Chapter A.4, Problem 3AYU

To determine

To explain: Whether the statement “In using synthetic division, the divisor is always a polynomial od degree 1, whose leading coefficient is 1.” is true or false.

Answer to Problem 3AYU

The statement “In using synthetic division, the divisor is always a polynomial od degree 1, whose leading coefficient is 1.” is true.

Explanation of Solution

Given information:

The statement “In using synthetic division, the divisor is always a polynomial od degree 1, whose leading coefficient is 1.”

Consider the provided statement “In using synthetic division, the divisor is always a polynomial od degree 1, whose leading coefficient is 1.”

The statement is true as divisor is taken which cannot be simplified further and also leading coefficient should be one.

For example: consider the provided polynomial $3x^3+2x^2-x+1$ and a factor of it $x-3$ .

To divide the polynomial $3x^3+2x^2-x+3$ by $x-3$ follow the steps provided below,

Here dividend is $3x^3+2x^2-x+3$ and divisor is $x-3$ .

Step 1: List down the coefficients of dividend in the descending powers of x, that are $3,2,-1,1$ .

Now, the divisor is of the form $x-c$ , where $c=3$ .

Step 2: To perform the synthetic division put the coefficients in the division sign along with 3 on the left side,

  $3\overline{)\begin{array}{cccc}3& 2& -1& 1\end{array}}$

Step 3: Bring down 3 to in third row,

  $3\overline{)\begin{array}{l}\begin{array}{cccc}3& 2& -1& 1\end{array}\\ \begin{array}{cccc}& & & \end{array}\\ \begin{array}{cccc}3& & & \end{array}\end{array}}$

Step 4: Multiply 3 with the first entry of row 3 and place the result in row 2 column 2.

  $3\overline{)\begin{array}{l}\begin{array}{cccc}3& 2& -1& 1\end{array}\\ \begin{array}{cccc}& 9& & \end{array}\\ \begin{array}{cccc}3& & & \end{array}\end{array}}$

Step 5: Now, add the elements of column 2 of row 1 and row 2.

  $3\overline{)\begin{array}{l}\begin{array}{cccc}3& 2& -1& 1\end{array}\\ \begin{array}{cccc}& 9& & \end{array}\\ \begin{array}{cccc}3& 11& & \end{array}\end{array}}$

Step 6: Multiply 3 with the second entry of row 3 and place the result in row 2 column 3.

  $3\overline{)\begin{array}{l}\begin{array}{cccc}3& 2& -1& 1\end{array}\\ \begin{array}{cccc}& 9& 33& \end{array}\\ \begin{array}{cccc}3& 11& & \end{array}\end{array}}$

Step 7: Now, add the elements of column 3 of row 1 and row 2.

  $3\overline{)\begin{array}{l}\begin{array}{cccc}3& 2& -1& 1\end{array}\\ \begin{array}{cccc}& 9& 33& \end{array}\\ \begin{array}{cccc}3& 11& 32& \end{array}\end{array}}$

Step 8: Multiply 3 with the third entry of row 3 and place the result in row 2 column 4.

  $3\overline{)\begin{array}{l}\begin{array}{cccc}3& 2& -1& 1\end{array}\\ \begin{array}{cccc}& 9& 33& 96\end{array}\\ \begin{array}{cccc}3& 11& 32& \end{array}\end{array}}$

Step 9: Now, add the elements of column 4 of row 1 and row 2.

  $3\overline{)\begin{array}{l}\begin{array}{cccc}3& 2& -1& 1\end{array}\\ \begin{array}{cccc}& 9& 33& 96\end{array}\\ \begin{array}{cccc}3& 11& 32& 97\end{array}\end{array}}$

Now, the first three elements of row 3 are coefficients of quotient in descending powers of x with degree one less than dividend and the last entry is remainder.

Therefore, quotient is $3x^2+11x+32$ and reminder is 97.

Thus, the statement “In using synthetic division, the divisor is always a polynomial od degree 1, whose leading coefficient is 1.” is true.