Consider this polynomial function:
f(x) = 4x3 - 3x2 + x - 4
Suppose that we evaluate it at an input of x = 2, like this:
f(2) = 4(23) - 3(22) + (2) - 4
f(2) = 32 - 12 = 2 - 4
f(2) = 18
In this process we raised the input to a power, as in 23.
Let us see that there is a way to evaluate this polynomial function using only multiplication and addition.
Start with the original polynomial and factor out an x. So, this:
4x3 - 3x2 + x - 4
Becomes:
x(4x2 - 3x + 1) - 4
Factor out another x from the parenthesized expression:
x(x(4x - 3) + 1) - 4
Now, imagine that you evaluate f(x) at x = 2. Begin with the inner most expression. Place a 2 for the input value of x, as in:
x(x(4(2) - 3) + 1) - 4
Now you would multiply 2 (the input) by 4 (the original coefficient of x3) and then add -3 (the original coefficient of x2). This would evaluate to 5. The expression now looks like:
x(x(5) + 1) - 4
Place a 2 for the next input value of x, as in:
x(2(5) + 1) - 4
Now you would multiply 2 (the input) by 5 and then add 1 (the original coefficient of x). This would evaluate to 11. The expression now looks like:
x(11) - 4
Place a 2 for the last input value of x, as in:
2(11) - 4
Now you would multiply 2 (the input) by 11 and then add -4 (the original final constant of the polynomial). This would evaluate to 18, the same value which we calculated at first, of course.
Now, this method of evaluating the output value of a polynomial can be captured in a type of short-hand notation by positioning the input value and the coefficients of the polynomial in a certain form.
Again, consider this polynomial:
f(x) = 4x3 - 3x2 + x - 4
We will picture it evaluated at the input value x = 2. Arrange the input value, the coefficients, and a line like this:
2) 4 -3 1 -4
-------------
Now drop down the 4:
2) 4 -3 1 -4
-------------
4
Multiply the input 2 times the 4. Place this product, 8, under the -3:
2) 4 -3 1 -4
8
-------------
4
Add the -3 and the 8. Place this sum, 5, under the line:
2) 4 -3 1 -4
8
-------------
4 5
Multiply the input 2 times the 5. Place this product, 10, under the 1:
2) 4 -3 1 -4
8 10
-------------
4 5
Add the 1 and the 10. Place this sum, 11, under the line:
2) 4 -3 1 -4
8 10
-------------
4 5 11
Multiply the input 2 times the 11. Place this product, 22, under the -4:
2) 4 -3 1 -4
8 10 22
-------------
4 5 11
Add the -4 and the 22. Place this sum, 18, under the line:
2) 4 -3 1 -4
8 10 22
-------------
4 5 11 18
We are done. This final value, 18, is the function output for f(x) evaluated with an input of 2.
This process is called synthetic division. At first this term may be confusing. What does a method of function evaluation have to do with division?
Consider dividing f(x) = 4x3- 3x2+ x - 4 by (x - 2). Standard polynomial division would look like this:
4x2 + 5x + 11
---------------------
x - 2 )4x3 - 3x2 + x - 4
4x3 - 8x2
---------
5x2 + x
5x2 - 10x
---------
11x - 4
11x - 22
--------
18
Compare the above with our previous synthetic division:
2) 4 -3 1 -4
8 10 22
-------------
4 5 11 18
Notice that the remainder from the standard polynomial long division is the last value calculated in the synthetic division.
Notice that the coefficients of the quotient from the standard polynomial long division are aligned before the remainder on the last row, under the line, of the synthetic division.
So, this synthetic division method can be used to determine the quotient and remainder when a polynomial function, f(x), is divided by a linear function of the form (x - c).
Specifically, this method can be used to determine if a certain value, c, is a root of the polynomial function f(x). If c is a root, then the synthetic division method will show a zero in the remainder position. This would be equivalent to determining that (x - c) is a linear factor of f(x).
An added benefit of this method is that it yields the coefficients of the quotient when f(x) is divided by (x - c). This quotient can be further examined and factored using synthetic division until all the linear factors of the original polynomial function, f(x), have been found.
Once all the linear factors of the polynomial function have been found, all the roots of the function are obvious.