Problem 53 - Algebra for College Students

Question

Factor completely each of the polynomials and indicate any that are not factorable using integers. $$18 n^{4}+25 n^{2}-3$$

Step-by-Step Solution

Answer

The polynomial \(18n^4 + 25n^2 - 3\) factors completely as \((2n^2 + 3)(3n - 1)(3n + 1)\).

Step 1: Identify the Type of Polynomial

We have the polynomial \(18n^4 + 25n^2 - 3\). Notice that it is a quadratic in terms of \(n^2\), if we let \(x = n^2\), then the polynomial becomes \(18x^2 + 25x - 3\). This substitution helps us recognize it as a quadratic trinomial.

Step 2: Search for Two Binomials

To factor \(18x^2 + 25x - 3\), find two numbers that multiply to \(18 \times (-3) = -54\) and add to \(25\). These numbers are \(27\) and \(-2\). We can thus rewrite the middle term as \(27x - 2x\).

Step 3: Rewrite and Group Terms

Rewriting the expression gives us \(18x^2 + 27x - 2x - 3\). Now, group the terms: \((18x^2 + 27x)\) and \((-2x - 3)\).

Step 4: Factor by Grouping

Factor out the greatest common factor from each group: \(9x(2x + 3) - 1(2x + 3)\). Notice \((2x + 3)\) is common in both terms.

Step 5: Factor out the Common Binomial

Now factor out the common binomial, \((2x + 3)\): \((2x + 3)(9x - 1)\). This is the factorization of the original trinomial.

Step 6: Substitute back \(x = n^2\)

Replace \(x\) with \(n^2\) to revert to the original variable: \((2n^2 + 3)(9n^2 - 1)\).

Step 7: Factor the Difference of Squares

Notice that \(9n^2 - 1\) is a difference of squares. It can be factored further into \((3n - 1)(3n + 1)\).

Step 8: Write the Complete Factorization

Combine all the factors to write the complete factorization: \((2n^2 + 3)(3n - 1)(3n + 1)\).

Key Concepts

Quadratic Trinomials

Quadratic trinomials are polynomial expressions consisting of three terms, typically in the form \(ax^2 + bx + c\). These types of polynomials are called "quadratic" because the highest degree of the variable, usually denoted as \(x\), is 2. When faced with a polynomial like \(18n^4 + 25n^2 - 3\), it's helpful to recognize it as a quadratic trinomial in terms of a different variable. By setting \(x = n^2\), it becomes \(18x^2 + 25x - 3\), making it easier to handle.

Understanding the structure and identifying quadratic trinomials is a critical first step in factorization. It allows us to apply familiar strategies, like factoring into the product of two binomials. The goal is to express the polynomial in a simplified, factored form, which is especially useful in problem-solving scenarios. Remember, the leading coefficient \(a\), the middle coefficient \(b\), and the constant term \(c\) play distinct roles in determining the factorization approach.

Factoring by Grouping

Factoring by grouping is an effective strategy used to factor certain kinds of polynomials, especially when dealing with quadratic trinomials after splitting the middle term. The idea is to break up a polynomial into groups where common factors can be identified and factored out.

In the exercise, we have the polynomial \(18x^2 + 25x - 3\) and need to factor it. We start by finding two numbers whose product equals \(-54\) (the product of \(18\) and \(-3\)) and whose sum is \(25\). These numbers are \(27\) and \(-2\).

This allows us to rewrite the polynomial as \(18x^2 + 27x - 2x - 3\) and group terms: \((18x^2 + 27x)\) and \((-2x - 3)\). From each group, we factor out the greatest common factors: \(9x(2x + 3)\) and \(-1(2x + 3)\). Since both terms share a common binomial \((2x + 3)\), we can factor this out, leaving us with \((2x + 3)(9x - 1)\).

This process highlights the importance of grouping and identifying common factors to systematically break down polynomials into simpler, factorable parts.

Difference of Squares

The concept of the difference of squares is based on the identity \(a^2 - b^2 = (a - b)(a + b)\). Recognizing when a polynomial can be expressed as a difference of squares is very helpful in further factorization. After initially factoring our quadratic trinomial as \((2n^2 + 3)(9n^2 - 1)\), we notice that \(9n^2 - 1\) fits this pattern.

Here, \(9n^2\) is a perfect square because it is \((3n)^2\), and \(1\) is also a perfect square \((1)^2\). Thus, the expression \(9n^2 - 1\) can be rewritten as \((3n - 1)(3n + 1)\). Applying the difference of squares technique allows us to factor our expression completely.

Factoring polynomials using the difference of squares is useful when simplifying expressions, solving equations, or integrating expressions in calculus. Look out for terms that can be rewritten as squares and spaces between them indicating subtraction. This can often be an efficient way to break down more complex polynomial expressions.