Chapter A.4, Problem 22AYU

To determine

To calculate:Whether $x+3$ is a factor of the polynomial $2x^6-18x^4+x^2-9$ with help of synthetic division.

Answer to Problem 22AYU

The polynomial $2x^6-18x^4+x^2-9$has$x+3$ as its factor.

Explanation of Solution

Given information:

The polynomial $2x^6-18x^4+x^2-9$ and a factor of it $x+3$ .

Formula used:

When a polynomial is divided by its factor then dividend is the product of divisor and quotient increased by remainder.

Calculation:

Consider the provided polynomial $2x^6-18x^4+x^2-9$ and a factor of it $x+3$ .

To divide the polynomial $2x^6-18x^4+x^2-9$ by $x+3$ follow the steps provided below,

Here dividend is $2x^6+0\cdot x^5-18x^4+0\cdot x^3+x^2+0\cdot x-9$ and divisor is $x-\left(-3\right)$ .

Step 1: List down the coefficients of dividend in the descending powers of x, that are $2,0,-18,0,1,0,-9$ .

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,

  $-3\overline{)\begin{array}{l}\begin{array}{ccccccc}2& 0& -18& 0& 1& 0& -9\end{array}\\ \begin{array}{ccccccc}& & & & & & \end{array}\\ \begin{array}{ccccccc}& & & & & & \end{array}\end{array}}$

Step 3: Bring down $2$ to in third row,

  $-3\overline{)\begin{array}{l}\begin{array}{ccccccc}2& 0& -18& 0& 1& 0& -9\end{array}\\ \begin{array}{ccccccc}& & & & & & \end{array}\\ \begin{array}{ccccccc}2& & & & & & \end{array}\end{array}}$

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

  $-3\overline{)\begin{array}{l}\begin{array}{ccccccc}2& 0& -18& 0& 1& 0& -9\end{array}\\ \begin{array}{ccccccc}& -6& & & & & \end{array}\\ \begin{array}{ccccccc}2& & & & & & \end{array}\end{array}}$

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

  $-3\overline{)\begin{array}{l}\begin{array}{ccccccc}2& 0& -18& 0& 1& 0& -9\end{array}\\ \begin{array}{ccccccc}& -6& & & & & \end{array}\\ \begin{array}{ccccccc}2& -6& & & & & \end{array}\end{array}}$

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

  $-3\overline{)\begin{array}{l}\begin{array}{ccccccc}2& 0& -18& 0& 1& 0& -9\end{array}\\ \begin{array}{ccccccc}& -6& 18& & & & \end{array}\\ \begin{array}{ccccccc}2& -6& & & & & \end{array}\end{array}}$

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

  $-3\overline{)\begin{array}{l}\begin{array}{ccccccc}2& 0& -18& 0& 1& 0& -9\end{array}\\ \begin{array}{ccccccc}& -6& 18& & & & \end{array}\\ \begin{array}{ccccccc}2& -6& 0& & & & \end{array}\end{array}}$

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

  $-3\overline{)\begin{array}{l}\begin{array}{ccccccc}2& 0& -18& 0& 1& 0& -9\end{array}\\ \begin{array}{ccccccc}& -6& 18& 0& & & \end{array}\\ \begin{array}{ccccccc}2& -6& 0& & & & \end{array}\end{array}}$

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

  $-3\overline{)\begin{array}{l}\begin{array}{ccccccc}2& 0& -18& 0& 1& 0& -9\end{array}\\ \begin{array}{ccccccc}& -6& 18& 0& & & \end{array}\\ \begin{array}{ccccccc}2& -6& 0& 0& & & \end{array}\end{array}}$

Repeat the steps 8 and 9,

  $-3\overline{)\begin{array}{l}\begin{array}{ccccccc}2& 0& -18& 0& 1& 0& -9\end{array}\\ \begin{array}{ccccccc}& -6& 18& 0& 0& -3& 9\end{array}\\ \begin{array}{ccccccc}2& -6& 0& 0& 1& -3& 0\end{array}\end{array}}$

Now, the first six 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 $2x^5-6x^4+x-3$ and reminder is 0.

Now, when a polynomial is divided by its divisor then dividend is the product of divisor and quotient increased by remainder.

  $2x^6-18x^4+x^2-9=\left(2x^5-6x^4+x-3\right)\left(x+3\right)+0$

Now, $\left(x-c\right)$ is a factor of a polynomial if remainder is 0. Here, remainder is 0 so, $x+3$ is a factor.

Thus, the polynomial $2x^6-18x^4+x^2-9$ has $x+3$ as its factor.