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# Addition and Subtraction of Polynomials

## Combining like terms in polynomial expressions

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Practice Addition and Subtraction of Polynomials
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What if you had two polynomials like 4x25\begin{align*}4x^2 - 5\end{align*} and 13x+2\begin{align*}13x + 2\end{align*}? How could you add and subtract them? After completing this Concept, you'll be able to perform addition and subtraction on polynomials like these.

### Try This

For more practice adding and subtracting polynomials, try playing the Battleship game at http://www.quia.com/ba/28820.html. (The problems get harder as you play; watch out for trick questions!)

### Guidance

To add two or more polynomials, write their sum and then simplify by combining like terms.

#### Example A

Add and simplify the resulting polynomials.

a) Add 3x24x+7\begin{align*}3x^2-4x+7\end{align*} and 2x34x26x+5\begin{align*}2x^3-4x^2-6x+5\end{align*}

b) Add x22xy+y2\begin{align*}x^2-2xy+y^2\end{align*} and 2y23x2\begin{align*}2y^2-3x^2\end{align*} and 10xy+y3\begin{align*}10xy+y^3\end{align*}

Solution

a)

Group like terms:Simplify:(3x24x+7)+(2x34x26x+5)=2x3+(3x24x2)+(4x6x)+(7+5)=2x3x210x+12

b)

Group like terms:Simplify:(x22xy+y2)+(2y23x2)+(10xy+y3)=(x23x2)+(y2+2y2)+(2xy+10xy)+y3=2x2+3y2+8xy+y3

To subtract one polynomial from another, add the opposite of each term of the polynomial you are subtracting.

#### Example B

a) Subtract x33x2+8x+12\begin{align*}x^3-3x^2+8x+12\end{align*} from 4x2+5x9\begin{align*}4x^2+5x-9\end{align*}

b) Subtract 5b22a2\begin{align*}5b^2-2a^2\end{align*} from 4a28ab9b2\begin{align*}4a^2-8ab-9b^2\end{align*}

Solution

a)

(4x2+5x9)(x33x2+8x+12)Group like terms:Simplify:=(4x2+5x9)+(x3+3x28x12)=x3+(4x2+3x2)+(5x8x)+(912)=x3+7x23x21

b)

(4a28ab9b2)(5b22a2)Group like terms:Simplify:=(4a28ab9b2)+(5b2+2a2)=(4a2+2a2)+(9b25b2)8ab=6a214b28ab

Note: An easy way to check your work after adding or subtracting polynomials is to substitute a convenient value in for the variable, and check that your answer and the problem both give the same value. For example, in part (b) above, if we let a=2\begin{align*}a=2\end{align*} and b=3\begin{align*}b=3\end{align*}, then we can check as follows:

Given(4a28ab9b2)(5b22a2)(4(2)28(2)(3)9(3)2)(5(3)22(2)2)(4(4)8(2)(3)9(9))(5(9)2(4))(113)37150Solution6a214b28ab6(2)214(3)28(2)(3)6(4)14(9)8(2)(3)2412648150

Since both expressions evaluate to the same number when we substitute in arbitrary values for the variables, we can be reasonably sure that our answer is correct.

Note: When you use this method, do not choose 0 or 1 for checking since these can lead to common problems.

Problem Solving Using Addition or Subtraction of Polynomials

One way we can use polynomials is to find the area of a geometric figure.

#### Example C

Write a polynomial that represents the area of each figure shown.

a)

b)

c)

d)

Solution

a) This shape is formed by two squares and two rectangles.

The blue square has area y×yThe yellow square has area x×xThe pink rectangles each have area x×y=y2.=x2.=xy.

To find the total area of the figure we add all the separate areas:

Total area=y2+x2+xy+xy=y2+x2+2xy

b) This shape is formed by two squares and one rectangle.

The yellow squares each have area a×aThe orange rectangle has area 2a×b=a2.=2ab.

To find the total area of the figure we add all the separate areas:

Total area=a2+a2+2ab=2a2+2ab

c) To find the area of the green region we find the area of the big square and subtract the area of the little square.

The big square has area:y×yThe little square has area:x×xArea of the green region=y2.=x2.=y2x2

d) To find the area of the figure we can find the area of the big rectangle and add the areas of the pink squares.

The pink squares each have area a×aThe blue rectangle has area 3a×a=a2.=3a2.

To find the total area of the figure we add all the separate areas:

Another way to find this area is to find the area of the big square and subtract the areas of the three yellow squares:

To find the total area of the figure we subtract:

Watch this video for help with the Examples above.

### Vocabulary

• A polynomial is an expression made with constants, variables, and positive integer exponents of the variables.
• In a polynomial, the number appearing in each term in front of the variables is called the coefficient.
• In a polynomial, the number appearing all by itself without a variable is called the constant.
• Like terms are terms in the polynomial that have the same variable(s) with the same exponents, but they can have different coefficients.

### Guided Practice

Subtract \begin{align*} 4t^2+7t^3-3t-5\end{align*} from \begin{align*}6t+3-5t^3+9t^2\end{align*}.

Solution:

When subtracting polynomials, we have to remember to subtract each term. If the term is already negative, subtracting a negative term is the same thing as adding:

The final answer is in standard form.

### Explore More

1. \begin{align*}(x+8)+(-3x-5)\end{align*}
2. \begin{align*}(-2x^2+4x-12)+(7x+x^2)\end{align*}
3. \begin{align*}(2a^2b-2a+9)+(5a^2b-4b+5)\end{align*}
4. \begin{align*}(6.9a^2-2.3b^2+2ab)+(3.1a-2.5b^2+b)\end{align*}
5. \begin{align*}\left ( \frac{3}{5}x^2-\frac{1}{4}x+4 \right )+ \left ( \frac{1}{10}x^2 + \frac{1}{2}x-2\frac{1}{5} \right )\end{align*}

Subtract and simplify.

1. \begin{align*}(-t+5t^2)-(5t^2+2t-9)\end{align*}
2. \begin{align*}(-y^2+4y-5)-(5y^2+2y+7)\end{align*}
3. \begin{align*}(-5m^2-m)-(3m^2+4m-5)\end{align*}
4. \begin{align*}(2a^2b-3ab^2+5a^2b^2)-(2a^2b^2+4a^2b-5b^2)\end{align*}
5. \begin{align*}(3.5x^2y-6xy+4x)-(1.2x^2y-xy+2y-3)\end{align*}

Find the area of the following figures.

### Vocabulary Language: English

distributive property

distributive property

The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
Polynomial

Polynomial

A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.