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9.2: Addition and Subtraction of Polynomials

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What if you had two polynomials like 4x^2 - 5 and 13x + 2 ? 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!)

Watch This

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 3x^2-4x+7 and 2x^3-4x^2-6x+5

b) Add x^2-2xy+y^2 and 2y^2-3x^2 and 10xy+y^3

Solution

a) & (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

b) & (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

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

Example B

a) Subtract x^3-3x^2+8x+12 from 4x^2+5x-9

b) Subtract 5b^2-2a^2 from 4a^2-8ab-9b^2

Solution

a) (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

b) (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

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 and b=3 , then we can check as follows:

& \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

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.

\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.

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

Total \ area &= y^2 + x^2 + xy + xy\\& = y^2 + x^2 + 2xy

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

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

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

Total \ area & = a^2 + a^2 + 2ab\\& = 2a^2 + 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.

\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

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.

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

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

Total \ area = a^2 + a^2 + a^2 + 3a^2 = 6a^2

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

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

To find the total area of the figure we subtract:

Area & = 9a^2 - (a^2 + a^2 + a^2)\\& = 9a^2 - 3a^2 \\& = 6a^2

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  4t^2+7t^3-3t-5 from 6t+3-5t^3+9t^2 .

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:

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\\

The final answer is in standard form.

Practice

Add and simplify.

  1. (x+8)+(-3x-5)
  2. (-2x^2+4x-12)+(7x+x^2)
  3. (2a^2b-2a+9)+(5a^2b-4b+5)
  4. (6.9a^2-2.3b^2+2ab)+(3.1a-2.5b^2+b)
  5. \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 )

Subtract and simplify.

  1. (-t+5t^2)-(5t^2+2t-9)
  2. (-y^2+4y-5)-(5y^2+2y+7)
  3. (-5m^2-m)-(3m^2+4m-5)
  4. (2a^2b-3ab^2+5a^2b^2)-(2a^2b^2+4a^2b-5b^2)
  5. (3.5x^2y-6xy+4x)-(1.2x^2y-xy+2y-3)

Find the area of the following figures.

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Difficulty Level:

At Grade

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Date Created:

Oct 01, 2012

Last Modified:

May 23, 2014
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