.

Difference of Squares

The difference of squares is a commonly used pattern in factorization. It's based on the formula \(a^2 - b^2 = (a - b)(a + b)\). This formula is useful when we have a binomial where each term is a square and they have a subtraction operation between them. In our example, once we've extracted the common factor, we're left with \(x^4 - 1\), which fits the pattern of a difference of squares where \(a = x^2\) and \(b = 1\).

Applying the difference of squares formula, we rewrite the polynomial as \(7((x^2 - 1)(x^2 + 1))\). Recognizing patterns like the difference of squares is essential for tackling complex factorizations, as it allows breaking down polynomials into simpler binomial or trinomial forms that are easier to work with.

Factorizing Quadratics

After recognizing a difference of squares, we arrive at the expression \(7((x^2 - 1)(x^2 + 1))\), with \(x^2 - 1\) still presenting an opportunity for further factorization. In algebra, when dealing with quadratics of the form \(x^2 - y^2\), we can apply the difference of squares rule again. In our case, \(x^2 - 1\) is another difference of squares since 1 is the square of 1 (\(1^2\)).

Thus, we factorize \(x^2 - 1\) as \((x - 1)(x + 1)\). However, \(x^2 + 1\) does not factor further since it does not meet the criteria for any special patterns and there are no real numbers that square to -1. The final factorization of the original polynomial is \(7(x-1)(x+1)(x^2+1)\). It's crucial to recognize which quadratic expressions can be further factorized and those which cannot, like the sum of squares \(x^2 + y^2\), to avoid common mistakes.

Algebraic Expressions

An algebraic expression consists of numbers, variables, and arithmetic operations. Polynomials are a type of algebraic expression that can have constants, variables, and exponents. The process of factorization involves breaking down these expressions into simpler, multiplicative components. The expression in our problem started as \(7x^4 - 7\) and concluded with its factors \(7(x-1)(x+1)(x^2+1)\).

Understanding algebraic expressions is the cornerstone of factorization. Recognizing the structure of a polynomial is essential to knowing which factorization methods to apply, such as common factor extraction, the difference of squares, or factoring quadratics. By mastering these techniques, solving and simplifying complex algebraic expressions become much more approachable, making it possible to solve a wide variety of mathematical problems.