Let us reconsider the polynomial function from exercise 2.2.3:

\[ f ( x ) = 2 x^3 - 3 x^2 - 11 x + 6 \]We are going to rewrite it in a peculiar way, called the **recursive form**.

Rewrite the polynomial function in recursive form:

\[ f ( x ) = 3 x^4 + x^3 - 2 x^2 - 5 x + 1 \] See SolutionLet 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 \(4\).

\[f ( 4 ) = ( ( 2 ( 4 ) - 3 ) ( 4 ) - 11 ) ( 4 ) + 6=42\]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 \(6\).
- The result is \(42\).

Notice the repetitive nature of this process.

We will now summarize the process by the use of the following table:

\(4\) | \(2\) | \(-3\) | \(-11\) | \(6\) |

\(8\) | \(20\) | \(36\) | ||

\(2\) | \(5\) | \(9\) | \(42\) |

The numbers are placed into this table in a particular order.

- First, the coefficients \(2,-3,-11,6\) of the polynomial are placed in the top row, beginning in the second column.
- Next, in order to find \(f(4)\) the input number \(4\) is placed in the top row, first column.
- Next, the leading coefficient \(2\) is copied to the third row.

\(4\) | \(2\) | \(-3\) | \(-11\) | \(6\) |

\(2\) |

After setting up the table in this fashion, the repetitive phase of the process begins. (Refer to the first table.)

- Multiply the input number \(4\) by the \(2\) in row three and place the product, \(8\), in row two of the next column. Add the two numbers in that column and place the result, \(5\), in the third row.
- Multiply the input number \(4\) by the \(5\) in row three and place the product \(20\) in in row two of the next column. Add the two numbers in that column and place the result \(9\) in the third row.
- Multiply the input number \(4\) by the \(9\) in row three and place the product \(36\) in in row two of the next column. Add the two numbers in that column and place the result \(42\) in the third row.

This process and the accompanying table is referred to as **synthetic division**.

Notice that after setting up the table, every number in the third row is multiplied by the input number and the product is placed in row two of the next column. Then the sum of the numbers in that column are placed in row three and the process repeats until there are no columns left. The number in row three of the last column will always be the output number corresponding to the input number in the first row of column one.

Given \(f ( x ) = 3 x^4 + x^3 - 2 x^2 - 5 x + 1\), use synthetic division to calculate \(f ( 3 )\).

See Solution
One might ask, why is this called synthetic division? What has this process to do with the operation of division? We are using the process to evaluate a polynomial functions for various input numbers. So why isn't this called **synthetic evaluation of polynomials**?

To see why, consider the following long division problem for polynomials.

Use **long** division to divide the polynomial in the synthetic division example \( 2 x^3 - 3 x^2 - 11 x + 6 \) by \( x - 4 \).

Using the result of this exercise we can now write \(f ( x )\) in the following form:

\[f ( x ) = ( x - 4 ) ( 2 x^2 + 5 x + 9 ) + 42\]The polynomial \( 2 x^2 + 5 x + 9 \) 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** above. 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 \(f ( x ) = ( x - 4 ) ( 2 x^2 + 5 x + 9 ) + 42.\)

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 \(c\), the output number \(f ( c )\) equals the remainder when \(f ( x )\) is divided by \( x - c \).

Let \(f ( x ) = 3 x^4 + x^3 - 2 x^2 - 5 x + 1\).

Use synthetic division to find the quotient \(q ( x )\) and the remainder \(r\) when \(f ( x )\) is divided by \(x - 1\).

Rewrite \(f ( x )\) in the form \(f ( x ) = ( x - 1 ) q ( x ) + r\).

What is the value of \(f ( 1 )\)?

See Solution