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# 12.2: Adding and Subtracting Polynomials

Difficulty Level: At Grade Created by: CK-12

## Introduction

The Pyramid Building

As the students rounded the corner on fifth street, they spotted a peculiar looking building. It was in the shape of a pyramid.

“Here we go again,” said Michael as he read the next problem to his friends.

“A pyramid-shaped building has rectangular floors that get increasingly smaller as you go higher up in the building. If the \begin{align*}87^{th}\end{align*} floor has a length of \begin{align*}6x+16\end{align*} and a width of 28, and each floor’s length and width decrease by 4 as you ascend, find the total area of the \begin{align*}87^{th}, 88^{th},\end{align*} and \begin{align*}89^{th}\end{align*} floor.”

This is the next dilemma.

To work on this problem, you will need to understand area as well as adding and subtracting polynomials. Pay close attention to the rules of this lesson and you will be able to figure out this problem by the end of the lesson.

What You Will Learn

In this lesson, you will learn the following skills.

• Subtract polynomials vertically.
• Subtract polynomials horizontally.
• Solve real – world problems by writing, simplifying, and evaluating polynomial expressions regarding rectangle areas.

Teaching Time

Now that you understand how to identify polynomials, we can work on adding and subtracting them. Let’s start with adding. To begin, let’s look at how we can add whole numbers with many digits.

Example

Add the numbers 5026 and 3210.

You might add it like this, right?

\begin{align*}5026 \\ \underline{+3210} \\ {8236}\end{align*}

If you think about it, you might notice that the same addition could be thought of in this way.

\begin{align*}& \qquad \qquad \qquad \text{thousands} \quad \qquad \text{hundreds} \quad \qquad \text{tens} \quad \qquad \text{ones} \\ & \quad 5026 \quad \rightarrow \qquad \ \ 5000 \qquad \qquad \qquad \qquad \qquad \ \ 20 \qquad \qquad 6\\ & \underline{+ \ 3210 \quad \rightarrow \quad \ + \ 3000 \qquad \qquad \quad \ \ 200 \qquad \ \ \quad10 \qquad \qquad \qquad}\\ & \quad 8236 \quad \leftarrow \qquad \ \ 8000 \quad \ + \qquad \quad 200 \quad + \quad \ 30 \quad \ + \quad \ 6 \end{align*}

Here we have shown 5026 to be \begin{align*}5000 + 20 + 6\end{align*}. The number 3210 is \begin{align*}3000 + 200 + 10\end{align*}.

Each of the similar places has been lined up vertically (one on top of the other) so that 3000 is beneath 5000 in the thousands place and 10 is beneath 20 in the tens place. Also, 200 is by itself because the first number had no digits in the hundreds place. Likewise, 6 is by itself because the second number had no digits in the ones place. Although this is not a practical way of writing a simple addition problem, it does demonstrate the technique we can use to add polynomials.

You learned in the previous lesson about like terms—terms are alike when they have exactly the same variable(s) to exactly the same power(s). When terms are alike, we can combine them by adding their coefficients.

Example: \begin{align*}5x^3+9x^3=14x^3\end{align*}

We also learned that polynomials are one or more terms in an expression.

Example of polynomials: \begin{align*}x^2+ 5 \qquad 3x-8+4x^5 \qquad -7a^2+9b-4b^3+6\end{align*}

Polynomials can be added in the same manner as we added 5026 and 3210.

Example

Add the polynomials \begin{align*}(7x^2+9x-5)\end{align*} and \begin{align*}(6x^2+3x+10)\end{align*}.

\begin{align*}& (7x^2+9x-5) \quad \rightarrow \quad 7x^2 \quad + \quad 9x \quad + \quad -5 \\ & \qquad \quad + \\ & \underline{(6x^2+3x+10) \ \ \rightarrow \quad 6x^2 \quad + \quad 3x \quad + \quad 10 \;} \\ & 13x^2+12x+5 \quad \leftarrow \ \ 13x^2 \ \ + \quad 12x \ \ + \quad \ 5\end{align*}

Each of the like terms was aligned vertically, one on top of the other. Notice that the negative sign on -5 was kept with the number 5. Be careful when you add the integers.

