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# 7.1: Operations with Polynomials

Difficulty Level: At Grade Created by: CK-12

Objectives

The lesson objectives for Operations with Polynomials are:

• Addition and subtraction of polynomials
• Multiplication of Polynomials
• Special products [(x+y)(xy);(x+y)2\begin{align*}\left [ (x+y)(x-y); (x+y)^2 \right .\end{align*} and (xy)2]\begin{align*}\left . (x-y)^2 \right ]\end{align*}

## Addition and Subtraction of Polynomials

Introduction

In this concept you will begin your study of polynomials by learning how to add and subtract polynomials. The word polynomial actually comes from the Greek word poly meaning “many” and the Latin word binomium meaning “binomial”. A monomial can be a number or a variable (like x\begin{align*}x\end{align*}) or can be the product of a number and a variable. With monomials, the exponents are always in whole numbers. So 3x2\begin{align*}3x^2\end{align*} is a monomial but 3x12\begin{align*}3x^{\frac{1}{2}}\end{align*} is not. A binomial has two terms. You remember binomials from a previous chapter. They are the mathematical expressions that have the form (x+a)\begin{align*}(x+a)\end{align*} such as (x+1)\begin{align*}(x+1)\end{align*} or (2x+3)\begin{align*}(2x+3)\end{align*}. A trinomial has three terms. The expression 2x2+3x4\begin{align*}2x^2+3x-4\end{align*} has three terms and therefore is a trinomial. A polynomial, by definition, is also a monomial or the sum of a number of monomials. So 3x2\begin{align*}3x^2\end{align*} can be considered a polynomial, 2x+3\begin{align*}2x+3\end{align*} can be considered a polynomial, and 2x2+3x4\begin{align*}2x^2+3x-4\end{align*} can be considered a polynomial.

In this lesson, you are going to learn how to add and subtract polynomials. Adding and subtracting polynomials uses a similar strategy used in chapter Can You Make it True. Once you have mastered this concept, you will multiplication of polynomials using the distributive property. This, again, is a property you have learned in earlier chapters.

Finally, in this lesson, you will have a chance to explore some special polynomials. These polynomials will have the form (x+y)(xy),(x+y)2\begin{align*}(x+y)(x-y),(x+y)^2\end{align*} and (xy)2\begin{align*}(x-y)^2\end{align*}. You will have a chance to see how these polynomials differ from the ones you worked with in earlier lessons. Now let’s begin!

Watch This

Guidance

You are going to build a rectangular garden in your back yard. The garden is 2 m more than 1.5 times as long as it is wide. Write an expression to show the area of the garden.

AreaAreaArea=l×w=(1.5x+2)x=1.5x2+2x\begin{align*}Area &= l \times w \\ Area &= (1.5x + 2) x \\ Area &= 1.5x^2 + 2x\end{align*}

Example A

Find the sum of (3x2+2x7)+(5x23x+3)\begin{align*}(3x^2+2x-7)+(5x^2-3x+3)\end{align*}

Remember in earlier chapters when you used algebra tiles to solve for equations such as equation with variables on one side of the equation. Algebra tiles can also be used when adding and subtracting polynomials. Look at the example below.

First, let’s put out the algebra tiles that match your polynomials.

Second, let’s rearrange them so that you can combine like terms (sound familiar?).

Lastly, remove the tiles that make zero pairs. This will leave you with the final polynomial, or your answer.

What’s left?

(3x2+2x7)+(5x23x+3)=8x2x4\begin{align*}(3x^2+2x-7) + (5x^2-3x+3) = 8x^2-x-4\end{align*}

Example B

Find the difference of (5x2+8x+6)(4x2+5x+4)\begin{align*}(5x^2+8x+6)-(4x^2+5x+4)\end{align*}

For this one, try a different method. Let’s try the vertical method where you line up the polynomials vertically and then subtract them.

Example C

Find the sum of (3x2+6x27x+5)(4x2+3x8)\begin{align*}(3x^2+6x^2-7x+5)-(4x^2+3x-8)\end{align*}

For this one, try another method. Let’s try the horizontal method where you line up the polynomials horizontally and then add them.

