# 2.4: Adding and Subtracting Polynomial Expressions

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$

In this section we concentrate on adding and subtracting polynomial expressions, based on earlier work combining like terms in Ascending and Descending Powers. Let’s begin with an addition example.

##### Example $$\PageIndex{1}$$

Simplify: $\left(a^{2}+3 a b-b^{2}\right)+\left(4 a^{2}+11 a b-9 b^{2}\right) \nonumber$

Solution

Use the commutative and associative properties to change the order and regroup. Then combine like terms.

\begin{aligned}\left(a^{2}+3 a b-b^{2}\right) &+\left(4 a^{2}+11 a b-9 b^{2}\right) \\ &=\left(a^{2}+4 a^{2}\right)+(3 a b+11 a b)+\left(-b^{2}-9 b^{2}\right) \\ &=5 a^{2}+14 a b-10 b^{2} \end{aligned} \nonumber

##### You Try $$\PageIndex{1}$$

Simplify: $$\left(3 s^{2}-2 s t+4 t^{2}\right)+\left(s^{2}+7 t s-5 t^{2}\right)$$

$$4 s^{2}+5 s t-t^{2}$$

If you are comfortable skipping a step or two, it is not necessary to write down all of the steps shown in Example $$\PageIndex{1}$$. Let’s try combining like terms mentally in the next example.

##### Example $$\PageIndex{2}$$

Simplify: $\left(x^{3}-2 x^{2} y+3 x y^{2}+y^{3}\right)+\left(2 x^{3}-4 x^{2} y-8 x y^{2}+5 y^{3}\right) \nonumber$

Solution

If we use the associative and commutative property to reorder and regroup, then combine like terms, we get the following result.

\begin{aligned}\left(x^{3}-2 x^{2} y\right.&+3 x y^{2}+y^{3} )+\left(2 x^{3}-4 x^{2} y-8 x y^{2}+5 y^{3}\right) \\ &=\left(x^{3}+2 x^{3}\right)+\left(-2 x^{2} y-4 x^{2} y\right)+\left(3 x y^{2}-8 x y^{2}\right)+\left(y^{3}+5 y^{3}\right) \\ &=3 x^{3}-6 x^{2} y-5 x y^{2}+6 y^{3} \end{aligned} \nonumber

However, if we can combine like terms mentally, eliminating the middle step, it is much more efficient to write:

$\begin{array}{l}{\left(x^{3}-2 x^{2} y+3 x y^{2}+y^{3}\right)+\left(2 x^{3}-4 x^{2} y-8 x y^{2}+5 y^{3}\right)} \\ {\quad \quad=3 x^{3}-6 x^{2} y-5 x y^{2}+6 y^{3}}\end{array} \nonumber$

##### You Try $$\PageIndex{2}$$

Simplify: $$\left(-5 a^{2} b+4 a b-3 a b^{2}\right)+\left(2 a^{2} b+7 a b-a b^{2}\right)$$

$$-3 a^{2} b+11 a b-4 a b^{2}$$

## Negating a Polynomial

Before attempting subtraction of polynomials, let’s first address how to negate or “take the opposite” of a polynomial. First recall that negating is equivalent to multiplying by $$−1$$.

##### Negating

If $$a$$ is any number, then

$−a =(−1)a. \nonumber$

That is, negating is equivalent to multiplying by $$−1$$.

We can use this property to simplify $$−(a + b)$$. First, negating is identical to multiplying by $$−1$$. Then we can distribute the $$−1$$.

\begin{aligned} -(a+b) &= (-1)(a+b) \qquad \color {Red} \text { Negating is equivalent to multiplying by } -1 \\ &= (-1) a+(-1) b \qquad \color {Red} \text { Distribute the }-1 . \\ &= -a+(-b) \qquad \color {Red} \text { Simplify: }(-1) a=-a \text { and }(-1) b=-b \\ &= -a-b \qquad \color {Red} \text { Subtraction means add the opposite. } \end{aligned} \nonumber

Thus, $$−(a + b)=−a −b$$. However, it is probably simpler to note that the minus sign in front of the parentheses simply changed the sign of each term inside the parentheses.

##### Negating a Sum

When negating a sum of terms, the effect of the minus sign is to change each term in the parentheses to the opposite sign. $−(a + b)=−a−b\nonumber$

Let’s look at this principle in the next example.

