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# Subtraction of Polynomials

## Subtract polynomials by combining like terms

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Subtracting Polynomials

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 7x\begin{align*}7x\end{align*} and a width of 14 feet. How many square units of concrete does he have?

In this concept, you will learn to subtract polynomials.

### Subtracting Polynomials

A polynomial is an algebraic expression that shows the sum of monomials. Just like you can add polynomials, you can subtract them too. You can perform this operation both vertically and horizontally. Let’s start with vertically.

One point to remember is 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*}. You can add the opposite of 8 instead of subtracting 8. You will use the same idea with polynomials.

Let’s look at an example.

Subtract the polynomials \begin{align*}(9x^2 + 4x - 7)\end{align*}  and  \begin{align*}(2x^2 + 6x - 4)\end{align*} .

First, line up the like terms so that you can subtract them vertically.

\begin{align*}\begin{array}{rcl} && \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 \ \underline{\text{Add the opposite} \quad \ \ } \\ && \quad 7x^2-2x-3 \quad \ \ \leftarrow \quad \ \ 7x^2 \quad + \ \quad -2x + \ -3 \quad \ \ \ \text{Combine lilke terms} \end{array}\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*}. You also changed the sign on the \begin{align*}6x\end{align*} to \begin{align*}-6x\end{align*} and the -4 to 4.

Now you can look at subtracting polynomials horizontally.

You just learned to subtract polynomials vertically. You can also subtract polynomials horizontally. First you will need to review the distributive property.

\begin{align*}x^2 - 8\end{align*}

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

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

Remember be careful with negative signs when using distributive property.

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 you see the term \begin{align*}-x\end{align*}, the coefficient is -1. Although you could write \begin{align*}-1x\end{align*}, you 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 unwritten but understood to be -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 given that each term will now be multiplied by -1.

Take a look at the following applications of the distributive property.

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

Here the -1 has been inserted and then the multiplication has been performed. As with adding the opposite, the sign changes on each of the terms in the polynomial.

You can now use this method to subtract polynomials horizontally. First you will distribute the negative sign to each of the terms in the subtracted polynomial and then you will combine like terms.

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

First, rewrite the polynomials without parentheses. Remember that subtracting is adding the opposite.

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

Next, combine like terms.

\begin{align*}\begin{array}{rcl} (5x + 3) - (2x - 8) &=& {\color{red}5x} + 3 - {\color{red}2x} + 8 \\ &=& {\color{red}3x} + 11 \end{array}\end{align*}

The answer is \begin{align*}3x+11\end{align*}.

### Examples

#### Example 1

Earlier, you were given a problem about the pool and the concrete.

First, find the area of the pool plus the walkway. The length of the large rectangle measures \begin{align*}7x + 5 + 5\end{align*}. Its width measures \begin{align*}14 + 5 + 5\end{align*}.

\begin{align*}\begin{array}{rcl} A &=& l \times w \\ A &=& (7x + 5 + 5) \times (14 + 5 + 5) \\ A &=& (7x + 10) \times (24) \\ A &=& 168x + 240 \end{array} \end{align*}

Next, find the area of just the swimming pool.

\begin{align*}\begin{array}{rcl} A &=& l \times w \\ A &=& (7x) \times(14) \\ A &=& 98x \end{array} \end{align*}

Then subtract the area of the pool from the area of the entire swimming area to find the area of the concrete walkway.

\begin{align*}\begin{array}{rcl} A_{\text{Concrete}} & = & A_ {\text{swimming area}} - A_{\text {swimming pool}} \\ A_{\text{Concrete}} & = & 168x + 240 - 98x\\ A_{\text{Concrete}} & = & 70x + 240 \end{array}\end{align*}

The answer is \begin{align*}70x + 240\end{align*}.

He needs \begin{align*}70x + 240\end{align*} square feet of concrete.

#### Example 2

Subtract the polynomials \begin{align*} (-7x^3 + 3x^2 - x + 4)\end{align*}  and \begin{align*} (-6x^2 + 9)\end{align*} .

First, line up the like terms so that you can subtract them vertically.

