Chapter A.4, Problem 24AYU

To determine

To calculate:Whether $x+4$ is a factor of the polynomial $x^6-16x^4+x^2-16$ with help of synthetic division.

Answer to Problem 24AYU

The polynomial $x^6-16x^4+x^2-16$has$x+4$ as its factor.

Explanation of Solution

Given information:

The polynomial $x^6-16x^4+x^2-16$ and a factor of it $x+4$ .

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 $x^6-16x^4+x^2-16$ and a factor of it $x+4$ .

To divide the polynomial $x^6-16x^4+x^2-16$ by $x+4$ follow the steps provided below,

Here dividend is $x^6+0\cdot x^5-16x^4+0\cdot x^3+x^2+0\cdot x-16$ and divisor is $x-\left(-4\right)$ .

Step 1: List down the coefficients of dividend in the descending powers of x, that are $1,0,-16,0,1,0,-16$ .

Now, the divisor is of the form $x-c$ , where $c=-4$ .

Step 2: To perform the synthetic division put the coefficients in the division sign along with $-4$ on the left side,

  $-4\overline{)\begin{array}{l}\begin{array}{ccccccc}1& 0& -16& 0& 1& 0& -16\end{array}\\ \begin{array}{ccccccc}& & & & & & \end{array}\\ \begin{array}{ccccccc}& & & & & & \end{array}\end{array}}$

Step 3: Bring down $1$ to in third row,

  $-4\overline{)\begin{array}{l}\begin{array}{ccccccc}1& 0& -16& 0& 1& 0& -16\end{array}\\ \begin{array}{ccccccc}& & & & & & \end{array}\\ \begin{array}{ccccccc}1& & & & & & \end{array}\end{array}}$

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

  $-4\overline{)\begin{array}{l}\begin{array}{ccccccc}1& 0& -16& 0& 1& 0& -16\end{array}\\ \begin{array}{ccccccc}& -4& & & & & \end{array}\\ \begin{array}{ccccccc}1& & & & & & \end{array}\end{array}}$

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

  $-4\overline{)\begin{array}{l}\begin{array}{ccccccc}1& 0& -16& 0& 1& 0& -16\end{array}\\ \begin{array}{ccccccc}& -4& & & & & \end{array}\\ \begin{array}{ccccccc}1& -4& & & & & \end{array}\end{array}}$

Step 6: Multiply $-4$ with the second entry of row 3 and place the result in row 2 column 3.

  $-4\overline{)\begin{array}{l}\begin{array}{ccccccc}1& 0& -16& 0& 1& 0& -16\end{array}\\ \begin{array}{ccccccc}& -4& 16& & & & \end{array}\\ \begin{array}{ccccccc}1& -4& & & & & \end{array}\end{array}}$

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

  $-4\overline{)\begin{array}{l}\begin{array}{ccccccc}1& 0& -16& 0& 1& 0& -16\end{array}\\ \begin{array}{ccccccc}& -4& 16& & & & \end{array}\\ \begin{array}{ccccccc}1& -4& 0& & & & \end{array}\end{array}}$

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

  $-4\overline{)\begin{array}{l}\begin{array}{ccccccc}1& 0& -16& 0& 1& 0& -16\end{array}\\ \begin{array}{ccccccc}& -4& 16& 0& & & \end{array}\\ \begin{array}{ccccccc}1& -4& 0& & & & \end{array}\end{array}}$

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

  $-4\overline{)\begin{array}{l}\begin{array}{ccccccc}1& 0& -16& 0& 1& 0& -16\end{array}\\ \begin{array}{ccccccc}& -4& 16& 0& & & \end{array}\\ \begin{array}{ccccccc}1& -4& 0& 0& & & \end{array}\end{array}}$

Repeat the steps 8 and 9,

  $-4\overline{)\begin{array}{l}\begin{array}{ccccccc}1& 0& -16& 0& 1& 0& -16\end{array}\\ \begin{array}{ccccccc}& -4& 16& 0& 0& -4& 16\end{array}\\ \begin{array}{ccccccc}1& -4& 0& 0& 1& -4& 0\end{array}\end{array}}$

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 $x^5-4x^4+x-4$ 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.

  $x^6-16x^4+x^2-16=\left(x^5-4x^4+x-4\right)\left(x+4\right)+0$

Now, $\left(x-c\right)$ is a factor of a polynomial if remainder is 0. Here, remainder is 0 so, $x+4$ is a factor.

Thus, the polynomial $x^6-16x^4+x^2-16$ has $x+4$ as its factor.