Chapter A.4, Problem 17AYU

To determine

To calculate:Whether $x-2$ is a factor of the polynomial $4x^3-3x^2-8x+4$ with help of synthetic division.

Answer to Problem 17AYU

The polynomial $4x^3-3x^2-8x+4$does not have$x-2$ as its factor.

Explanation of Solution

Given information:

The polynomial $4x^3-3x^2-8x+4$ 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 $4x^3-3x^2-8x+4$ and a factor of it $x-2$ .

To divide the polynomial $4x^3-3x^2-8x+4$ by $x-2$ follow the steps provided below,

Here dividend is $4x^3-3x^2-8x+4$ and divisor is $x-2$ .

Step 1: List down the coefficients of dividend in the descending powers of x, that are $4,-3,-8,4$ .

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,

  $2\overline{)\begin{array}{cccc}4& -3& -8& 4\end{array}}$

Step 3: Bring down 4 to in third row,

  $2\overline{)\begin{array}{l}\begin{array}{cccc}4& -3& -8& 4\end{array}\\ \begin{array}{cccc}& & & \end{array}\\ 4\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{cccc}4& -3& -8& 4\end{array}\\ \begin{array}{cccc}& 8& & \end{array}\\ 4\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{cccc}4& -3& -8& 4\end{array}\\ \begin{array}{cccc}& 8& & \end{array}\\ \begin{array}{cccc}4& 5& & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{cccc}4& -3& -8& 4\end{array}\\ \begin{array}{cccc}& 8& 10& \end{array}\\ \begin{array}{cccc}4& 5& & \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{cccc}4& -3& -8& 4\end{array}\\ \begin{array}{cccc}& 8& 10& \end{array}\\ \begin{array}{cccc}4& 5& 2& \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{cccc}4& -3& -8& 4\end{array}\\ \begin{array}{cccc}& 8& 10& 4\end{array}\\ \begin{array}{cccc}4& 5& 2& \end{array}\end{array}}$

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

  $2\overline{)\begin{array}{l}\begin{array}{cccc}4& -3& -8& 4\end{array}\\ \begin{array}{cccc}& 8& 10& 4\end{array}\\ \begin{array}{cccc}4& 5& 2& 8\end{array}\end{array}}$

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 $4x^2+5x+2$ and reminder is 8.

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

  $4x^3-3x^2-8x+4=\left(4x^2+5x+2\right)\left(x-2\right)+8$

Now, $\left(x-c\right)$ is a factor of a polynomial if remainder is 0. Here, remainder is 8 so, $x-2$ is not a factor.

Thus, the polynomial $4x^3-3x^2-8x+4$ does not have $x-2$ as its factor.