Polynomial Calculations: How To Add, Subtract, Multiply And Divide Polynomials

Polynomials are a set of terms that consist of a constant and a variable raised to the power of any non-negative integer, connected through operators (e.g., addition, subtraction). We are all familiar with using operators and how they work. All we now need to do is to learn how to perform polynomial calculations.

Like and unlike terms

When the variables of two or more terms are the same (have the same power, have the same alphabet denoting the variable), then those terms are called like terms. Unlike terms have different variables or have the same variable, but are raised to a different power.

You can add, subtract, multiply, and divide like terms into a one-variable answer. However, you cannot add or subtract unlike terms into a simple form with just one variable.

With a know-how of like and unlike terms, we can move forward and start with how the four basic operators are used in polynomial calculation.

Polynomial addition and subtraction

We can only add or subtract like terms in a polynomial. For example, let’s find the sum of the following polynomials:

3x3+4x2-12x+5

18x3+7x+1

Step 1: Group the like terms together. Here 3x3 and 18x3, 12x and 7x, and 5 and 1 are like terms

(3x3+18x3)+(-12x+7x)+(5+1)+4x2

Step 2: Add the like terms

21x3-5x+6+4x2

Step 3: Arrange in decreasing order of degree

21x3+4x2-5x+6. This is our answer.

Polynomial multiplication

Polynomial multiplication is also known as expanding. We can multiply both like and unlike polynomials. It follows the distributive property of multiplication, which looks like this:

(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd=(a+b)(c+d)=(a+b)c+(a+b)d

Here, a, b, c, and d represent the terms of a polynomial.

However, there is a rule that you need to remember. Whenever you multiply two variables with a power, the product is the variable raised to the sum of the powers.

For example, if we multiply x2 with x5, we get x(2+5), which is x7

With our concepts in place, let’s multiply (x5 + 10x3) and (5x2-4x)

Step 1: Arrange the terms in the form: a(c+d) + b(c +d) or (a+b)c + (a+b)d

x5 (5x2– 4x) + 10x3(5x2– 4x)

Step 2: Multiply the terms, keeping in mind that the powers are added during multiplication

x5.5x2 – x5.4x + 10x3.5x2 – 10x3.4x

= 5x7 – 4x6  + 50x5 – 40x4 .

Let’s look at another example:

Expand: (4x4y + 3xy)(y3 + xy)

Step 1: Arrange the terms according to the distributive property of multiplication

4x4y(y3 + xy) + 3xy(y3 + xy)

Step 2: Multiply the terms

4x4y4 + 4x5y2+ 3xy4 + 3x2y2.

Polynomial division

Dividing a polynomial with another polynomial is nothing different from normal division. If you have mastered multiplying polynomials, you can use those principles to divide polynomials as well.

For example, if we divide 4x4+3x3-11x2 -x +2 by (x+2)

Step 1: First, the leading term of the dividend, here 4x4, is divided with the leading term of the binomial divisor, here x. This is the leading term of the quotient.

Polynomials 1

Step 2: Next, we multiply the divisor with the leading term of the quotient and subtract the product from the dividend.

Polynomials 2

Step 3: The next number of the polynomial is dropped down to form a new polynomial

Polynomials 3

Step 4: Repeat steps 1 and 2. This time, you’ll divide the leading term -5x2 by x, the leading term of the divisor. This becomes the second term of the quotient. Multiply the divisor with this term and subtract.

Polynomials 4

Step 5: This is the complete solution.

Polynomials 5

(Follow this format while making the illustration)

Polynomials are used in equations to solve for variables. Knowing how to operate polynomials is very important to progress in solving polynomial equations.

  1. (3x2 + 2x) + (5x2 – x + 1)

Solution: Combine like terms: (3x2 + 5x2) + (2x – x) + 1 = 8x2 + x + 1

  1. (4x3 – 2x2 + 5x) – (x3 – 3x2 + 7x)

Solution: Distribute the negative sign and combine like terms: 4x3 – 2x2 + 5x – x3 + 3x2 – 7x = (4x3 – x3) + (-2x2 + 3x2) + (5x – 7x) = 3x3 + x2 – 2x

  1. (2x + 3) × (x2 – 1)

Solution: Use the distributive property: (2x × x2) + (2x × -1) + (3 × x2) + (3 × -1) = 2x3 – 2x + 3x2 – 3

(x2 + 3x + 2) / (x + 1)

Solution: x+2

1. Adding flowerbeds

A rectangular flowerbed has a length of 2x meters and a width of 3x + 1 meters. Next to it, there’s a square flowerbed with a side length of x meters. Find the total area of both flowerbeds.

Solution:

First, find the area of the rectangular flowerbed: area = length x width = (2x) × (3x + 1) = 6x2 + 2x

Then, find the area of the square flowerbed: area = side x side = (x) × (x) = x^2

Now, add the areas of both flowerbeds: total area = (6x2 + 2x) + x2 = 7x2 + 2x

Therefore, the total area of both flowerbeds is 7x2 + 2x square meters.

2. Building a fence

A rectangular garden has a length of (x + 2) meters and a width of (x – 1) meters. You want to build a fence around the entire garden. If each meter of fencing costs 5 dollars, how much will the entire fence cost?

Solution:

The perimeter of the garden is the total length of the fence needed. To find the perimeter, add the lengths of all sides: perimeter = length + width + length + width

Substitute the expressions for length and width: perimeter = (x + 2) + (x – 1) + (x + 2) + (x – 1)

Combine like terms: perimeter = 4x + 2

The total cost of the fence is the product of the perimeter and the cost per meter: cost = perimeter x price/meter = (4x + 2) x $5

Therefore, the total cost of the fence is 20x + 10 dollars.

3. Filling a pool

A rectangular pool has a length of 3x2 meters and a width of 2x meters. It needs to be filled with water. If a water truck can deliver 6x3 liters of water at once, how many truckloads are needed to fill the pool completely?

Solution:

First, find the area of the pool, which represents the volume of water needed: area (volume) = length x width = (3x2) × (2x) = 6x3 square meters (Since volume is a 3D measurement, we can keep the square meters unit for now)

Then, divide the total volume needed by the volume delivered per truckload: number of truckloads = total volume/volume per truckload = 6x3 liters / 6x3 liters/truckload

Since we’re dividing by the same variable with the same exponent, the terms cancel out, leaving us with: the number of truckloads = 1

Therefore, only 1 truckload is needed to fill the pool completely.

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