(3x3+6x27x+5)+(4x2+3x8)=3x3+(6x2+4x2)+(7x+3x)+(58)=3x3+2x24x3\begin{align*}(3x^3+6x^2-7x+5)+(4x^2+3x-8)&=3x^3+(6x^2+{\color{blue}4x^2})+(-7x+{\color{blue}3x})+(5-{\color{blue}8}) \\ &=3x^3+2x^2-4x-3\end{align*}

Vocabulary

Binomial
A binomial has two terms that are added or subtracted from each other. The terms of a binomial is a variable (x)\begin{align*}(x)\end{align*}, a product of a number and a variable (4x)\begin{align*}(4x)\end{align*}, or the product of multiple variables with or without a number (4x2y+3)\begin{align*}(4x^2y + 3)\end{align*}. One of the terms in the binomial can be a number.
Monomial
A monomial can be a number or a variable (like x\begin{align*}x\end{align*}) or can be the product of a number and a variable (like 3x\begin{align*}3x\end{align*} or 3x2\begin{align*}3x^2\end{align*}). A monomial has only one term.
Polynomial
A polynomial, by definition, is also a monomial or the sum of a number of monomials. So 3x2\begin{align*}3x^2\end{align*} can be considered a polynomial, 2x+3\begin{align*}2x+3\end{align*} can be considered a polynomial, and 2x2+3x4\begin{align*}2x^2+3x-4\end{align*} can be considered a polynomial.
Trinomial
A trinomial has three terms (4x2+3x7)\begin{align*}(4x^2+3x-7)\end{align*}. The terms of a trinomial can be a variable (x)\begin{align*}(x)\end{align*}, a product of a number and a variable (3x)\begin{align*}(3x)\end{align*}, or the product of multiple variables with or without a number (4x2)\begin{align*}(4x^2)\end{align*}. One of the terms in the trinomial can be a number (7)\begin{align*}(-7)\end{align*}.
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often referred to as the literal coefficient.

Guided Practice

1. Use algebra tiles to add the polynomials (2x2+4x+3)+(x23x2)\begin{align*}(2x^2+4x+3) + (x^2-3x-2)\end{align*}.

2. Use the vertical method to subtract the polynomials (5x29x+7)(3x25x+6)\begin{align*}(5x^2-9x+7) - (3x^2-5x+6)\end{align*}.

3. Use the horizontal method to add the polynomial (8x3+5x24x+2)+(4x3+7x5)\begin{align*}(8x^3+5x^2-4x+2) + (4x^3+7x-5)\end{align*}.

1. (2x2+4x+3)+(x23x2)\begin{align*}(2x^2+4x+3) + (x^2-3x-2)\end{align*}

Step 1: Set up the algebra tiles

Step 2: rearrange the tiles to combine like terms

Step 3: Remove the zero pairs

Step 4: See what’s left!

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

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

\begin{align*}& \quad \ \ 5x^2 -9x +7 \quad \Longrightarrow \quad 5x^2-9x+7 \\ &-\underline{(3x^2-5x+6)} \qquad \quad \underline{-3x^2+5x-6} \\ & \qquad \qquad \qquad \qquad \qquad \quad \ 2x^2-4x+1\end{align*}

Therefore: \begin{align*}(5x^2-9x+7) - (3x^2-5x+6) = 2x^2-4x+1\end{align*}

3. \begin{align*}(8x^3+5x^2-4x+2) + (4x^3+7x-5)\end{align*}

\begin{align*}(8x^3+5x^2-4x+2) + (4x^3+7x-5) &= (8x^3+{\color{blue}4x^3}) + (5x^2) + (-4x+{\color{blue}7x}) + (2-{\color{blue}5}) \\ (8x^3+5x^2-4x+2) + ({\color{blue}4x^3}+{\color{blue}7x}-{\color{blue}5}) &= 12x^3 + 5x^2 + 3x -3 \end{align*}

Summary

When adding and subtracting polynomials, you can use algebra tiles to help visualize the problem or you could use the vertical or horizontal methods. To use algebra tiles, simply place the tiles down that match your polynomial and eliminate the pairs that add up to zero (called the zero pairs). Whatever is left over is the answer to your problem. To use the vertical method, simply line the polynomials up in a vertical column and add or subtract like terms. To use the horizontal method, simply line the polynomials up in a horizontal row and add or subtract like terms.