##### Example $$\PageIndex{3}$$

Simplify: $$-\left(-3 x^{2}+4 x-8\right)$$

Solution

First, negating is equivalent to multiplying by $$−1$$. Then distribute the $$−1$$.

\begin{aligned} -&\left(-3 x^{2}+4 x-8\right) \\ &=(-1)\left(-3 x^{2}+4 x-8\right) \qquad \color {Red} \text{Negating is equivalent to multiplying by } -1\\ &=(-1)\left(-3 x^{2}\right)+(-1)(4 x)-(-1)(8) \qquad \color {Red} \text {Distribute the } -1 \\ &=3 x^{2}+(-4 x)-(-8) \qquad \color {Red} \text {Simplify: } (-1)(-3 x^{2})=3 x^{2}, (-1)(4 x)=-4 x, \text {and} (-1)(8)=-8\\ &=3 x^{2}-4 x+8 \qquad \color {Red} \text {Subtraction means add the opposite.} \end{aligned} \nonumber

Alternate solution:

As we saw above, a negative sign in front of a parentheses simply changes the sign of each term inside the parentheses. So it is much more efficient to write$−(−3x^2 +4x−8) = 3x^2 −4x +8 \nonumber$simply changing the sign of each term inside the parentheses.

##### You Try $$\PageIndex{3}$$

Simplify: $$-\left(2 x^{2}-3 x+9\right)$$

$$-2 x^{2}+3 x-9$$

## Subtracting Polynomials

Now that we know how to negate a polynomial (change the sign of each term of the polynomial), we’re ready to subtract polynomials.

##### Example $$\PageIndex{4}$$

Simplify: $$\left(y^{3}-3 y^{2} z+4 y z^{2}+z^{3}\right)-\left(2 y^{3}-8 y^{2} z+2 y z^{2}-8 z^{3}\right)$$

Solution

First, distribute the minus sign, changing the sign of each term of the second polynomial.

$\left(y^{3}-3 y^{2} z+4 y z^{2}+z^{3}\right)-\left(2 y^{3}-8 y^{2} z+2 y z^{2}-8 z^{3}\right)=y^{3}-3 y^{2} z+4 y z^{2}+z^{3}-2 y^{3}+8 y^{2} z-2 y z^{2}+8 z^{3} \nonumber$

Regroup, combining like terms. You may perform this next step mentally if you wish.

$\begin{array}{l}{=\left(y^{3}-2 y^{3}\right)+\left(-3 y^{2} z+8 y^{2} z\right)+\left(4 y z^{2}-2 y z^{2}\right)+\left(z^{3}+8 z^{3}\right)} \\ {=-y^{3}+5 y^{2} z+2 y z^{2}+9 z^{3}}\end{array} \nonumber$

##### You Try $$\PageIndex{4}$$

Simplify: $$\left(4 a^{2} b+2 a b-7 a b^{2}\right)-\left(2 a^{2} b-a b-5 a b^{2}\right)$$

$$2 a^{2} b+3 a b-2 a b^{2}$$

## Some Applications

Recall that the area of a rectangle having length $$L$$ and width $$W$$ is found using the formula $$A = LW$$. The area of a square having side s is found using the formula $$A = s^2$$ (see Figure $$\PageIndex{1}$$).

##### Example $$\PageIndex{5}$$

Find the area of the square in Figure $$\PageIndex{2}$$ by summing the area of its parts.

Solution

Let’s separate each of the four pieces and label each with its area (see Figure $$\PageIndex{3}$$).

The two shaded squares in Figure $$\PageIndex{3}$$ have areas $$A_1 = x^2$$ and $$A_3 = 9$$, respectively. The two unshaded rectangles in Figure $$\PageIndex{3}$$ have areas $$A_2 =3 x$$ and $$A_4 =3x$$. Summing these four areas gives us the area of the entire figure.

\begin{aligned} A &=A_{1}+A_{2}+A_{3}+A_{4} \\ &=x^{2}+3 x+9+3 x \\ &=x^{2}+6 x+9 \end{aligned} \nonumber

##### You Try $$\PageIndex{5}$$

Find the area of the square shown below by summing the area of its parts.

$$x^{2}+8 x+16$$