Example

Add the polynomials \begin{align*}(-2x^3+9x^2-3)\end{align*} and \begin{align*}(8x^2+5x-14)\end{align*}.

\begin{align*}& \quad (-2x^3+9x^2-3) \qquad \ \rightarrow \quad -2x^3 \quad + \quad 9x^2 \quad \ + \quad \quad \quad + \ \ -3\\ & \underline{\ +(8x^2+5x-14) \qquad \quad \rightarrow \ \qquad \qquad \qquad 8x^2 \ \ \ \ + \quad 5x \quad + \ -14 \ \ } \\ & -2x^3+17x^2+5x-17 \ \leftarrow \quad -2x^3 \quad + \quad 17x^2 \ \ + \quad 5x \quad + \ \ -17\end{align*}

Each of the terms was again aligned vertically. Notice this time that the space underneath \begin{align*}-2x^3\end{align*} is empty. That is because there was no like term in the second polynomial that could be combined with \begin{align*}-2x^3\end{align*}. There was a like term of \begin{align*}9x^2\end{align*}. The \begin{align*}8x^2\end{align*} is the like term that could be combined with \begin{align*}9x^2\end{align*} so it was placed beneath. Their sum was \begin{align*}17x^2\end{align*}. There was no like term of \begin{align*}5x\end{align*}, either. But -3 and -14, the constants, were like terms so were aligned together and combined. The result was \begin{align*}-2x^3+17x^2+5x-17\end{align*}. Unneeded addition signs are removed.

Be sure that you have the definitions for polynomials and like terms down in your notebooks.

A second method for adding polynomials is horizontally—in a single line. Just as you might add \begin{align*}6 + 19 = 25\end{align*} without placing them one on top of the other, so can a polynomial be added to another.

Example

\begin{align*}& =(7x^2+3x-11)+(3x^2-9x+5)\\ &=7x^2+3x-11+3x^2-9x+5 \\ &=10x^2-6x-6 \end{align*}

Step 1: rewrite without parenthesis

Step 2: combine like terms

In Step 1, the polynomial can be rewritten without parentheses because the parentheses serve only to show the separation of the polynomials. Notice, we do not align them vertically by like terms as before. However, we had to take care to recognize and combine like terms correctly.

This method can be a little trickier. If you find yourself getting confused, then go back and add the polynomials horizontally.

III. Subtract Polynomials Vertically

Just like we could add polynomials, we can subtract them too. We can perform this operation both vertically and horizontally. Let’s start with vertically.

When we subtract polynomials, we can use a similar procedure as with addition—we can subtract vertically. However, remember that subtraction is the same as “adding the opposite.” In other words, \begin{align*}5 - 8\end{align*} is the same as \begin{align*}5 + (-8)\end{align*}. We can add the opposite of 8 instead of subtracting 8. We will use the same idea with polynomials.

Remember that subtracting is the same as adding the opposite. Write this down in your notebook.

Example

\begin{align*}(9x^2+4x-7)-(2x^2+6x-4)\end{align*}

Set up the problem vertically by aligning the like terms.

\begin{align*}& \quad (9x^2+4x-7) \quad \rightarrow \ \ \quad 9x^2 \quad + \ \ \quad 4x \quad + \quad -7 \quad \text{Lines up like terms.} \\ & \underline{-(2x^2+6x-4) \quad \rightarrow \quad -2x^2 \quad + \quad -6x \quad + \ \quad 4 \quad \ \text{Add the opposite.} \quad \ \ } \\ & \quad 7x^2-2x-3 \quad \ \ \leftarrow \quad \ \ 7x^2 \quad + \ \quad -2x \ \ + \quad -3 \quad \text{Combine lilke terms.}\end{align*}

When you add the opposite, the sign changes on each of the terms in the subtracted polynomial. Inside the parentheses, the coefficient of \begin{align*}2x^2\end{align*} is positive. But when you add the opposite, the sign changes to negative, or \begin{align*}-2x^2\end{align*}. We also changed the sign on the \begin{align*}6x\end{align*} to \begin{align*}-6x\end{align*} and the -4 to 4.