Problem Set

Use algebra tiles to solve the following problems.

1. \begin{align*}(x^2+4x+5) + (2x^2+3x+7)\end{align*}
2. \begin{align*}(2r^2+6r+7) + (3r^2+5r+8)\end{align*}
3. \begin{align*}(3t^2-2t+4) + (2t^2+5t-3)\end{align*}
4. \begin{align*}(4s^2-2s-3) + (5s^2+7s-6)\end{align*}
5. \begin{align*}(5y^2+7y-3) + (-2y^2-5y+6)\end{align*}

Use the vertical method to solve each of the following problems.

1. \begin{align*}(6x^2+36x+13) + (4x^2+13x+33)\end{align*}
2. \begin{align*}(12a^2+13a+7) + (9a^2+15a+8)\end{align*}
3. \begin{align*}(9y^2-17y-12) + (5y^2+12y+4)\end{align*}
4. \begin{align*}(11b^2+7b-12) - (15b^2-19b-21)\end{align*}
5. \begin{align*}(25x^2+17x-23) - (-14x^3-14x-11)\end{align*}

Use the horizontal method to solve each of the following problems.

1. \begin{align*}(-3y^2+10x-5) + (5y^2+5y+8)\end{align*}
2. \begin{align*}(-7x^2-5x+11) + (5x^2+4x-9)\end{align*}
3. \begin{align*}(9a^3-2a^2+7) + (3a^2+8a-4)\end{align*}
4. \begin{align*}(3x^2-2x+4) - (x^2+x-6)\end{align*}
5. \begin{align*}(4s^3+4s^2-5s-2) - (-2s^2-5s+6)\end{align*}

Using the algebra tiles...

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

Step 1: Set up the algebra tiles

Step 2: rearrange the tiles to combine like terms

Step 3: Remove the zero pairs (there are none)

Step 4: See what’s left!

\begin{align*}(x^2+4x+5) + (2x^2+3x+7) = 3x^2+7x+12\end{align*}

1. \begin{align*}(3t^2-2t+4) + (2t^2+5t-3)\end{align*}

Step 1: Set up the algebra tiles

Step 2: rearrange the tiles to combine like terms

Step 3: Remove the zero pairs

Step 4: See what’s left!

\begin{align*}(3t^2-2t+4) + (2t^2+5t-3) = 5t^2+3t+1\end{align*}

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

Step 1: Set up the algebra tiles

Step 2: Rearrange the tiles to combine like terms

Step 3: Remove the zero pairs

Step 4: See what’s left!

\begin{align*}(5y^2+7y-3) + (-2y^2-5y+3) = 3y^2+2y+3\end{align*}

Using the vertical method...

1. \begin{align*}(6x^2+36x+13) + (4x^2+13x+33)\end{align*}

\begin{align*}& \quad 6x^2 + 36x +13 \\ & \ \underline{+4x^2 + 13x +33} \\ & \ \ 10x^2 + 49x + 46\end{align*}

1. \begin{align*}(9y^2-17y-12) + (5y^2+12y+4)\end{align*}

\begin{align*}& \ \ 9y^2 - 17y - 12 \\ & \underline{+5y^2 + 12y + 4 \;} \\ & \ 14y^2 - 5y \ - 8\end{align*}

1. \begin{align*}(25x^2+17x-23) - (-14x^3-14x-11)\end{align*}

\begin{align*}& \qquad 25x^2 + 17x - 23 \quad \Longrightarrow \qquad \quad \ \ 25x^2 + 17x - 23 \\ &\underline{-(-14x^3 - 14x - 11)} \qquad \quad \underline{-14x^3 \qquad \ \ + 14x + 11} \\ & \qquad \qquad \qquad \qquad \qquad \qquad \ \underline{\; 14x^3 + 25x^2 + 31x -12}\end{align*}

Using the horizontal method...

