# Operations with Polynomials

## Understanding Arithmetic Operations with Polynomials

Polynomials are mathematical expressions that involve variables raised to whole-number exponents and coefficients. These expressions can be added, subtracted, multiplied, and divided. In this article, we will explore different methods for solving arithmetic operations with polynomials.

The two methods for adding or subtracting polynomials are the horizontal and vertical methods.

### Horizontal Method

To add horizontally, we use the distributive property to combine like terms, resulting in one term for each exponent.

Result:

### Vertical Method

The vertical method involves stacking the polynomials on top of each other and completing any missing terms with a coefficient of zero. This method provides a more organized way to identify like terms and prevent confusion.

When subtracting polynomials, we use the same methods, paying close attention to the signs. To avoid mistakes, we change the signs of all terms in the second polynomial and then add like terms, also known as finding the additive inverse.

Subtract: and

#### Horizontal Method

Change the signs of all terms in the second polynomial and then combine like terms.

## Multiplying Polynomials

For multiplication, we can use both horizontal and vertical methods.

Multiply: and

#### Horizontal Method

Multiply each term in the first polynomial by the second polynomial using the distributive property. Then combine like terms.

#### Vertical Method

____________
Multiply by
Multiply by
____________

The result should be the same regardless of the method used!

## Dividing Polynomials

To divide polynomials, we can use long division or synthetic division methods.

### Long Division Method

Here is an example of using the long division method to divide by :

Divide: by

Before we begin, we need to identify the parts of the division. is the dividend, is the divisor, and the result is the quotient. The remainder is what's left at the end.

To start, we complete missing terms in the dividend with a coefficient of zero to have the polynomial in decreasing order of exponents. Then we follow these steps:

1. Divide the first term of the dividend by the first term of the divisor.
2. Multiply the result by both terms in the divisor and place them under the dividend lined up with their corresponding exponent.
3. Subtract like terms, being careful with the signs.
4. Bring down the next term in the polynomial (dividend).
5. Repeat steps 1-4 until the degree of the remainder is lower than that of the divisor.

Complete missing terms:
Subtract like terms.
Bring down 0x.
Subtract like terms.
Bring down -36.
Subtract like terms.

The remainder is: 0

The result can be expressed as:

### Synthetic Division Method

Another way to solve this example is using the synthetic division method, as the divisor has a degree of 1. According to the Remainder Theorem, if a polynomial is divided by , the remainder is . We can use this to evaluate the polynomial using synthetic substitution, where . For a refresher on evaluating polynomials using this method, check out our article on Evaluating and Graphing Polynomials.

Divide: by

## How to Efficiently Divide Polynomials Using Horner's Method

The Horner's Method is a useful way to divide polynomials quickly. This method involves evaluating the dividend for a fixed value of x, bringing down the leading coefficient, and then simplifying the expression by multiplying and adding terms. The result will be the quotient and remainder of the division.

To start, evaluate the dividend for a fixed value of x, which is the divisor. Then, bring down the leading coefficient below the horizontal line.

## How to Factor Polynomials Using Different Techniques

Polynomial factoring involves breaking down the expression into simpler terms. The approach taken will depend on the degree of the polynomial and the coefficient of the term with the highest power.

For a quadratic polynomial with a leading coefficient of 1: To find the solutions, you need to identify the factors that make the polynomial equal to 0. For example, in x² + 5x + 6, the factors are 2 and 3, giving the solutions x = -2 and x = -3.

For a quadratic polynomial with a leading coefficient greater than 1: An extra step is needed. For instance, in 2x² + 11x + 12, after multiplying the leading coefficient (2) and constant (12) to get 24, you need to find two numbers that add up to 11 (the coefficient of x) and also multiply to 24. In this case, the factors are 3 and 8. The polynomial can then be factored as (2x + 3)(x + 4), giving the solutions x = -3/2 and x = -4.

For a polynomial with a degree greater than 2: In such cases, you may have to extract any common factors before proceeding. For example, to factor 3x³ + 2x² - 15x, you first pull out the common factor x to get x(3x² + 2x - 15). You can then follow the same steps as for a quadratic polynomial to get the solutions x = -3, x = 1/3, and x = -5.