Example

\begin{align*}(-7x^3+3x^2-x+4)-(-6x^2+9)\end{align*}

\begin{align*}& \quad (-7x^3+3x^2-x+4) \quad \rightarrow \ \ \quad -7x^3 \quad + \ \ \quad 3x^2 \quad + \quad -x \quad + \quad \quad 4 \quad \ \text{Lines up like terms.} \\ & \underline{\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; -(-6x^2+9) \quad \rightarrow \quad \quad \quad \quad \quad \quad \quad \ \ 6x^2 \quad + \ \quad \quad \quad \ + \quad -9 \ \ \ \text{Add the opposite.}\; \; \; \; \; \;} \\ & \quad -7x^3+9x^2-x-5 \quad \ \leftarrow \quad \quad -7x^3 \quad + \quad \ 9x^2 \quad + \quad -x \quad + \quad -5 \quad \ \text{Combine lilke terms.}\end{align*}

Now we can look at subtracting polynomials horizontally.

IV. Subtract Polynomials Horizontally

When we added polynomials, we used two methods—adding vertically and adding horizontally. You just learned to subtract polynomials vertically. As you have guessed, we can also subtract polynomials horizontally. First we will review the distributive property.

The distributive property: For all real numbers \begin{align*}a, b,\end{align*} and \begin{align*}c, \ a(b + c) = ab + ac\end{align*}.

Examples:

\begin{align*}5(3x+7)&=15x+35 \\ 4(2y-7)&=8y-28 \\ -2(9x+3)&=-18x-6 \\ -3 (-2y-4)&=6y+12\end{align*}

Remember to be careful with negative signs when using distributive property.

Now, let us remember that coefficients are the numerical factors of variables. The coefficient of \begin{align*}3x\end{align*} is 3. The coefficient of \begin{align*}9x^2\end{align*} is 9. When we see the term \begin{align*}-x\end{align*}, the coefficient is -1. Although you could write \begin{align*}-1x\end{align*}, we normally do not because the 1 is considered unnecessary. How does this relate to the distributive property? The negative sign could be in front of the parentheses, like this: \begin{align*}-(3x - 2)\end{align*}. This is similar to \begin{align*}-x\end{align*} where the coefficient is the unwritten -1. Just like you could write \begin{align*}-1x\end{align*}, you could also write \begin{align*}-1(3x - 2)\end{align*}. The distributive property is now more apparent in that each term will now be multiplied by -1.

Example

\begin{align*}& -(7x+5)=-1(7x+5)=-7x-5 \\ & -(x^2-3x+14)=-1(x^2-3x+14)=-x^2+3x-14 \\ & \qquad \qquad \quad \uparrow \end{align*}

Here you can insert the -1 and then multiply. As with adding the opposite, the sign changes on each of the terms in the polynomial.

We can now use this method to subtract polynomials horizontally. First we’ll distribute the negative sign to each of the terms in the subtracted polynomial and then we will combine like terms just as we did when we added polynomials.

Example

\begin{align*}& (5x+3)-(2x-8)\\ &=(5x+3)-1(2x-8) \\ &=5x+3-2x+8 \\ &=3x+11 \end{align*}

It may seem that way, but if you go step by step and remember that subtracting is adding the opposite, then you will be able to subtract polynomials vertically and horizontally.

V. Solve Real – World Problems by Writing, Simplifying and Evaluating Polynomial Expressions Regarding Rectangle Areas

As seen in previous lessons, polynomials can be applied to geometrical shapes. Keep in mind some basic geometrical ideas.

Area refers to the space inside an object or space. It is measured in square units. The area of land is generally measured with units such as acres, square miles, or square kilometers. The area formula of a rectangle is \begin{align*}A = lw\end{align*} or area equals length times width. Shapes that are not rectangular are often broken up into rectangular parts. Their total area is calculated by adding the areas of the separate parts.

Example

A concrete walkway surrounds a rectangular swimming pool. In order to know how much weather treatment to buy, the owner must know how many square units of concrete he has. The walkway is 5 feet wide on all sides. The swimming pool has a length of \begin{align*}7x\end{align*} and a width of 14 feet. How many square units concrete does he have?

In order to find the area of the concrete, we must find the area of the large rectangle and then subtract the area of the swimming pool.

The length of the large rectangle measures \begin{align*}7x + 5 + 5\end{align*}. Its width measures \begin{align*}14 + 5 + 5\end{align*}.

So its area will be \begin{align*}(7x + 5 + 5) \cdot (14 + 5 + 5)\end{align*}

The area of the swimming pool will be its length times its width or \begin{align*}7x \cdot 14\end{align*}.

Area of the swimming pool is \begin{align*}98x\end{align*}.