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

\begin{align*} (-3y^2 + 10x - 5) + (5y^2 + 5y + 8) &= (-3y^2 + 5y^2) + (10x + 5y) + (-5 + 8)\\ &= 2y^2 + 15y + 3 \end{align*}

1. \begin{align*}(9a^3 - 2a^2 + 7) + (3a^2 + 8a -4)\end{align*}

\begin{align*}(9a^3 - 2a^2 + 7) + (3a^2 + 8a -4) &= (9a^3) + (-2a^2 + 3a^2) + (8a) + (7 - 4) \\ &= 9a^3 + a^2 + 8a + 3\end{align*}

1. \begin{align*}(4s^3 + 4s^2 - 5s - 2) - (-2s^2 - 5s + 6)\end{align*}

\begin{align*} (4s^3 + 4s^2 - 5s - 2) - (-2s^2 - 5s + 6) &= (4s^3 + 4s^2 - 5s -2) + (+2s^2 + 5s -6) \\ &= (4s^3) + (4s^2 + 2s^2) + (-5s + 5s) + (-2 -6) \\ &= 4s^3 + 6s^2 -8 \end{align*}

## Multiplication of Polynomials

Introduction

In this second concept of lesson Operations with Polynomials, you will learn how to multiply polynomials. When multiplying polynomials, you will use the distributive property. Remember that the distributive property states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. In other words if you were to multiply \begin{align*}3 \times(5x+2)\end{align*} you would get \begin{align*}15x+6\end{align*}. You have seen the distributive property before but in this lesson you will apply it to multiplying binomials to get larger polynomials or multiplying trinomials and binomials to get polynomials with degrees greater than 2.

Watch This

Guidance

Jack was asked to frame a picture. He was told that the ration of height to width of the frame was to be 5 inches longer than the glass width and 7 inches longer than the glass height. Jack measures the glass and finds the height to width ratio is 4:3. Write the expression to determine the area of the picture frame.

What is known?

The width is 5 inches longer than the glass

The height is 7 inches longer than the glass

The glass has a height to width ratio of 4:3

The equations:

The height of the picture frame is \begin{align*}4x + 7\end{align*}

The width of the picture frame is \begin{align*}3x + 5\end{align*}

The formula:

\begin{align*}\text{Area} &= w \times h \\ \text{Area} &= (3x + 5) (4x + 7) \\ \text{Area} &= 12x^2 + 21x + 20x + 35 \\ \text{Area} &= 12x^2 + 41x + 35\end{align*}

Example A

Use the distributive property to find the product of \begin{align*}(x+6)(x+5)\end{align*}

Remember to answer this question; you will use the distributive property. The distributive property would tell you to multiply \begin{align*}x\end{align*} in the first set of brackets by everything inside the second set of brackets, then multiply 6 in the first set of brackets by everything in the second set of brackets. Let's see what that looks like.

Example B

Use the distributive property to find the product of \begin{align*}(2x+5)(x-3)\end{align*}

To answer this question; you will also use the distributive property. The distributive property would tell you to multiply \begin{align*}2x\end{align*} in the first set of brackets by everything inside the second set of brackets, then multiply 5 in the first set of brackets by everything in the second set of brackets. Let's see what that looks like.

Example C

Use the distributive property to find the product of \begin{align*}(4x+3)(2x^2+3x-5)\end{align*}

At first, this question may seem different but you still use the distributive property to find the polynomial. The distributive property would tell you to multiply \begin{align*}4x\end{align*} in the first set of brackets by everything inside the second set of brackets, then multiply 3 in the first set of brackets by everything in the second set of brackets. Let's see what that looks like.

Vocabulary

Distributive Property
The distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, in the expression: \begin{align*}{\color{red}\frac{2}{3}} ({\color{blue}x + 5})\end{align*}, the distributive property states that the product of a number \begin{align*}\left({\color{red}\frac{2}{3}}\right)\end{align*} and a sum \begin{align*}({\color{blue}x+5})\end{align*} is equal to the sum of the individual products of the number \begin{align*}\left({\color{red}\frac{2}{3}}\right)\end{align*} and the addends \begin{align*}({\color{blue}x}\end{align*} and \begin{align*}{\color{blue}5})\end{align*}.
Like Terms
Like terms refer to terms in which the degrees match and the variables match. For example \begin{align*}3x\end{align*} and \begin{align*}4x\end{align*} are like terms.