## How to Simplify Polynomials

When simplifying fractional algebraic expressions that involve polynomials, the key is to factor the numerator and denominator and then cancel out any common factors. For instance, in (x² + 6x + 8)/(x² + 4x), you first notice that both polynomials share a common factor of x, which can be canceled out. Then, factor the remaining expressions to get (x + 2)/(x)(x + 4). In some cases, you may need to use the factor theorem to determine if a specific value can be substituted to make the expression equal to 0.

## Understanding the Factor Theorem for Polynomials

The factor theorem is a useful tool for quickly factoring polynomials. It states that if a value p can be substituted into a polynomial function f(x) to make f(p) = 0, then (x - p) is a factor of f(x). This can be applied to cubic polynomials in three steps:

• Substitute different values into f(x) until you find one that makes f(p) = 0
• Divide the polynomial f(x) by (x - p). If the remainder is 0, then (x - p) is a factor.
• Factor the remaining quadratic expression to write f(x) as the product of three linear factors.

For example, given f(x) = x³ + 5x² + 6x, if we substitute x = -2, we get f(-2)= 0, meaning (x + 2) is a factor. We can then divide f(x) by (x + 2) to get x² + 3x, which can be further factored to (x + 3)(x). The final result is x(x + 2)(x + 3) = 0, giving the solutions x = 0, x = -2, and x = -3.

## Key Concepts for Working with Polynomials

When dealing with polynomials, it is important to understand the essential concepts. Let's explore the fundamental operations of addition, subtraction, multiplication, and division, as well as factoring and simplifying, and the methods used to solve them.

When adding, subtracting, or dividing polynomial expressions, it is crucial to fill in any missing terms by assuming they have a coefficient of 0. This allows for proper alignment and calculation. You can use either the horizontal or vertical method to make the process easier. Both methods involve utilizing the distributive property, expanding brackets, and combining like terms. Alternatively, long division or synthetic division can be used to divide polynomials.

Multiplication

The two methods for multiplying polynomials are the horizontal and vertical methods.

## The Two Methods of Multiplying Polynomials: Horizontal and Vertical

In mathematics, there are two main methods for multiplying polynomials: the horizontal method and the vertical method. Both approaches result in the same answer, but they differ in the sequence and structure of their steps.

Horizontal Method

In the horizontal method, the first polynomial is multiplied by the second using the distributive property. The resulting expression is then expanded into brackets and simplified by combining like terms. This method is straightforward and easy to follow.

Vertical Method

The vertical method involves stacking the polynomials on top of each other and multiplying each term in the first polynomial by every term in the second polynomial, starting from the term with the smallest exponent and working up to the highest. Once all terms have been multiplied, like terms are aligned and combined, resulting in the final answer. This method may take longer, but it can be helpful for visual learners as it shows the step-by-step process.

Examples of Operations with Polynomials

• Addition: (2x^2 + x - 3) + (x^3 + 3x + 5) = x^3 + 2x^2 + 4x + 2
• Subtraction: (x^3 + 3x + 5) - (2x^2 + x - 3) = -x^3 + 2x^2 - 2x - 8
• Multiplication: (2x^2 + x - 3) * (x + 1) = 2x^3 + 3x^2 - 2x - 3

Word Problems

When solving word problems involving polynomials, it is important to follow a few steps. First, carefully examine the given information and identify the known and unknown values. Then, create corresponding polynomials for each value. Next, determine which basic operation is needed (addition, subtraction, multiplication, or division) and use the appropriate method to solve the problem. Lastly, simplify the resulting polynomial if necessary.

Factoring and Simplifying Polynomials

Factoring is a useful technique for simplifying polynomials. It involves rewriting a polynomial as the product of simpler terms. This can be helpful when simplifying fractional algebraic expressions, as common factors can be cancelled. Additionally, the factor theorem can be applied to speed up the factoring process, particularly with cubic polynomials.

With a solid understanding of the fundamental concepts and methods of operations with polynomials, you can confidently tackle problems and equations involving them. Just remember to pay attention to the details and complete all necessary steps for accurate results.