Problem 56 - Algebra for College Students

Question

Factor completely each of the polynomials and indicate any that are not factorable using integers. $$x^{4}-13 x^{2}+36$$

Step-by-Step Solution

Answer

The polynomial factors as \((x - 2)(x + 2)(x - 3)(x + 3)\).

Step 1: Recognize the form

The polynomial given is \( x^4 - 13x^2 + 36 \). Notice it can be viewed as a quadratic in form, where \( x^2 \) acts as the variable in a quadratic trinomial.

Step 2: Substitution

To simplify the factoring process, substitute \( y = x^2 \). The polynomial becomes \( y^2 - 13y + 36 \).

Step 3: Factor the quadratic

Factor \( y^2 - 13y + 36 \). Look for two integers whose product is 36 and whose sum is -13. The factors are -4 and -9, thus the quadratic factors as \((y - 4)(y - 9)\).

Step 4: Substitute back

Substitute \( y = x^2 \) back into the factors to get \((x^2 - 4)(x^2 - 9)\).

Step 5: Factor further using difference of squares

Both \( x^2 - 4 \) and \( x^2 - 9 \) can be factored further using the difference of squares formula. \( x^2 - 4 = (x - 2)(x + 2) \) and \( x^2 - 9 = (x - 3)(x + 3) \).

Step 6: Write the complete factorization

The complete factorization of the polynomial \( x^4 - 13x^2 + 36 \) is \((x - 2)(x + 2)(x - 3)(x + 3)\).

Key Concepts

Understanding Quadratic Trinomial

A quadratic trinomial is a polynomial with three terms, generally in the form \( ax^2 + bx + c \). Even though it is termed 'quadratic', it always assumes the variable raised only to the second power. In the context of our exercise, we initially reframe the original polynomial \( x^4 - 13x^2 + 36 \) as a quadratic trinomial by substituting \( x^2 \) with another variable \( y \). This allows it to appear as \( y^2 - 13y + 36 \).

By doing this substitution, we cleverly manipulate the expression into a recognizable quadratic form, which is much easier to handle. The key in working with quadratic trinomials is to identify the proper approach to factorization. In this case, we consider the factors of \( c \) that sum to \( b \). This technique provides a simple pathway toward simplifying complex polynomials into smaller, more manageable parts.

Exploring the Difference of Squares

The difference of squares is a crucial algebraic form that appears often in polynomial factorization. This occurs in situations where we have a term squared minus another term squared: \( a^2 - b^2 \). The beauty of the difference of squares is that it can be neatly factored into \( (a - b)(a + b) \).

In our example, after substituting back \( y = x^2 \) into \( (y - 4)(y - 9) \), we achieve expressions like \( x^2 - 4 \) and \( x^2 - 9 \), both of which are differences of squares. By applying the difference of squares formula:

• \( x^2 - 4 = (x - 2)(x + 2) \)
• \( x^2 - 9 = (x - 3)(x + 3) \)

The elegance of the difference of squares is its ability to transform a quadratic expression into two linear factors, making the problem much more approachable, ultimately simplifying the polynomial to its complete factorized form.

The Role of the Substitution Method

Substitution is a highly effective mathematical technique used to simplify expressions and equations, especially when they take a complex form. In our exercise, we use a substitution method to make the polynomial \( x^4 - 13x^2 + 36 \) easier to manage by letting \( y = x^2 \).

By redefining part of the polynomial in terms of \( y \), we convert it to \( y^2 - 13y + 36 \), which is far simpler to factor as a quadratic trinomials. The substitution allows us to avoid directly working with very high degrees and instead allows us to handle expressions we're more familiar with.

Once we factor the simplified form, we substitute back to recompute in terms of \( x \), eventually revealing the full solution path. Substitution, therefore, serves as a powerful tool, reducing complicated mathematical processes into fundamental concepts we can easily solve step-by-step.