Guided Practice

1. Use the distributive property to find the product of \begin{align*}(x+3)(x-6)\end{align*}.

2. Use the distributive property to find the product of \begin{align*}(2x+5)(3x^2-2x-7)\end{align*}.

3. An average football field has the dimensions of 160 ft by 360 ft. If the expressions to find these dimensions were \begin{align*}(3x+7)\end{align*} and \begin{align*}(7x+3),\end{align*} what value of \begin{align*}x\end{align*} would give the dimensions of the football field?

1. \begin{align*}(x+3)(x-6)\end{align*}

2. \begin{align*}(2x + 5)(3x^2 - 2x - 7)\end{align*}

3. \begin{align*}\text{Area} = l \times w\end{align*}

\begin{align*}\text{Area} &= 360 \times 160 \\ (7x+3) &= 360 \\ 7x &= 360 - 3 \\ 7x &= 357 \\ x &= 51 \\ \\ (3x +7) &= 160 \\ 3x &= 160 - 7 \\ 3x &= 153 \\ x &= 51 \end{align*}

Therefore the value of \begin{align*}x\end{align*} that satisfies these expressions is 51.

Summary

With multiplying binomials and polynomials to get even larger polynomials, you can use the distributive property. Remember that when using the distributive property, everything in the first set of brackets must first be multiplied by everything in the second set of brackets. After you have done this, you can simply combine like terms and then you have your final answer.

Problem Set

Use the distributive property to find the product of each of the following polynomials:

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

Use the distributive property to find the product of each of the following polynomials:

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

Use the distributive property to find the product of each of the following polynomials:

1. \begin{align*}9a^2(6a^3+3a+7)\end{align*}
2. \begin{align*}-4s^2(3s^3+7s^2+11)\end{align*}
3. \begin{align*}(x+5)(5x^3+2x^2+3x+9)\end{align*}
4. \begin{align*}(t-3)(6t^3+11t^2+22)\end{align*}
5. \begin{align*}(2g-5)(3g^3+9g^2+7g+12)\end{align*}

Using the distributive property...

1. \begin{align*}(x+4)(x+6)\end{align*}
1. \begin{align*}(x+7)(x-8)\end{align*}
1. \begin{align*}(x-4)(x-7)\end{align*}

Using the distributive property...

1. \begin{align*}(x+3)(x^2+x+5)\end{align*}
1. \begin{align*}(2x+5)(x^2-8x+3)\end{align*}
1. \begin{align*}(5x-4)(4x^2-8x+5)\end{align*}

Using the distributive property...

1. \begin{align*}9a^2(6a^3+3a+7)\end{align*}
1. \begin{align*}(x+5)(5x^3+2x^2+3x+9)\end{align*}
1. \begin{align*}(2g-5)(3g^3+9g^2+7g+12)\end{align*}

## Special Cases

Introduction

With multiplying binomials, there are special cases that when you learn to recognize them, it makes the multiplication faster and more efficient. In this concept you are going to learn the pattern that is involved with special cases of polynomials. Once you start to see this pattern in the math you are doing, your problem-solving will most likely become quicker.

There are three different special products that will be looked at in this concept. These include the following:

1. \begin{align*}(x+y)^2 = x^2 + 2xy + y^2\end{align*}
2. \begin{align*}(x-y)^2 = x^2 - 2xy + y^2\end{align*}
3. \begin{align*}(x+y)(x-y) = x^2 - y^2\end{align*}

The strategy to solve these special cases is the same as what we used for the previous concept. In other words, you again are going to use the distributive property. However, as you go through this concept, make sure you pay particular attention to the patterns that result when you multiply the two binomials.

Watch This

Guidance

A flower is homozygous blue (RR) and another flower is homozygous white (rr). Use a Punnett square to show that a mixture of the two can produce a white flower.

Each flower will have one-half of the blue genes and one-half of the white genes. Therefore the equation formed will be:

\begin{align*}0.5B + 0.5W\end{align*}

The offspring will have the genetic makeup (the mixture produced) using the equation:

\begin{align*}(0.5B + 0.5W)^2\end{align*}

Notice that this is one of the special products (special product numbered 1 above).

\begin{align*}(x+y)^2=x^2+2xy+y^2\end{align*}

So you can expand the offspring genetic makeup equation to find out the percentage of offspring (or flowers) that will be blue, white, or light blue.