To find the area of the concrete, subtract the area of the swimming pool from the total area:

\begin{align*} &(168x + 240)-98x \\ &=70x + 240 \end{align*}

Now let’s go back to the problem from the introduction.

## Real-Life Example Completed

The Pyramid Building

Here is the problem from the introduction. Reread it and then figure out the areas of the other floors of the building.

As the students rounded the corner on fifth street, they spotted a peculiar looking building. It was in the shape of a pyramid.

“Here we go again,” said Michael as he read the next problem to his friends.

“A pyramid-shaped building has rectangular floors that get increasingly smaller as you go higher up in the building. If the \begin{align*}87^{th}\end{align*} floor has a length of \begin{align*}6x + 16\end{align*} and a width of 28, and each floor’s length and width decrease by 4 as you ascend, find the total area of the \begin{align*}87^{th}, 88^{th},\end{align*} and \begin{align*}89^{th}\end{align*} floor.”

This is the next dilemma.

Solution to Real – Life Example

Here is the work to figure out the area of the other three floors and the total area too.

\begin{align*}87^{th} \ \text{floor} \quad \text{Area} &= 28(6x + 16) = 168x + 448 \\ 88^{th} \ \text{floor} \quad \text{Area} &= (28 - 4)(6x + 16 - 4) = 24(6x + 12) = 144x + 288\\ 89^{th} \ \text{floor} \quad \text{Area} &= (28 - 4 - 4)(6x + 16 - 4 - 4) = 20(6x + 8) = 120x + 160 \\ \text{Total area} &= (168x + 448) + (144x + 288) + (120x + 160) \\ &=168x + 448 + 144x + 288 + 120x + 160 \\ &=432x + 896 \end{align*}

## Vocabulary

Here are the words that are found in this lesson.

Polynomial
one or more terms in an expression, often referred to specifically in situations where there are more than three terms.
Like Terms
terms that have the same variable and power.
Area
the space inside an object or area. It is measured in square units.

## Time to Practice

Directions: Add the following polynomials vertically. Be sure to align like terms.

1. \begin{align*}(4x^2+7x-2)+(3x-17)\end{align*}
2. \begin{align*}(-4x^4-x^3+8)+(-2x^3+5x+6)\end{align*}
3. \begin{align*}(10x^3-4x^2-2x+5)+(-x^2+9x-5)\end{align*}
4. \begin{align*}(6x^2+5x+9)+(4x^2+3x+6)\end{align*}
5. \begin{align*}(9x^2-3x+4)+(6x^2-9x+2)\end{align*}
6. \begin{align*}(3y^2+4x-9)+(-5y^2-6x+10)\end{align*}

Directions: Add the following polynomials horizontally.

1. \begin{align*}(-3x-8)+(15x+5)\end{align*}
2. \begin{align*}(x^4+7x^3-2x+7)+(-8x^3+9x^2-4)\end{align*}
3. \begin{align*}(4x^2y-3x^2y^2+7xy)+(9x^2y^2-5xy+3x^2)\end{align*}
4. \begin{align*}(5xy-3x+19)+ (4xy-9x-22)\end{align*}

Subtract the following polynomials vertically.

1. \begin{align*}(x^2+5x)-(3x-14)\end{align*}
2. \begin{align*}(7y^3+4y^2-3y-1)-(y^3+6y^2-4)\end{align*}

Subtract the following polynomials horizontally.

1. \begin{align*}(m^2+17m-11)-(3m^2+8m+12)\end{align*}
2. \begin{align*}(z^2+3z)-(3z^2+7z+16)-(4z-13)\end{align*}
3. Find the sum of the area of the two rectangles.
4. Divide this shape into rectangles. Then find its total area.
5. A wooden rectangular patio has a length of 27 and a width of \begin{align*}6x - 2\end{align*}. Cut into the patio are three equal rectangular planters of length \begin{align*}4x + 1\end{align*} and width 3. What is the total area of the wooden patio without the planters? Use a sketch to help you.
6. An ancient pyramid has a walkway about half-way up. The pyramid has four sides. On each side of the pyramid, the rectangular walkway is 10 feet wide and \begin{align*}40x + 8\end{align*} long. There is a second walkway closer to the top which is half as wide and half as long. What is the area of the long walkway?
7. What is the area of the short walkway?
8. What is the total area of the two walkways?

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