\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 \ \\ & \underline{\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; -(-6x^2+9) \quad \rightarrow \quad \quad \quad \quad \quad \quad \quad \ \ 6x^2 \qquad \qquad \qquad \qquad -9 \ \; \; \; \; \; \;} \end{align*}

Next, combine like terms.

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

The answer is \begin{align*}-7x^3 + 9x^2 - x -5\end{align*} .

#### Example 3

Subtract the polynomials: \begin{align*}(8x^2 + 4x- 7) - (2x^2 + 9x + 3)\end{align*}.

First, rewrite the polynomials without parentheses. Remember that subtracting is adding the opposite.

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

Next, combine like terms.

\begin{align*}\begin{array}{rcl} (8x^2 + 4x - 7) - (2x^2 + 9x + 3) &=& {\color{blue}8x^2} + {\color{red}4x} -7 {\color{blue} - 2x^2} {\color{red} - 9x} - 3 \\ &=& 6{\color{blue}x^2} - {\color{red}5x} - 10 \end{array}\end{align*}

The answer is \begin{align*}6x^2 - 5x - 10\end{align*}.

#### Example 4

Subtract the polynomials: \begin{align*}(10xy + 4x- 7) - (3x- 4)\end{align*}.

First, rewrite the polynomials without parentheses. Remember that subtracting is adding the opposite.

\begin{align*}(10xy + 4x - 7) - (3x - 4) = 10xy + 4x - 7 - 3x + 4\end{align*}

Next, combine like terms.

\begin{align*}\begin{array}{rcl} (10xy + 4x - 7) - (3x - 4) &=& 10xy + {\color{red}4x} {\color{blue}- 7} {\color{red}- 3x} + {\color{blue}4} \\ &=& 10xy + {\color{red}x} {\color{blue} -3} \end{array}\end{align*}

The answer is \begin{align*}10xy + x - 3\end{align*}.

#### Example 5

Subtract the polynomials: \begin{align*}(14x^2 + 8x- 7y + 1) - (2x^2 + 2x- 4y + 2)\end{align*}.

First, rewrite the polynomials without parentheses. Remember that subtracting is adding the opposite.

\begin{align*}(14x^2 + 8x - 7y + 1) - (2x^2 + 2x -4y + 2) = 14x^2 + 8x - 7y + 1 - 2x^2 -2x + 4y - 2\end{align*}

Next, combine like terms.

\begin{align*}\begin{array}{rcl} (14x^2 + 8x - 7y + 1) - (2x^2 + 2x -4y + 2) &=& {\color{purple}14x^2} + {\color{red}8x} {\color{blue}- 7y} + 1 {\color{purple}- 2x^2} {\color{red} -2x} + {\color{blue}4y} - 2 \\ &=& {\color{purple}12x^2} + {\color{red}6x} {\color{blue}-3y} - 1 \end{array}\end{align*}

The answer is \begin{align*}12x^2 + 6x - 3y - 1\end{align*} .

### Review

Subtract the following polynomials vertically.

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

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

3. \begin{align*}(5xy + 5x + 3) - (12xy - 4x - 8)\end{align*}

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

5. \begin{align*}(8x + 5y + 1) - (9x + 2y + 5)\end{align*}

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

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

8. \begin{align*}(18x^3 + 2x^2 + 8x + 2) - (3x^2 - 4x - 9)\end{align*}

9. \begin{align*}(8x + 9y - 20) - (3x - 14)\end{align*}

10. \begin{align*}(16x^2 + 5x - 3y + 7) - (3x - 14y + 10)\end{align*}

11. \begin{align*}(18x^2 + 5xy - 6x + 21) - (3x^2 - 14xy - 9x + 1)\end{align*}

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

Subtract the following polynomials horizontally.

13. \begin{align*}(m^2 + 17m - 11) - (3m^2 + 8m + 12)\end{align*}

14. \begin{align*}(z^2 + 3z) - (3z^2 + 7z + 16) - (4z - 13)\end{align*}

15. \begin{align*}(5x^2 + 3xy) - (3x^2 + 7xy + 16) - (4xy - 13)\end{align*}

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### Vocabulary Language: English

Area

Area is the space within the perimeter of a two-dimensional figure.

like terms

Terms are considered like terms if they are composed of the same variables with the same exponents on each variable.

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.