Therefore 25% of the offspring flowers will be blue, 50% will be light blue, and 25% will be white.

Example A

Find the product: \begin{align*}(x+11)^2\end{align*}

This is one of the special products similar to the previous example. The first step is to expand the binomials before applying the distributive property.

Example B

Find the product: \begin{align*}(x-7)^2\end{align*}

This is a second example of the special products of polynomials. Again, the first step is to expand the binomials before applying the distributive property.

Example C

Find the product: \begin{align*}(x+9)(x-9)\end{align*}

This is the third type of the special products of polynomials. For this type, the binomials are already expanded and you just need to apply the distributive property.

Notice in this last example, when the binomials have the same numerical terms and opposite signs, that when you expand there is no middle term. In other words:

Special Product: \begin{align*}(x+y)(x-y)= x^2 - y^2\end{align*}

Example: \begin{align*}(x+9)(x-9)= x^2 - 81\end{align*}

Vocabulary

Distributive Property
The distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, in the expression: \begin{align*}{\color{red}3}({\color{blue}x + 5})\end{align*}, the distributive property states that the product of a number \begin{align*}({\color{red}3})\end{align*} and a sum \begin{align*}({\color{blue}x + 5})\end{align*} is equal to the sum of the individual products of the number \begin{align*}({\color{red}3})\end{align*} and the addends \begin{align*}({\color{blue}x}\end{align*} and \begin{align*}{\color{blue}5})\end{align*}.

Guided Practice

1. Expand the following binomial: \begin{align*}(x+4)^2\end{align*}.

2. Expand the following binomial: \begin{align*}(5x -3)^2\end{align*}.

3. Determine whether each of the following is a difference of two squares:

a) \begin{align*}a^2-16\end{align*}

b) \begin{align*}9b^2-49\end{align*}

c) \begin{align*}c^2-60\end{align*}

1. \begin{align*}(x+4)^2\end{align*}.

2. \begin{align*}(5x-3)^2\end{align*}

3. a) \begin{align*}&\qquad a^2 - 16 \\ &\quad \nearrow \qquad \ \nwarrow \\ & a^2 = a \times a \quad \text{Factors of} \ 16: \\ &\qquad \qquad \quad \ \ 1 \times 16 \\ &\qquad \qquad \quad \ \ 2 \times 8 \\ &\qquad \qquad \quad \ \ {\color{red}4 \times 4}\end{align*}

Therefore: \begin{align*}a^2 - 16 = (a+4)(a-4)\end{align*} Yes, \begin{align*}a^2 - 16\end{align*} is a difference of two squares.

b) \begin{align*}&\qquad 9b^2 - 49 \\ &\quad \nearrow \qquad \ \ \nwarrow \\ & b^2 = b \times b \quad \text{Factors of} \ 49: \\ & \quad \mathbf{and} \qquad \qquad 1 \times 49 \\ &\text{Factors of } 9: \quad {\color{red}7 \times 7}\\ & 1 \times 9\\ & {\color{red}3 \times 3}\end{align*}

Therefore: \begin{align*}9b^2 - 49 = (3b +7)(3b - 7)\end{align*}

Yes, \begin{align*}9b^2 - 49\end{align*} is a difference of two squares.

c) \begin{align*}&\qquad c^2 - 60 \\ &\quad \nearrow \qquad \ \ \nwarrow \\ & c^2 = c \times c \quad \text{Factors of} \ 60: \\ &\qquad \qquad \quad \ \ 1 \times 60 \\ &\qquad \qquad \quad \ \ 2 \times 30 \\ &\qquad \qquad \quad \ \ 3 \times 20 \\ &\qquad \qquad \quad \ \ 4 \times 15 \\ &\qquad \qquad \quad \ \ 5 \times 12 \\ &\qquad \qquad \quad \ \ 6 \times 10 \end{align*}

Since there are no factors of 60 that are matches (like \begin{align*}4 \times 4 = 16\end{align*} and \begin{align*}7 \times 7 = 49\end{align*}), the expression \begin{align*}c^2 - 60\end{align*} cannot be said to be a difference of two squares.

