Problem 22 - Algebra A Combined Function

Question

Factor each trinomial completely. $$ y^{3}+12 y^{2}+36 y $$

Step-by-Step Solution

Answer

The complete factorization is \( y(y + 6)^2 \).

Step 1: Identify Common Factors

First, look for any common factors in all the terms of the polynomial. Notice that each term has at least one y. This means we can factor out a y from the entire polynomial. Doing this gives us:\[ y(y^2 + 12y + 36) \]

Step 2: Check the Quadratic

The remaining polynomial \( y^2 + 12y + 36 \) is a quadratic trinomial. We will now focus on factoring this trinomial.

Step 3: Find the Factors of the Quadratic

Look for two numbers that multiply to 36 and add to 12. These numbers are 6 and 6, because: - 6 multiplied by 6 equals 36 - 6 plus 6 equals 12

Step 4: Write the Quadratic as a Square

Since 6 and 6 are the same, we can write the quadratic trinomial as a perfect square:\[ y^2 + 12y + 36 = (y + 6)^2 \]

Step 5: Write the Complete Factored Form

Now we will include the y (original common factor) in our factored result, which makes the complete factorization:\[ y(y + 6)^2 \]

Key Concepts

Common Factors

The concept of common factors is one of the foundational principles of factoring expressions, especially polynomials. Common factors are elements that are repeated across all terms in a polynomial. In the original example, we have the polynomial expression \( y^3 + 12y^2 + 36y \).

By examining each term, we see that all terms have at least one 'y'. This indicates that the common factor here is 'y'.

1. Always begin by identifying the highest common factor among the terms for simplification.
2. Once identified, factor it out to simplify the polynomial, making it easier to work on the remaining expression.

This approach paves the way to simplifying a complex polynomial into a more manageable form.

Quadratic Trinomials

After factoring out the common factor from a polynomial, the next step often involves dealing with a quadratic trinomial. A quadratic trinomial is a three-term polynomial with the general form \( ax^2 + bx + c \).

In our example, after factoring out the common factor \( y \), the resulting polynomial is \( y^2 + 12y + 36 \), which is a typical quadratic trinomial.

1. The process of factoring quadratic trinomials typically involves finding two numbers that multiply to the constant term (c) and add up to the linear coefficient (b). For \( y^2 + 12y + 36 \), we look for numbers that multiply to 36 and add to 12.
2. These numbers help break down the trinomial further, leading to its factorization.

Emphasizing these steps simplifies addressing quadratic trinomials effectively.

Perfect Square Trinomials

Perfect square trinomials are a special type of quadratic trinomial where two identical binomials multiply to give the trinomial. Recognizing perfect square trinomials can streamline the factoring process considerably.

In the polynomial \( y^2 + 12y + 36 \), the numbers 6 and 6 satisfy the condition for both the product and sum (36 and 12 respectively).

• This allows us to express \( y^2 + 12y + 36 \) as \( (y + 6)^2 \), highlighting its structure as a perfect square trinomial.
• Understanding how to identify and factor perfect square trinomials can significantly reduce complexity in polynomial factorization.

Factoring Polynomials

Factoring polynomials is a crucial skill in algebra that involves breaking down complex expressions into simpler, multipliable components. The primary objective is to simplify the polynomial into its constituent factors.

In the example provided, the initial polynomial \( y^3 + 12y^2 + 36y \) is first reduced by recognizing the common factor 'y', resulting in \( y(y^2 + 12y + 36) \).

• The subsequent step involves refactoring the quadratic trinomial \( y^2 + 12y + 36 \) as a perfect square trinomial \( (y + 6)^2 \).
• Finally, combining the common factor and the simplified trinomial yields the completely factored form \( y(y + 6)^2 \).

This methodical approach to breaking down polynomials into simpler, understandable parts is essential for solving algebraic equations efficiently.