### Addition and Subtraction of Polynomials

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

#### Adding Polynomials

Add and simplify the resulting polynomials.

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

\begin{align*}& (3x^2-4x+7)+(2x^3-4x^2-6x+5)\\ \text{Group like terms:} & = 2x^3+(3x^2-4x^2)+(-4x-6x)+(7+5)\\ \text{Simplify:} & = 2x^3-x^2-10x+12\end{align*}

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

\begin{align*}& (x^2-2xy+y^2)+(2y^2-3x^2)+(10xy+y^3)\\
\text{Group like terms:} & = (x^2-3x^2)+(y^2+2y^2)+(-2xy+10xy)+y^3\\
\text{Simplify:} & = -2x^2+3y^2+8xy+y^3\end{align*}

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

#### Subtracting Polynomials

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

\begin{align*}(4x^2+5x-9)-(x^3-3x^2+8x+12) & = (4x^2+5x-9)+(-x^3+3x^2-8x-12)\\ \text{Group like terms:} & = -x^3+(4x^2+3x^2)+(5x-8x)+(-9-12)\\ \text{Simplify:} & = -x^3+7x^2-3x-21\end{align*}

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

\begin{align*}(4a^2-8ab-9b^2)-(5b^2-2a^2) & = (4a^2-8ab-9b^2)+(-5b^2+2a^2)\\
\text{Group like terms:} & = (4a^2+2a^2)+(-9b^2-5b^2)-8ab\\
\text{Simplify:} & = 6a^2-14b^2-8ab\end{align*}

**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 \begin{align*}a=2\end{align*} and \begin{align*}b=3\end{align*}, then we can check as follows:

\begin{align*}& \text{Given} && \text{Solution}\\ & (4a^2-8ab-9b^2)-(5b^2-2a^2) && 6a^2-14b^2-8ab\\ & (4(2)^2-8(2)(3)-9(3)^2)-(5(3)^2-2(2)^2) && 6(2)^2-14(3)^2-8(2)(3)\\ & (4(4)-8(2)(3)-9(9))-(5(9)-2(4)) && 6(4)-14(9)-8(2)(3)\\ & (-113)-37 && 24-126-48\\ & -150 && -150\end{align*}

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.

#### Writing a Polynomial

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

a)

This shape is formed by two squares and two rectangles.

\begin{align*}\text{The blue square has area} \ y \times y & = y^2.\\ \text{The yellow square has area} \ x \times x & = x^2.\\ \text{The pink rectangles each have area} \ x \times y & = xy.\end{align*}

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

\begin{align*}Total \ area &= y^2 + x^2 + xy + xy\\ & = y^2 + x^2 + 2xy\end{align*}

b)

This shape is formed by two squares and one rectangle.

\begin{align*}\text{The yellow squares each have area} \ a \times a & = a^2.\\ \text{The orange rectangle has area} \ 2a \times b & = 2ab.\end{align*}

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

\begin{align*}Total \ area & = a^2 + a^2 + 2ab\\ & = 2a^2 + 2ab\end{align*}

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.

\begin{align*}\text{The big square has area}: y \times y & = y^2.\\ \text{The little square has area}: x \times x & = x^2.\\ Area \ of \ the \ green \ region & = y^2 - x^2\end{align*}

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.

\begin{align*}\text{The pink squares each have area} \ a \times a & = a^2.\\ \text{The blue rectangle has area} \ 3a \times a & = 3a^2.\end{align*}

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

\begin{align*}Total \ area = a^2 + a^2 + a^2 + 3a^2 = 6a^2\end{align*}

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

\begin{align*}\text{The big square has area} \ 3a \times 3a & = 9a^2.\\ \text{The yellow squares each have area} \ a \times a & = a^2.\end{align*}

To find the total area of the figure we subtract:

\begin{align*}Area & = 9a^2 - (a^2 + a^2 + a^2)\\ & = 9a^2 - 3a^2 \\ & = 6a^2 \end{align*}

### Example

#### Example 1

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

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:

\begin{align*}6t+3-5t^3+9t^2-(4t^2+7t^3-3t-5)&= \\ 6t+3-5t^3+9t^2-(4t^2)-(7t^3)-(-3t)-(-5)&=\\ 6t+3-5t^3+9t^2-4t^2-7t^3+3t+5&=\\ (6t+3t)+(3+5)+(-5t^3-7t^3)+(9t^2-4t^2)&=\\ 9t+8-12t^3+5t^2&=\\ -12t^3+5t^2+9t+8\\ \end{align*}

The final answer is in standard form.

### Review

Add and simplify.

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

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

Find the area of the following figures.

### Review (Answers)

To view the Review answers, open this PDF file and look for section 9.2.