Problem 36 - Introductory Algebra for College Students

Question

Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$2 y^{4}+28 y^{3}+98 y^{2}$$

Step-by-Step Solution

Answer

The completely factorized form of the polynomial \(2y^4 + 28y^3 + 98y^2\) is \(2y^2(y + 7)^2\).

Step 1: Inspection and Identifying Common Factors

The given polynomial is \(2y^4 + 28y^3 + 98y^2\). Start by inspecting the terms to identify any common factors. We can see that all terms share a common factor of \(2y^2\).

Step 2: Factor out the Common Factor

When we take out the common factor \(2y^2\) from each term in the polynomial, the polynomial becomes \(2y^2(y^2 + 14y + 49)\).

Step 3: Factorize the Quadratic Polynomial

The expression inside the brackets, \(y^2 + 14y + 49\), is a quadratic polynomial, which can be factorized further. This quadratic can be written as \((y + 7)^2\) because \(7^2 = 49\) and \(2*7 = 14\).

Step 4: Write the Completely Factorized Form

Substitute the factored form of the quadratic back into the polynomial. The completely factorized form of the given polynomial is \(2y^2(y + 7)^2\).

Key Concepts

Common Factors in Polynomials

Understanding the concept of common factors in polynomials is crucial for simplifying complex expressions and solving algebraic problems. A common factor is an expression that can be divided evenly into each term of the polynomial. Identifying common factors is the first step in the factorization process and can greatly simplify the expression.

In the exercise example, the polynomial given was \(2y^4 + 28y^3 + 98y^2\). Here, every term is divisible by \(2y^2\), making it the common factor. To factor it out, we divide each term of the polynomial by \(2y^2\), which simplifies the polynomial and sets the stage for further factorization. This process is imperative because it reduces the complexity of the polynomial, making it easier to handle in subsequent steps of the factorization.

Factorization Techniques

Factorization techniques are the methods used to break down polynomials into simpler, irreducible factors. There are various techniques to achieve this, but some of the most common include factoring by grouping, using special formulas like the difference of squares or perfect square trinomials, and factoring quadratic polynomials. Mastering these techniques is key to solving polynomials effectively.

For instance, after extracting the common factor in our exercise, we're left with a quadratic polynomial inside the brackets, which can be handled using techniques relevant to quadratic equations. In this case, the trinomial \(y^2 + 14y + 49\) was identified to be a perfect square and factored as \((y + 7)^2\). Recognizing patterns such as this and applying the right technique simplifies the problem and leads to a clearer understanding of how to tackle similar factorization challenges.

Quadratic Polynomials

Quadratic polynomials are second-degree polynomials, the most recognizable form being \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. Solving quadratics is a fundamental skill in algebra. These polynomials can be factorized into simpler binomials when possible, revealing the polynomial's roots. Strategies for factorization may include finding two numbers that multiply to \(ac\) and add up to \(b\) when \(a\) is 1, or using the quadratic formula when factorization is not straightforward.

In the provided exercise, the quadratic was a perfect square trinomial - a special form where the expression can be written as the square of a binomial, \((y + 7)^2\). Recognizing and factoring these types of quadratics not only provides the solution but also assists in understanding the structure and properties of polynomial expressions.