Problem 35 - 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. $$7 y^{4}+14 y^{3}+7 y^{2}$$

Step-by-Step Solution

Answer

The polynomial \(7 y^{4}+14 y^{3}+7 y^{2}\) is completely factored as \(7y^2(y+1)^2\).

Step 1: Recognize the Common Factor

Firstly, recognize that every term in the polynomial has a common factor of \(7y^2\).

Step 2: Apply the distributive property

Using the distributive property which states that \(a(b + c) = ab+ ac\), we remove the common factor by dividing every term by \(7y^2\) to achieve: \(y^2 + 2y + 1\). The operation results in \(7y^2(y^2 + 2y + 1)\).

Step 3: Recognize a perfect square trinomial

The trinomial \(y^2 + 2y + 1\) is a perfect square trinomial since it can be written as \((y+1)^2\). Therefore, the factored form of the polynomial is \(7y^2(y+1)^2\).

Step 4: Verification

Verify the result through multiplication: \(7y^2(y+1)^2 = 7y^2(y^2 + 2y + 1) = 7y^4 + 14y^3 + 7y^2\), which verifies that the factorization is correct.

Key Concepts

Common Factor

When dealing with polynomial factorization, the first step is to identify any common factors among the terms. A common factor is a number or expression that uniformly divides each term of the polynomial without leaving a remainder. In the exercise, we have the polynomial \(7y^4 + 14y^3 + 7y^2\).

Here, each term contains a common factor of \(7y^2\), because:

• \(7y^4\) is divisible by \(7y^2\)
• \(14y^3\) is divisible by \(7y^2\)
• \(7y^2\) is divisible by \(7y^2\)

Identifying and factoring out this common factor simplifies our task, reducing the polynomial to a simpler form, \(y^2 + 2y + 1\). Recognizing common factors is crucial as it helps transform complex expressions into simpler, more manageable ones.

Perfect Square Trinomial

A perfect square trinomial is a special type of polynomial that arises from squaring a binomial. It generally takes the form \(a^2 + 2ab + b^2\), which can be rewritten as \((a + b)^2\).

In the simplified polynomial \(y^2 + 2y + 1\) from the exercise, we recognize it as a perfect square trinomial:

• \(y^2\) is \(a^2\), implying \(a = y\)
• \(2y\) corresponds to \(2ab\), where both \(a\) and \(b\) are 1
• \(1\) is \(b^2\), meaning \(b = 1\)

This recognition allows us to factor it as \((y + 1)^2\). Recognizing perfect square trinomials is beneficial because they allow polynomials to be factored neatly and verified easily through expansion.

Distributive Property

The distributive property is a fundamental algebraic property that allows us to multiply a single term across terms within parentheses. It is expressed as \(a(b + c) = ab + ac\).

In the exercise, once we identify the common factor \(7y^2\), we use the distributive property to factor it from each term:

• Dividing \(7y^4 + 14y^3 + 7y^2\) by \(7y^2\) simplifies to \(y^2 + 2y + 1\)

This operation transforms the original polynomial to \(7y^2(y^2 + 2y + 1)\). The distributive property is crucial because it optimally re-arranges polynomials to allow for further factorization and simplification.