Summary

The previous three lessons began your study of polynomial expansion. Specifically you have had the opportunity to add, subtract and multiply polynomials using algebra tiles, the horizontal and vertical methods, and the distributive property. In this final lesson of the section, you were working with special cases of polynomials. There are three special cases that when learned help you to be able to find the product of binomials a little easier and quicker.

The three special cases involve squaring a binomial such as in \begin{align*}(x + y)^2\end{align*}, and \begin{align*}(x - y)^2\end{align*}, and one case involving the multiplying the sum and difference binomials \begin{align*}(x + y)(x - y)\end{align*}. The three special cases along with an example are shown for you below.

Special Case 1: \begin{align*}(x+y)^2 = x^2 + 2xy + y^2\end{align*}

Example: \begin{align*}(x+5)^2 = x^2 + 10x +25\end{align*}

Special Case 2: \begin{align*}(x-y)^2 = x^2 - 2xy + y^2\end{align*}

Example: \begin{align*}(2x-8)^2 = 4x^2 -32x + 64\end{align*}

Special Case 3: \begin{align*}(x+y)(x-y) = x^2 - y^2\end{align*}

Example: \begin{align*}(5x+10)(5x-10) =25x^2 - 100\end{align*}

Problem Set

Expand the following binomials:

1. \begin{align*}(t+12)^2\end{align*}
2. \begin{align*}(w+15)^2\end{align*}
3. \begin{align*}(2e+7)^2\end{align*}
4. \begin{align*}(3z+2)^2\end{align*}
5. \begin{align*}(7m+6)^2\end{align*}

Expand the following binomials:

1. \begin{align*}(g-6)^2\end{align*}
2. \begin{align*}(d-15)^2\end{align*}
3. \begin{align*}(4x-3)^2\end{align*}
4. \begin{align*}(2p-5)^2\end{align*}
5. \begin{align*}(6t-7)^2\end{align*}

Find the product of the following binomials:

1. \begin{align*}(x+13)(x-13)\end{align*}
2. \begin{align*}(x+6)(x-6)\end{align*}
3. \begin{align*}(2x+5)(2x-5)\end{align*}
4. \begin{align*}(3x+4)(3x-4)\end{align*}
5. \begin{align*}(6x+7)(6x-7)\end{align*}

Expand...

1. \begin{align*}(t+12)^2\end{align*}
1. \begin{align*}(2e+7)^2\end{align*}
1. \begin{align*}(7m+6)^2\end{align*}

Expand...

1. \begin{align*}(g-6)^2\end{align*}
1. \begin{align*}(4x-3)^2\end{align*}
1. \begin{align*}(6t-7)^2\end{align*}

Find the product...

1. \begin{align*}(x+13)(x-13)\end{align*}
1. \begin{align*}(2x+5)(2x-5)\end{align*}
1. \begin{align*}(6x+7)(6x-7)\end{align*}

## Summary

In this first lesson of chapter Many Terms Means Much Work you have been introduced to operations of polynomials. You have learned to add and subtract polynomials in the first concept. In the second concept, you took a look at multiplication of polynomials. You had the opportunity to use algebra tiles once again for expanding and then simplifying polynomials. As you had found out in previous chapters, algebra tiles are an effective tool for visualizing mathematical concepts. However, sometimes these tools can become cumbersome and inefficient when numbers become too large or the equations too complex. Alternate methods for adding, subtracting, and multiplying polynomials were shown, namely the vertical and horizontal methods. These two methods can be quite efficient to use when expanding and or simplifying polynomials.

You also used the distributive property. This property you have used in many chapters throughout this text and a very valuable method to learn in mathematics. For expanding polynomials, the distributive property is the key to the solution process. Using the distributive property and then combining like terms, the expansion of polynomials, including the special cases learned in the final concept of this lesson, the solution process becomes clear.

Remember as well that the three special cases \begin{align*}\{(x+y)^2,(x-y)^2,\end{align*} and \begin{align*}(x+y)(x-y)\}\end{align*} should be studied so that they can be easily and quickly identified. This will save you time (and energy) when solving polynomial problems as their products can be rapidly determined.

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