Lesson 7-5 Adding and Subtracting Polynomials
Example 1 Add Polynomials
Find each sum.
a. (r2 + 11 + 7r) + (–2r + 10 + 8r2)
Horizontal Method
(r2 + 11 + 7r) + (–2r + 10 + 8r2)
= (r2 + 8r2) + [7r + (–2r)] + (11 + 10)
= 9r2 + 5r + 21
Group like terms.
Combine like terms.
Vertical Method
Align like terms in columns and add like terms together.
r2 + 7r + 11
(+) 8r2 + (–2r) + 10
9r2 + 5r + 21
Align and combine like terms.
b. (9 – 5a2 + a) + (a2 + 3a3 – 4a)
Horizontal Method
(9 – 5a2 + a) + (a2 + 3a3 – 4a)
= 3a3 + (-5a2 + a2) + [a + (-4a)] + 9
= 3a3 – 4a2 – 3a + 9
Group like terms.
Combine like terms.
Vertical Method
Align like terms in columns and add like terms together.
0a3 – 5a2 + a + 9
(+) 3a3 + a2 – 4a + 0
3a3 – 4a2 – 3a + 9
Insert placeholders to help align the terms.
Align and combine like terms.
Example 2 Subtract Polynomials
Find each difference.
a. (n3 + 3n – 9) – (n – 2n3 + 1)
Horizontal Method
Subtract n – 2n3 + 1 by adding its additive inverse.
(n3 + 3n – 9) – (n – 2n3 + 1)
= (n3 + 3n – 9) + (–n + 2n3 – 1)
= (n3 + 2n3) + [3n + (–n)] + [– 9 + (–1)]
= 3n3 + 2n – 10
The additive inverse of n – 2n + 1 is n + 2n – 1.
Group like terms.
Combine like terms.
3
Vertical Method
Align like terms in columns and subtract by adding the additive inverse.
n3 + 3n – 9
(-) –2n3 + n + 1
Add the opposite.
n3 + 3n – 9
(+) 2n3 – n – 1
3n3 + 2n – 10
Thus, (n3 + 3n – 9) – (n – 2n3 + 1) = 3n3 + 2n – 10.
3
b. (6x2 – 17x3 + 7) – (2x2 – x3 + x)
Horizontal Method
Subtract 2x2 – x3 + x by adding its additive inverse.
(6x2 – 17x3 + 7) – (2x2 – x3 + x)
= (6x2 – 17x3 + 7) + (-2x2 + x3 – x)
= (-17x3 + x3) + [6x2 + (-2x2)] + (–x) + 7
= -16x3 + 4x2 – x + 7
2
3
2
3
The additive inverse of 2x – x + x is -2x + x – x.
Group like terms.
Combine like terms.
Vertical Method
Align like terms in columns and subtract by adding the additive inverse.
–17x3 + 6x2 + 7
(-) –x3 + 2x2 + x
Add the opposite.
–17x3 + 6x2 + 7
(+)
x3 – 2x2 – x
–16x3 + 4x2 – x + 7
Thus, (6x2 – 17x3 + 7) – (2x2 – x3 + x) = -16x3 + 4x2 – x + 7.
Real-World Example 3 Add and Subtract Polynomials
SCHOOLS The total number of students in a school district includes the number in public
schools and the number in private schools. The equations below represent the total number of
students (in thousands) S and the number in private schools R, where n is the number of years
since 1989.
S = 716.4n + 45,711
R = 82n + 5198
a. Write an equation that represents the number of students in public schools U.
Subtract the polynomial for R from the polynomial for S.
Total
– Private
Public
716.4n + 45,711
(–) 82n + 5198
Add the opposite.
716.4n + 45,711
(+) -82n – 5198
634.4n + 40,513
An equation is U = 634.4n + 40,513.
b. Use the equation to predict the number of public school students in the year 2010.
The year 2010 is 2010 – 1989 or 21 years after the year 1989.
U = 634.4n + 40,513
= 634.4(21) + 40,513
= 53,835.4
Original equation
Substitute 21 for n.
Simplify.
So, if this trend continues, the number of public school students in 2010 would be about
53,835.4 thousand or 53,835,400.