Have you ever seen a building shaped like a pyramid? Take a look at this dilemma.

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*}

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 Concept and you will be able to figure out this problem by the end of it.**

### Guidance

**A** *polynomial***is an algebraic expression that shows the sum of** *monomials***.**

Since the prefix *mono* means one, a monomial is a single piece or *term*. The prefix *poly* means many. So the word *polynomial* refers to one or more than one term in an expression. The relationship between these terms may be sums or difference.

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

**We call an expression with a single term a** *monomial***, an expression with two terms is a** *binomial***, and an expression with three terms is a** *trinomial***. An expression with more than three terms is named simply by its number of terms—“five-term polynomial.”**

Now we are going to add polynomials, but first let's review how to add whole numbers with many digits.

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*} 5000+20+6. The number 3210 is \begin{align*}3000 + 200 + 10\end{align*}3000+200+10.**

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.

**Do you remember how to identify like terms?**

**Like Terms** have exactly the same variable(s) to exactly the same power(s). **When terms are alike, we can combine them by adding their coefficients.**

**\begin{align*}5x^3+9x^3=14x^3\end{align*} 5x3+9x3=14x3**

We also learned that *polynomials***are defined as having one or more terms in an expression**.

**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.**

Take a look.

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

A second method for adding polynomials is horizontally—in a single line. Just as you might add \begin{align*}6 + 19 = 25\end{align*}

\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 vertically.**

Add the following polynomials.

#### Example A

\begin{align*}(4x^2+7x-2)+(3x^2+2x-1)\end{align*}

**Solution: \begin{align*}7x^2+9x-3\end{align*} 7x2+9x−3**

#### Example B

\begin{align*}(-4x^2+7x-2)+(-7x^2+3x-17)\end{align*}

**Solution: \begin{align*}-11x^2+10x-19\end{align*} −11x2+10x−19**

#### Example C

\begin{align*}(4xy+7x-2)+(-19xy-17x-9)\end{align*}

**Solution: \begin{align*}-15xy-10x-11\end{align*} −15xy−10x−11**

Now let's go back to the dilemma from the beginning of the Concept.

**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

- 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.

### Guided Practice

Here is one for you to try on your own.

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

**Solution**

\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*}.**

### Video Review

### Practice

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

- \begin{align*}(4x^2+7x-2)+(3x-17)\end{align*}
- \begin{align*}(-4x^4-x^3+8)+(-2x^3+5x+6)\end{align*}
- \begin{align*}(10x^3-4x^2-2x+5)+(-x^2+9x-5)\end{align*}
- \begin{align*}(6x^2+5x+9)+(4x^2+3x+6)\end{align*}
- \begin{align*}(9x^2-3x+4)+(6x^2-9x+2)\end{align*}
- \begin{align*}(3y^2+4x-9)+(-5y^2-6x+10)\end{align*}
- \begin{align*}(14x^2+6x-2)+(9x-1)\end{align*}
- \begin{align*}(-2x^2+7x-2)+(-3x^2-17)\end{align*}
- \begin{align*}(9x^2+7x-2y)+(3x^2-x+9y)\end{align*}
- \begin{align*}(4xy+7x-21)+(-12xy+4x-8)\end{align*}
- \begin{align*}(11x^2+9x-2y)+(3x^2-8x-5y-2)\end{align*}

Directions: Add the following polynomials horizontally.

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