Answer :
We are performing synthetic division with the divisor [tex]$1$[/tex] and the polynomial with coefficients [tex]$4$[/tex], [tex]$6$[/tex], and [tex]$-1$[/tex]. The process is as follows:
1. Write the coefficients:
[tex]$$4,\quad 6,\quad -1$$[/tex]
2. Bring down the first coefficient:
[tex]$$4$$[/tex]
3. Multiply the divisor [tex]$1$[/tex] by [tex]$4$[/tex] (the value just brought down) and add this to the second coefficient:
[tex]$$4 \times 1 = 4$$[/tex]
[tex]$$6 + 4 = 10$$[/tex]
4. Multiply the new value [tex]$10$[/tex] by the divisor [tex]$1$[/tex] and add this to the third coefficient:
[tex]$$10 \times 1 = 10$$[/tex]
[tex]$$-1 + 10 = 9$$[/tex]
The final value [tex]$9$[/tex] is the remainder.
Thus, the remainder in the synthetic division is [tex]$$\boxed{9}.$$[/tex]
1. Write the coefficients:
[tex]$$4,\quad 6,\quad -1$$[/tex]
2. Bring down the first coefficient:
[tex]$$4$$[/tex]
3. Multiply the divisor [tex]$1$[/tex] by [tex]$4$[/tex] (the value just brought down) and add this to the second coefficient:
[tex]$$4 \times 1 = 4$$[/tex]
[tex]$$6 + 4 = 10$$[/tex]
4. Multiply the new value [tex]$10$[/tex] by the divisor [tex]$1$[/tex] and add this to the third coefficient:
[tex]$$10 \times 1 = 10$$[/tex]
[tex]$$-1 + 10 = 9$$[/tex]
The final value [tex]$9$[/tex] is the remainder.
Thus, the remainder in the synthetic division is [tex]$$\boxed{9}.$$[/tex]
