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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