Problem 71 - Elementary Algebra

Question

Use any of the factoring methods to factor. Identify any prime polynomials. $$ v^{2}+18 v+81 $$

Step-by-Step Solution

Answer

The polynomial \( v^{2} + 18v + 81 \) factors to \( (v + 9)^2 \).

Step 1: Identify the Quadratic Polynomial

Write down the given quadratic polynomial which is \[ v^{2} + 18v + 81 \]

Step 2: Check for Perfect Square Trinomial

Determine if the polynomial is a perfect square trinomial. A perfect square trinomial is of the form \[ a^2 + 2ab + b^2 \] where \[ a = v \] and \[ b = 9 \].

Step 3: Verify the Structure

Verify if \[v^2 + 2 \times v \times 9 + 9^2 \] matches the given polynomial. Calculate and check: \[ v^2 + 2 \times 9 \times v + 81 = v^2 + 18v + 81 \]. Since the structure matches, it is a perfect square trinomial.

Step 4: Write as a Square of a Binomial

Factor the polynomial into the square of a binomial: \[ (v + 9)^2 \]. Hence, the factorization of \[ v^{2} + 18v + 81 \] is \[ (v + 9)(v + 9) \] or simply \[ (v + 9)^2 \].

Key Concepts

Perfect Square Trinomial

A Perfect Square Trinomial is a specific type of polynomial. It looks actually like a perfect square when expanded. This type of trinomial has the form \( a^2 + 2ab + b^2 \).
In our exercise, we have: \( v^2 + 18v + 81 \).
To check if this is a perfect square trinomial, we need to see if it fits the pattern **a^2 + 2ab + b^2**.
Here, **a** is the square root of the first term \(v^2\), so **a** = **v**.
Then, **b** should be the square root of the last term \(81\), so **b** = **9**.
Notice the middle term \( 18v \). It should equal **2ab**, and indeed it matches since \( 2 \times v \times 9 = 18v \).
This verifies that \(v^2 + 18v + 81 \) is a perfect square trinomial.
Once identified, it can be factored easily!

Quadratic Polynomial

A Quadratic Polynomial is any polynomial that can be written in the form \[ ax^2 + bx + c \], where **a**, **b**, and **c** are constants, and **x** represents a variable.
The polynomial given in the exercise is: \( v^2 + 18v + 81 \).
In this example, **a** = **1**, **b** = **18**, and **c** = **81**. Here, **v** is the variable.
What makes it quadratic is the highest power of the variable, which is 2 in **v^2**.
Quadratic polynomials often describe parabolic shapes when graphed. They appear everywhere in mathematics, from physics to engineering.
Understanding how to work with them, including factoring, is essential for many applications!

Factoring Methods

Factoring Methods are different approaches used to break down a polynomial into simpler terms or products.
It simplifies solving equations and identifying specific characteristics of the polynomial.
For the polynomial in our example \( v^2 + 18v + 81 \), recognizing it as a perfect square trinomial helped us factor it quickly.
Other common factoring methods include:

• **Factoring by Grouping**: Useful when you can group terms to find common factors.
• **Using the Quadratic Formula**: When the polynomial is hard to factor by other means.
• **Prime Factorization**: Breaking down into prime polynomial factors.

Here, we identified \(v^2 + 18v + 81 \) as a perfect square trinomial, so we directly used that pattern.
We factored it as \( (v + 9)(v + 9) \) or simply \( (v + 9)^2 \).
Knowing these methods and practicing helps make the process simpler and more intuitive!