Let us reconsider the polynomial function from exercise 2.2.3:
f ( x ) = 2 x3 - 3 x2 - 11 x + 6
We are going to rewrite it in a peculiar way, called the recursive form.
f ( x ) = ( ( 2 x - 3 ) x - 11 ) x + 6
Exercise 2.4.1
Verify that the two forms of the equation for f are equivalent.
Exercise 2.4.2
Rewrite the polynomial function in recursive form:
f ( x ) = 3 x4 + x3 - 2 x2 - 5 x + 1
Let us go back to the original f ( x ) = ( ( 2 x - 3 ) x - 11 ) x + 6 written in recursive form and evaluate the function at some particular number, say 4.
f ( 4 ) = ( ( 2 ( 4 ) - 3 ) ( 4 ) - 11 ) ( 4 ) + 6
In other words,
Multiply 2 by 4 then add - 3.
Multiply the result by 4 then add - 11.
Multiply the result by 4 then add 6.
Notice the repetitive nature of this process.
We will now summarize the process by the use of the following table:
The numbers are placed into this table in a particular order. First, the coffecients of the polynomial are placed in the top row, beginning in the second column. Next the input number ( in this case, 4 ) is placed in the top row, first column. Next, the leading coefficient is copied to the third row. Next, begins the repetive phase of the process.
Multiply 4 by the 2 and place the product, 8, in the second column. Add the two numbers in the column and place the result, 5, in the third row.
Multiply 4 by the 5 and place the product, 20, in the second column. Add the two numbers in the column and place the result, 9, in the third row.
Multiply 4 by the 9 and place the product, 36, in the second column. Add the two numbers in the column and place the result, 42, in the third row.
The resulting number, 42, is the output corresponding to the input 4.
This process and the accompanying table is referred to as synthetic division.
Exercise 2.4.3
Use synthetic division to calculate f ( 3 ).
Now, one might ask, why is this called synthetic division? What has this process to do with the operation of division?
To see, consider the following long division problem for polynomials.
Exercise 2.4.4
Use long division to find ( 2 x3 - 3 x2 - 11 x + 6 ) / ( x - 4 ).
Using the result of this exercise we can now write f ( x ) in the following form:
f ( x ) = ( x - 4 ) ( 2 x2 + 5 x + 9 ) + 42.
The polynomial ( 2 x2 + 5 x + 9 ) is the quotient of the division process and the number 42 is the remainder.
Notice that the coefficients of the quotient are found on the third row of the synthetic division table. This is why this process and the acompanying table are referred to as synthetic division. The process produces both the quotient and the remainder of the long division process.
Now, let us return, briefly, to
f ( x ) = ( x - 4 ) ( 2 x2 + 5 x + 9 ) + 42.
Do you now see why f ( 4 ) must equal 42 ?
This illustrates the remainder principle: When applying the synthetic division process to an input number of c, the output number f ( c ) equals the remainder when f ( x ) is divided by ( x - c ).
Exercise 2.4.5
Let f ( x ) = 3 x4 + x3 - 2 x2 - 5 x + 1. Use synthetic division to find the quotient q ( x ) and the remainder r when f ( x ) is divided by x - 1. Then rewrite f ( x ) in the form
f ( x ) = ( x - 1 ) q ( x ) + r. Find the value of f ( 1 ).