Chapter A.4, Problem 19AYU

To determine

To calculate:Whether $x-2$ is a factor of the polynomial $3x^4-6x^3-5x+10$ with help of synthetic division.

Answer to Problem 19AYU

The polynomial $3x^4-6x^3-5x+10$has$x-2$ as its factor.

Explanation of Solution

Given information:

The polynomial $3x^4-6x^3-5x+10$ and a factor of it $x-2$ .

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 $3x^4-6x^3-5x+10$ and a factor of it $x-2$ .

To divide the polynomial $3x^4-6x^3-5x+10$ by $x-2$ follow the steps provided below,

Here dividend is $3x^4-6x^3+0\cdot x^2-5x+10$ and divisor is $x-2$ .

Step 1: List down the coefficients of dividend in the descending powers of x, that are $3,-6,0,-5,10$ .

Now, the divisor is of the form $x-c$ , where $c=2$ .

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}3& -6& 0& -5& 10\end{array}\\ \begin{array}{ccccc}& & & & \end{array}\\ \begin{array}{ccccc}& & & & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}3& -6& 0& -5& 10\end{array}\\ \begin{array}{ccccc}& & & & \end{array}\\ \begin{array}{ccccc}3& & & & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}3& -6& 0& -5& 10\end{array}\\ \begin{array}{ccccc}& 6& & & \end{array}\\ \begin{array}{ccccc}3& & & & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}3& -6& 0& -5& 10\end{array}\\ \begin{array}{ccccc}& 6& & & \end{array}\\ \begin{array}{ccccc}3& 0& & & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}3& -6& 0& -5& 10\end{array}\\ \begin{array}{ccccc}& 6& 0& & \end{array}\\ \begin{array}{ccccc}3& 0& & & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}3& -6& 0& -5& 10\end{array}\\ \begin{array}{ccccc}& 6& 0& & \end{array}\\ \begin{array}{ccccc}3& 0& 0& & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}3& -6& 0& -5& 10\end{array}\\ \begin{array}{ccccc}& 6& 0& 0& \end{array}\\ \begin{array}{ccccc}3& 0& 0& & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}3& -6& 0& -5& 10\end{array}\\ \begin{array}{ccccc}& 6& 0& 0& \end{array}\\ \begin{array}{ccccc}3& 0& 0& -5& \end{array}\end{array}}$

Repeat the above steps 8 and 9 for remaining elements of first row,

  $2\overline{)\begin{array}{l}\begin{array}{ccccc}3& -6& 0& -5& 10\end{array}\\ \begin{array}{ccccc}& 6& 0& 0& -10\end{array}\\ \begin{array}{ccccc}3& 0& 0& -5& 0\end{array}\end{array}}$

Now, the first four 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 $3x^3-5$ 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.

  $3x^4-6x^3-5x+10=\left(3x^3-5\right)\left(x-2\right)+0$

Now, $\left(x-c\right)$ is a factor of a polynomial if remainder is 0. Here, remainder is 0 so, $x-2$ is a factor.

Thus, the polynomial $3x^4-6x^3-5x+10$ has $x-2$ as its factor.