Problem 105 - Algebra and Trigonometry

Question

Factor Completely. $$6 x^{4}+35 x^{2}-6$$

Step-by-Step Solution

Answer

\((6x^2 - 1)(x^2+6)\)

Step 1: Recognize the form and make the substitution

Identify that the problem can be viewed as a quadratic where \(u = x^2\). Substituting \(u\) for \(x^2\), the equation becomes \(6u^2 + 35u - 6\).

Step 2: Factor the Quadratic

Factor the equation to form two binomial terms. We are looking for two numbers that multiply together to give \(ac = -36\), and add together to give \(b = 35\). We found the numbers 36 and -1. So the factored form is: \(6u^2 + 36u - u - 6 = 6u(u+6) - 1(u+6) = (6u - 1)(u+6)\).

Step 3: Substitute back

Replace \(u\) with \(x^2\). So the completely factored form of the original expression becomes \(6x^2 - 1)(x^2+6)\).

Key Concepts

Quadratic Substitution

Quadratic substitution is a helpful technique used to simplify polynomials that resemble quadratic equations. A quadratic has a form similar to \(ax^2 + bx + c\). When you have an equation like \(6x^4 + 35x^2 - 6\), notice that it can be tricky to factor directly because of the higher power. The key here is realizing that you can make a substitution to turn it into a simpler expression.
To do this, identify the recurring variable part \(x^2\) and substitute it with a single variable, such as \(u\). Here, you let \(u = x^2\).

• This turns the original expression into a more familiar quadratic form, \(6u^2 + 35u - 6\).
• This method effectively reduces the power of the polynomial, making it manageable to work with.
• After you work with the new form, remember to substitute back to express the final answer in terms of the original variable.

This technique isn't limited to just squared terms; it can apply to any polynomial expression that can be reimagined in terms of quadratic-like terms.

Polynomial Factoring

Polynomial factoring is the process of decomposing an expression into a product of simpler expressions or factors. In this case, we began with the expression \(6u^2 + 35u - 6\).
The immediate goal is to find two binomials that multiply to give the original quadratic expression. You achieve this by searching for two numbers that:

• Multiply together to equal the product of the leading coefficient (the number in front of \(u^2\)) and the constant term. Here, \(ac = -36\).
• Add together to equal the middle coefficient, which is \(b = 35\).

In this problem, 36 and -1 fit the criteria because \(36 \times (-1) = -36\) and \(36 + (-1) = 35\). Use these numbers to split the middle term, \(35u\), into two separate terms: \(36u\) and \(-u\).
The decomposition process results in the expression \(6u^2 + 36u - u - 6\). Continue by factoring each part individually and then combine the results to get the fully factored form: \((6u - 1)(u + 6)\).

Factoring by Grouping

Factoring by grouping is a technique used when a polynomial is structured to allow pairs of terms to be factored out separately. This method is especially effective once you add terms strategically, as seen with \(6u^2 + 36u - u - 6\).
Here's how factoring by grouping is applied:

• Start by grouping pairs of terms together: \((6u^2 + 36u)\) and \((-u - 6)\).
• Within each group, factor out the greatest common factor. From \(6u^2 + 36u\), factor out \(6u\), obtaining \(6u(u + 6)\). From \(-u - 6\), factor out \(-1\) to get \(-1(u + 6)\).
• With both groups factored, note that \(u + 6\) is a common factor, which allows it to be factored out from the entire expression.

This gives the factored form \((6u - 1)(u + 6)\). Finally, replace \(u\) back with \(x^2\) to express the polynomial in the original variable: \((6x^2 - 1)(x^2 + 6)\).
Factoring by grouping elegantly simplifies complex polynomials and demonstrates how strategic restructuring can lead to clean solutions.