Learning Outcomes
- Perform algebraic operations on polynomial functions
Just as we have performed algebraic operations on polynomials, we can do the same with polynomial functions. In this section, we will show you how to perform algebraic operations on polynomial functions and introduce related notation.
The four basic operations on functions are adding, subtracting, multiplying, and dividing. The notation for these functions is as follows:
Addition (f+g)(x)=f(x)+g(x)
Subtraction (f−g)(x)=f(x)−g(x)
Multiplication (f⋅g)(x)=f(x)g(x)
Division fg(x)=f(x)g(x)
We will focus on applying these operations to polynomial functions in this section.
Add and Subtract Polynomial Functions
Adding and subtracting polynomial functions is the same as adding and subtracting polynomials. When you evaluate a sum or difference of functions, you can either evaluate first or perform the operation on the functions first as we will see. Our next examples describe the notation used to add and subtract polynomial functions.
Example
Let f(x)=2x3−5x+3 and h(x)=x−5,
Find the following:
(f+h)(x) and (h−f)(x)
Show Solution
(f+h)(x)=f(x)+h(x)=(2x3−5x+3)+(x−5)=2x3−5x+3+x−5combine like terms=2x3−4x−2simplify
(h−f)(x)=h(x)−f(x)=(x−5)−(2x3−5x+3)=x−5−2x2+5x−3combine like terms=−2x2+6x−8simplify
In our next example, we will evaluate a sum and difference of functions and show that you can get to the same result in one of two ways.
Example
Let f(x)=2x3−5x+3 and h(x)=x−5
Evaluate: (f+h)(2)
Show that you get the same result by
1) Evaluating the functions first then performing the indicated operation on the result.
2) Performing the operation on the functions first then evaluating the result.
Show Solution
1) First, we will evaluate the functions separately:
f(2)=2(2)3−5(2)+3=16−10+3=9
h(2)=(2)−5=−3
Now we will perform the indicated operation using the results:
(f+h)(2)=f(2)+h(2)=9+(−3)=6
2) We can get the same result by adding the functions first and then evaluating the result at x=2.
(f+h)(x)=f(x)+h(x)=2x3−4x−2 from the previous example.
Now we can evaluate this result at x=2
(f+h)(2)=2(2)3−4(2)−2=16−8−2=6
Both methods give the same result, and both require about the same amount of work.
Multiply and Divide Polynomial Functions
We saw that multiplying polynomials often required the use of the distributive property and that the algebra of dividing polynomials could get messy fast! The same techniques can be used to multiply and divide polynomial functions. Additionally, the same idea applies to evaluating a product or quotient of functions as we discovered in the previous example. We can either evaluate the function and then perform the indicated operation or vice-versa. You may already be thinking it will be a lot less work to evaluate the polynomials and then divide the results!
Example
Let g(t)=2t3−t2+7 and f(t)=5t2−3
Find (g⋅f)(t), and evaluate (g⋅f)(−1)
Show Solution
(g⋅f)(t)=(2t3−t2+7)(5t2−3)=(2t3⋅(5t2)−t2⋅(5t2)+7⋅(5t2))+(2t3⋅(−3)−t2⋅(−3)+7⋅(−3))apply the distributive property=(10t5−5t4+35t2)+(−6t3+3t2−21)simplify=10t5−5t4−6t3+38t2−21combine like terms
Evaluate (g⋅f)(−1)
(g⋅f)(t)=10t5−5t4−6t3+38t2−21(g⋅f)(−1)=10(−1)5−5(−1)4−6(−1)3+38(−1)2−21=−10−5+6+38−21=8
In the next example, we will divide polynomial functions and then evaluate the new function.
Example
Given p(x)=2x2+x−15 and r(x)=x+3
Find pr(x) and evaluate pr(2)
Show Solution
We can use synthetic division for this polynomial division since the coefficient on r(x)=x+3 is 1.
This result means that pr(x)=2x2+x−15x+3=2x−5
Now evaluate this quotient for x=−3 both ways as we did in a previous example.
First, we will evaluate the result after polynomial division:
pr(x)=2x−5pr(2)=2(2)−5=4−5=−1
Next, we will evaluate each function for x=2, then we will divide the results.
p(2)=2(2)2+(2)−15=8+2−15=−5
r(2)=(2)+3=5
Divide the results:
pr(2)=−55=−1
