Chapter A.4, Problem 25AYU

To determine

To calculate:Whether $x-\frac{1}{2}$ is a factor of the polynomial $2x^4-x^3+2x-1$ with help of synthetic division.

Answer to Problem 25AYU

The polynomial $2x^4-x^3+2x-1$has$x-\frac{1}{2}$ as its factor.

Explanation of Solution

Given information:

The polynomial $2x^4-x^3+2x-1$ and a factor of it $x-\frac{1}{2}$ .

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 $2x^4-x^3+2x-1$ and a factor of it $x-\frac{1}{2}$ .

To divide the polynomial $2x^4-x^3+2x-1$ by $x-\frac{1}{2}$ follow the steps provided below,

Here dividend is $2x^4-x^3+0\cdot x^2+2x-1$ and divisor is $x-\frac{1}{2}$ .

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

Now, the divisor is of the form $x-c$ , where $c=\frac{1}{2}$ .

Step 2: To perform the synthetic division put the coefficients in the division sign along with $\frac{1}{2}$ on the left side,

  $\frac{1}{2}\overline{)\begin{array}{l}\begin{array}{ccccc}2& -1& 0& 2& -1\end{array}\\ \begin{array}{ccccc}& & & & \end{array}\\ \begin{array}{ccccc}& & & & \end{array}\end{array}}$

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

  $\frac{1}{2}\overline{)\begin{array}{l}\begin{array}{ccccc}2& -1& 0& 2& -1\end{array}\\ \begin{array}{ccccc}& & & & \end{array}\\ \begin{array}{ccccc}2& & & & \end{array}\end{array}}$

Step 4: Multiply $\frac{1}{2}$ with the first entry of row 3 and place the result in row 2 column 2.

  $\frac{1}{2}\overline{)\begin{array}{l}\begin{array}{ccccc}2& -1& 0& 2& -1\end{array}\\ \begin{array}{ccccc}& 1& & & \end{array}\\ \begin{array}{ccccc}2& & & & \end{array}\end{array}}$

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

  $\frac{1}{2}\overline{)\begin{array}{l}\begin{array}{ccccc}2& -1& 0& 2& -1\end{array}\\ \begin{array}{ccccc}& 1& & & \end{array}\\ \begin{array}{ccccc}2& 0& & & \end{array}\end{array}}$

Step 6: Multiply $\frac{1}{2}$ with the second entry of row 3 and place the result in row 2 column 3.

  $\frac{1}{2}\overline{)\begin{array}{l}\begin{array}{ccccc}2& -1& 0& 2& -1\end{array}\\ \begin{array}{ccccc}& 1& 0& & \end{array}\\ \begin{array}{ccccc}2& 0& & & \end{array}\end{array}}$

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

  $\frac{1}{2}\overline{)\begin{array}{l}\begin{array}{ccccc}2& -1& 0& 2& -1\end{array}\\ \begin{array}{ccccc}& 1& 0& & \end{array}\\ \begin{array}{ccccc}2& 0& 0& & \end{array}\end{array}}$

Step 8: Multiply $\frac{1}{2}$ with the third entry of row 3 and place the result in row 2 column 4.

  $\frac{1}{2}\overline{)\begin{array}{l}\begin{array}{ccccc}2& -1& 0& 2& -1\end{array}\\ \begin{array}{ccccc}& 1& 0& 0& \end{array}\\ \begin{array}{ccccc}2& 0& 0& 2& \end{array}\end{array}}$

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

  $\frac{1}{2}\overline{)\begin{array}{l}\begin{array}{ccccc}2& -1& 0& 2& -1\end{array}\\ \begin{array}{ccccc}& 1& 0& 0& 1\end{array}\\ \begin{array}{ccccc}2& 0& 0& 2& \end{array}\end{array}}$

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

  $\frac{1}{2}\overline{)\begin{array}{l}\begin{array}{ccccc}2& -1& 0& 2& -1\end{array}\\ \begin{array}{ccccc}& 1& 0& 0& 1\end{array}\\ \begin{array}{ccccc}2& 0& 0& 2& 0\end{array}\end{array}}$

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 $2x^3+2$ 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.

  $2x^4-x^3+2x-1=\left(2x^3+2\right)\left(x-\frac{1}{2}\right)+0$

Now, $\left(x-c\right)$ is a factor of a polynomial if remainder is 0. Here, remainder is 0 so, $x-\frac{1}{2}$ is a factor.

Thus, the polynomial $2x^4-x^3+2x-1$ has $x-\frac{1}{2}$ as its factor.