Read: Algebra of Polynomial Functions
[latex]\frac{f}{ g} (x) = \frac{f(x)}{g(x)}[/latex] 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
For [latex]f(x)=2x^3-5x+3[/latex] and [latex]h(x)=x-5[/latex], Find the following: [latex-display](f+h)(x)[/latex] and [latex](h-f)(x)[/latex-display]Answer: [latex]\begin{array}{ccc}(f+h)(x)=f(x)+ h(x)(2x^3-5x+3)+(x-5)\\=2x^3-5x+3+x-5\,\,\,\,\,\text{combine like terms}\\=2x^3-4x-2\,\,\,\,\,\text{simplify}\end{array}[/latex] [latex]\begin{array}{ccc}(h-f)(x)=h(x)-f(x)=(x-5)-(2x^3-5x+3)\\=x-5-2x^2+5x-3\,\,\,\,\,\,\text{combine like terms}\\=-2x^2+6x-8\,\,\,\,\,\text{simplify}\end{array}[/latex]
Example
For [latex]f(x)=2x^3-5x+3[/latex] and [latex]h(x)=x-5[/latex] Evaluate: [latex](f+h)(2)[/latex] Show that you get the same result by 1)evaluating the functions first, then performing the indicated operation on the result and 2) performing the operation on the functions first, then evaluating the resultAnswer: 1)[latex](f+h)(2)[/latex] First, we will evaluate the functions separately: [latex]f(2)=2(2)^3-5(2)+3=16-10+3=9[/latex] [latex]h(2)=(2)-5=-3[/latex] Now we will perform the indicated operation using the results: [latex](f+h)(2)=f(2)+h(2)=9+(-3)=6[/latex] 2) We can get the same result by adding the functions first, then evaluating the result at [latex]x=2[/latex] [latex](f+h)(x)=f(x)+h(x)=2x^3-4x-2[/latex] from above. Now we can evaluate this result at [latex]x=2[/latex] [latex](f+h)(2)=2(2)^3-4(2)-2=16-8-2=6[/latex] 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
Given: [latex]g(t)=2t^3-t^2+7[/latex] and [latex]f(t)=5t^2-3[/latex] Find: [latex](g · f)(t)[/latex], and evaluate [latex](g · f)(-1)[/latex]Answer: [latex-display]\begin{array}{ccc}(g · f)(t)=(2t^3-t^2+7)(5t^2-3)\\=(2t^3\cdot(5t^2)-t^2\cdot(5t^2)+7\cdot(5t^2))+(2t^3\cdot(-3)-t^2\cdot(-3)+7\cdot(-3))\,\,\,\,\,\text{apply the distributive property}\\=(10t^5-5t^4+35t^2)+(-6t^3+3t^2-21)\,\,\,\,\text{simplify}\\=10t^5-5t^4-6t^3+38t^2-21\,\,\,\,\,\text{combine like terms}\end{array}[/latex-display] Evaluate [latex](g · f)(-1)[/latex] [latex-display]\begin{array}{ccc}(g · f)(t)=10t^5-5t^4-6t^3+38t^2-21\\(g · f)(-1)=10(-1)^5-5(-1)^4-6(-1)^3+38(-1)^2-21\\=-10-5+6+38-21\\=8\end{array}[/latex-display]
Example
Given [latex]p(x)=2x^2+x-15[/latex] and [latex]r(x)=x+3[/latex] Find [latex]\frac{p}{r}(x)[/latex] and evaluate [latex]\frac{p}{r}(2)[/latex]Answer:
We can use synthetic division for this polynomial division since the coefficient on [latex]r(x)=x+3[/latex] =[latex]1[/latex].
This result means that [latex]\frac{p}{r}(x)=\frac{2x^2+x-15}{x+3}=2x-5[/latex]
Let's evaluate this quotient for [latex]x = -3[/latex] both ways as we did in a previous example.
First, we will evaluate the result after polynomial division: [latex]\begin{array}{ccc}\frac{p}{r}(x)=2x-5\\\frac{p}{r}(2)=2(2)-5=4-5=-1\end{array}[/latex]
Next, we will evaluate each function for [latex]x = 2[/latex], then we will divide the results. [latex]p(2)=2(2)^2+(2)-15=8+2-15=-5[/latex] [latex]r(2)=(2)+3=5[/latex]
Divide the results: [latex]\frac{p}{r}(2)\frac{-5}{5}=1[/latex]
