1. Examples
1. Find synthetic division of `x^2-10x+25` and `x-5`
Solution:
`(x^2-10x+25)/(x-5)` using synthetic division
To determine root divisor, we have to solve divisor equation `x-5=0`
`:.` our root becomes `x=5`
Write coefficients of the dividend `x^2-10x+25` to the right and our root `5` to the left
Step-1 : Write down the first coefficient `1`
Step-2 : Multiply our root `5` by our last result `1` to get `5` [ `5` × `1`=`5` ]
| `5` | `1` | `-10` | `25` | | `` | `5` | `` | | `1` | | |
Step-3 : Add new result `5` to the next coefficient of the dividend `-10`, and write down the sum `-5`, [ `(-10)` + `5`=`-5` ]
| `5` | `1` | `-10` | `25` | | `` | `5` | `` | | `1` | `-5` | |
Step-4 : Multiply our root `5` by our last result `-5` to get `-25` [ `5` × `(-5)`=`-25` ]
| `5` | `1` | `-10` | `25` | | `` | `5` | `-25` | | `1` | `-5` | |
Step-5 : Add new result `-25` to the next coefficient of the dividend `25`, and write down the sum `0`, [ `25` + `(-25)`=`0` ]
| `5` | `1` | `-10` | `25` | | `` | `5` | `-25` | | `1` | `-5` | `0` |
We have completed the table and have obtained the following coefficients
`1,-5,0`
All coefficients, except last one, are coefficients of quotient, last coefficient is remainder.
Thus quotient is `x-5` and remainder is `0`
2. Find synthetic division of `x^3+4x^2-4x-16` and `x-2`
Solution:
`(x^3+4x^2-4x-16)/(x-2)` using synthetic division
To determine root divisor, we have to solve divisor equation `x-2=0`
`:.` our root becomes `x=2`
Write coefficients of the dividend `x^3+4x^2-4x-16` to the right and our root `2` to the left
| `2` | `1` | `4` | `-4` | `-16` | | `` | `` | `` | `` | | | | | |
Step-1 : Write down the first coefficient `1`
| `2` | `1` | `4` | `-4` | `-16` | | `` | `` | `` | `` | | `1` | | | |
Step-2 : Multiply our root `2` by our last result `1` to get `2` [ `2` × `1`=`2` ]
| `2` | `1` | `4` | `-4` | `-16` | | `` | `2` | `` | `` | | `1` | | | |
Step-3 : Add new result `2` to the next coefficient of the dividend `4`, and write down the sum `6`, [ `4` + `2`=`6` ]
| `2` | `1` | `4` | `-4` | `-16` | | `` | `2` | `` | `` | | `1` | `6` | | |
Step-4 : Multiply our root `2` by our last result `6` to get `12` [ `2` × `6`=`12` ]
| `2` | `1` | `4` | `-4` | `-16` | | `` | `2` | `12` | `` | | `1` | `6` | | |
Step-5 : Add new result `12` to the next coefficient of the dividend `-4`, and write down the sum `8`, [ `(-4)` + `12`=`8` ]
| `2` | `1` | `4` | `-4` | `-16` | | `` | `2` | `12` | `` | | `1` | `6` | `8` | |
Step-6 : Multiply our root `2` by our last result `8` to get `16` [ `2` × `8`=`16` ]
| `2` | `1` | `4` | `-4` | `-16` | | `` | `2` | `12` | `16` | | `1` | `6` | `8` | |
Step-7 : Add new result `16` to the next coefficient of the dividend `-16`, and write down the sum `0`, [ `(-16)` + `16`=`0` ]
| `2` | `1` | `4` | `-4` | `-16` | | `` | `2` | `12` | `16` | | `1` | `6` | `8` | `0` |
We have completed the table and have obtained the following coefficients
`1,6,8,0`
All coefficients, except last one, are coefficients of quotient, last coefficient is remainder.
Thus quotient is `x^2+6x+8` and remainder is `0`
3. Find synthetic division of `x^3+6x^2+12x+10` and `x+2`
Solution:
`(x^3+6x^2+12x+10)/(x+2)` using synthetic division
To determine root divisor, we have to solve divisor equation `x+2=0`
`:.` our root becomes `x=-2`
Write coefficients of the dividend `x^3+6x^2+12x+10` to the right and our root `-2` to the left
| `-2` | `1` | `6` | `12` | `10` | | `` | `` | `` | `` | | | | | |
Step-1 : Write down the first coefficient `1`
| `-2` | `1` | `6` | `12` | `10` | | `` | `` | `` | `` | | `1` | | | |
Step-2 : Multiply our root `-2` by our last result `1` to get `-2` [ `(-2)` × `1`=`-2` ]
| `-2` | `1` | `6` | `12` | `10` | | `` | `-2` | `` | `` | | `1` | | | |
Step-3 : Add new result `-2` to the next coefficient of the dividend `6`, and write down the sum `4`, [ `6` + `(-2)`=`4` ]
| `-2` | `1` | `6` | `12` | `10` | | `` | `-2` | `` | `` | | `1` | `4` | | |
Step-4 : Multiply our root `-2` by our last result `4` to get `-8` [ `(-2)` × `4`=`-8` ]
| `-2` | `1` | `6` | `12` | `10` | | `` | `-2` | `-8` | `` | | `1` | `4` | | |
Step-5 : Add new result `-8` to the next coefficient of the dividend `12`, and write down the sum `4`, [ `12` + `(-8)`=`4` ]
| `-2` | `1` | `6` | `12` | `10` | | `` | `-2` | `-8` | `` | | `1` | `4` | `4` | |
Step-6 : Multiply our root `-2` by our last result `4` to get `-8` [ `(-2)` × `4`=`-8` ]
| `-2` | `1` | `6` | `12` | `10` | | `` | `-2` | `-8` | `-8` | | `1` | `4` | `4` | |
Step-7 : Add new result `-8` to the next coefficient of the dividend `10`, and write down the sum `2`, [ `10` + `(-8)`=`2` ]
| `-2` | `1` | `6` | `12` | `10` | | `` | `-2` | `-8` | `-8` | | `1` | `4` | `4` | `2` |
We have completed the table and have obtained the following coefficients
`1,4,4,2`
All coefficients, except last one, are coefficients of quotient, last coefficient is remainder.
Thus quotient is `x^2+4x+4` and remainder is `2`
This material is intended as a summary. Use your textbook for detail explanation.
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