Answer :
It seems there might be a misunderstanding in how the problem is presented. Typically, synthetic division involves dividing a polynomial by a linear divisor of the form[tex]\( x - c \)[/tex], where [tex]\( c \)[/tex] is a constant. The expression[tex]\( 1\sqrt{4} 6 -1 \)[/tex] doesn't directly fit into the standard format for synthetic division.
However, if we interpret the problem as dividing the polynomial [tex]\( x^3 + 6x - 1 \)[/tex] by [tex]\( x - \sqrt{4} \)[/tex], which simplifies to [tex]\( x - 2 \)[/tex]since [tex]\( \sqrt{4} = 2 \),[/tex] then we can proceed with the synthetic division as follows:
1. Write down the coefficients of the polynomial [tex]\( x^3 + 6x - 1 \):[/tex] [tex]\( 1, 0, 6, -1 \).[/tex]
2. Since we are dividing by[tex]\( x - 2 \)[/tex], use [tex]\( 2 \)[/tex]as the divisor.
3. Set up the synthetic division table:
\[
\begin{array}{r|rrr}
2 & 1 & 0 & 6 & -1 \\
& & 2 & 4 & 10 \\
\hline
& 1 & 2 & 10 & 9 \\
\end{array}
\]
4. Perform the synthetic division:
- Bring down the first coefficient [tex]\( 1 \).[/tex]
- Multiply [tex]\( 1 \)[/tex]by [tex]\( 2 \)[/tex]and write the result under the next coefficient: [tex]\( 2 \).[/tex]
- Add [tex]\( 0 + 2 = 2 \).[/tex]
- Multiply [tex]\( 2 \)[/tex] by [tex]\( 2 \)[/tex] and write the result under the next coefficient: [tex]\( 4 \).[/tex]
- Add [tex]\( 6 + 4 = 10 \).[/tex]
- Multiply [tex]\( 10 \) by \( 2 \)[/tex] and write the result under the next coefficient: [tex]\( 20 \).[/tex]
- Add [tex]\( -1 + 20 = 19 \).[/tex]
5. The resulting coefficients from synthetic division are [tex]\( 1, 2, 10, 19 \).[/tex]
Therefore, the quotient when dividing [tex]\( x^3 + 6x - 1 \)[/tex]by [tex]\( x - \sqrt{4} \)[/tex] (or [tex]\( x - 2 \))[/tex] is[tex]\( x^2 + 2x + 10 \)[/tex] with a remainder of [tex]\( 19 \).[/tex]
