9.2: Addition and Subtraction of Polynomials
What if you had two polynomials like \begin{align*}4x^2 - 5\end{align*} and \begin{align*}13x + 2\end{align*}? How could you add and and subtract them? After completing this Concept, you'll be able to perform addition and subraction 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!)
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CK-12 Foundation: 0902S Lesson Addition and Subtraction of Polynomials
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 \begin{align*}3x^2-4x+7\end{align*} and \begin{align*}2x^3-4x^2-6x+5\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*}
Solution
a) \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) \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.
Example B
a) Subtract \begin{align*}x^3-3x^2+8x+12\end{align*} from \begin{align*}4x^2+5x-9\end{align*}
b) Subtract \begin{align*}5b^2-2a^2\end{align*} from \begin{align*}4a^2-8ab-9b^2\end{align*}
Solution
a) \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) \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.
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.
\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*}
Watch this video for help with the Examples above.
CK-12 Foundation: Addition and Subtraction of Polynomials
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:
\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.
Practice
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.
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, .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.Image Attributions
Here you'll learn how to add and subtract polynomials and simplify your answers. You'll also solve real-world problems using addition and subtraction of polynomials.
Concept Nodes:
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, .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.