Problem 52 - Algebra for College Students

Question

Factor completely each of the polynomials and indicate any that are not factorable using integers. $$x^{4}-x^{2}-12$$

Step-by-Step Solution

Answer

The polynomial \(x^4 - x^2 - 12\) factors to \((x - 2)(x + 2)(x^2 + 3)\).

Step 1: Identify the polynomial structure

The given polynomial is \( x^4 - x^2 - 12 \). Notice that it has terms involving \( x^4 \) and \( x^2 \), making it look like a quadratic in terms of \( x^2 \). Let \( y = x^2 \), so the polynomial becomes \( y^2 - y - 12 \).

Step 2: Factor as a quadratic in terms of y

We now need to factor \( y^2 - y - 12 \) as a quadratic equation. To factor this, look for two numbers that multiply to \(-12\) and add to \(-1\). These numbers are \(-4\) and \(3\). Thus, the factorization is \((y - 4)(y + 3)\).

Step 3: Substitute back the variable

Substitute \( y = x^2 \) back into the factored expression. Thus, \( (y - 4) \) becomes \( (x^2 - 4) \) and \( (y + 3) \) becomes \( (x^2 + 3) \). Thus, the expression becomes \( (x^2 - 4)(x^2 + 3) \).

Step 4: Factor completely

The term \( x^2 - 4 \) is a difference of squares and can be further factored into \( (x - 2)(x + 2) \). The term \( x^2 + 3 \) is not factorable using integers (as there are no integer solutions for such a sum of squares). Therefore, the completely factored form of the polynomial is \((x - 2)(x + 2)(x^2 + 3)\).

Key Concepts

Quadratic Equations

Quadratic equations are a foundational concept in algebra, typically represented in the standard form as \( ax^2 + bx + c = 0 \). These equations have variable terms up to the second degree. In our exercise, we began with the polynomial \( x^4 - x^2 - 12 \). By recognizing the structure, we rewrote the polynomial in terms of another variable, \( y = x^2 \), shifting the perspective to look like a quadratic equation, \( y^2 - y - 12 \).This change allows us to use techniques developed for solving quadratics. Specifically, we factor by finding two numbers that multiply to give the constant term (-12) and add up to the linear coefficient (-1). This step is crucial as it helps in simplifying complex expressions and finding solutions more efficiently. Understanding this method lays the groundwork for addressing more challenging algebraic equations.

Difference of Squares

The difference of squares is a special factoring technique used when an expression can be written as \( a^2 - b^2 \). It allows us to factor the expression into \( (a - b)(a + b) \). In the context of our example, after substituting back \( y = x^2 \), we identified \( x^2 - 4 \) as a difference of squares.This is because \( x^2 - 4 \) can be seen as \( x^2 - 2^2 \). By applying the difference of squares formula, we factored it into \( (x - 2)(x + 2) \). Recognizing patterns like these simplifies the process of factoring polynomials. This technique is very useful when working with polynomials, ensuring expressions are in their simplest form.

Sum of Squares

While factoring polynomials, it's essential to recognize when certain terms can't be factored using integers. The sum of squares \( x^2 + b^2 \) is a common scenario where standard integer factoring doesn't apply. In our problem, the term \( x^2 + 3 \) was identified as such a case.This term can't be factored into real-number integers, because no real numbers square to give a negative ( due to the plus sign). However, it's left as is in the factorization process unless working over complex numbers, where other techniques might be utilized. Knowing this limitation helps avoid unnecessary complication in polynomial factoring and keeps the solution clear and correct.