Problem 72 - Algebra A Combined Function

Question

Factor. $$ 2 y^{4}+11 y^{2}-6 $$

Step-by-Step Solution

Answer

The factors are \((2y^2 - 1)(y^2 + 6)\).

Step 1: Identify Polynomial Structure

We need to factor the polynomial \(2y^4 + 11y^2 - 6\). Notice that this is a quadratic in terms of \(y^2\). Let's substitute \(u = y^2\), making our expression \(2u^2 + 11u - 6\). Our task is to factor this quadratic expression.

Step 2: Use the AC Method to Factor

The AC method helps factor quadratics. Here, \(A = 2\) and \(C = -6\). First, calculate \(A \times C = 2 \times -6 = -12\). We need factors of \(-12\) that add up to \(B = 11\). The factors are \(12\) and \(-1\).

Step 3: Rewrite and Group

Rewrite the quadratic equation using the factors: \(2u^2 + 12u - u - 6\). Group the terms: \((2u^2 + 12u) + (-u - 6)\).

Step 4: Factor by Grouping

Factor out the greatest common factor from each group: \(2u(u + 6) - 1(u + 6)\). Notice \(u + 6\) is common in both terms, so factor it out: \((2u - 1)(u + 6)\).

Step 5: Re-substitute \(u = y^2\)

Now substitute back \(u = y^2\) into the factored expression: \((2y^2 - 1)(y^2 + 6)\). These are the factors of the original polynomial.

Key Concepts

Quadratic Expressions

A quadratic expression is a polynomial of degree two, generally in the form of \( ax^2 + bx + c \). In our problem, the polynomial \( 2y^4 + 11y^2 - 6 \) can be rewritten as a quadratic by substituting \( u = y^2 \), which transforms it into \( 2u^2 + 11u - 6 \).
What's unique here is recognizing the structure that allows us to make this substitution. Understanding quadratic expressions in this way helps us simplify polynomials that at first glance seem more complicated. By making clever substitutions, we turn a more complex problem into one we know how to solve.

AC Method

The AC method is a key technique for factoring quadratic expressions, particularly when the coefficient \( A \) of \( u^2 \) is not 1. Here's how it works:

• Identify \( A \) and \( C \) from the quadratic expression \( Au^2 + Bu + C \). In our example, \( A = 2 \) and \( C = -6 \).
• Calculate \( A \times C \). Here, \( 2 \times -6 = -12 \).
• Find two numbers that multiply to \( -12 \) and add up to \( B = 11 \). These numbers are \( 12 \) and \( -1 \).

The AC method turns the original expression into a sum of terms that can be split into groups, which makes it easier to factor.

Polynomial Structure

Understanding the structure of a polynomial helps make seemingly complex expressions easier to handle. Polynomials are algebraic expressions that consist of variables and coefficients, connected by addition, subtraction, multiplication, and non-negative integer exponents.
In our case, recognizing that \( 2y^4 + 11y^2 - 6 \) fits into a quadratic structure when expressed in terms of \( y^2 \), as \( 2u^2 + 11u - 6 \), simplifies the problem. This insight is essential because it allows us to utilize familiar factoring techniques to solve the problem.
Every polynomial has a unique structure, and understanding this structure can significantly reduce the complexity of algebraic manipulation.

Factor by Grouping

Factor by grouping is a method often used when dealing with four-term polynomials, after the initial rewriting of the expression using the AC method. Here's how it works:

• Rewrite the quadratic in a way that it can be separated into two groups. For \( 2u^2 + 11u - 6 \), it becomes \( 2u^2 + 12u - u - 6 \).
• Group the terms: \( (2u^2 + 12u) + (-u - 6) \).
• Factor out the greatest common factor from each group, yielding \( 2u(u + 6) - 1(u + 6) \).
• The expression \( (u + 6) \) is common, so factor it out to get \( (2u - 1)(u + 6) \).

Finally, re-substituting \( u = y^2 \) results in the factored form \( (2y^2 - 1)(y^2 + 6) \). This method highlights how breaking down a problem into smaller, simpler parts can make factoring manageable and straightforward.