Remainder theorem Examples

1. Find remainder theorem of `x^2-10x+25` and `x-5`

Solution:
`(x^2-10x+25)/(x-5)` using remainder theorem

Remainder theorem : If an expression `f(x)` is divided by `(x-a)` then the remainder is `f(a)`.

To determine root divisor, we have to solve divisor equation `x-5=0`

`:.x=5`

So `f(x)=x^2-10x+25`

we have to find `f(5)`

`f(5)= 5^2-10 xx 5+25`

`=25-50+25`

`=0`

`:.` The remainder is `0`.

Method-2: solution using synthetic division method

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

`5`	`1`	`-10`	`25`
	``	``	``

Step-1 : Write down the first coefficient `1`

`5`	`1`	`-10`	`25`
	``	``	``
	`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 remainder theorem of `x^3+4x^2-4x-16` and `x-2`

Solution:
`(x^3+4x^2-4x-16)/(x-2)` using remainder theorem

Remainder theorem : If an expression `f(x)` is divided by `(x-a)` then the remainder is `f(a)`.

To determine root divisor, we have to solve divisor equation `x-2=0`

`:.x=2`

So `f(x)=x^3+4x^2-4x-16`

we have to find `f(2)`

`f(2)= 2^3+4 xx 2^2-4 xx 2-16`

`=8+16-8-16`

`=0`

`:.` The remainder is `0`.

Method-2: solution using synthetic division method

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 remainder theorem of `x^3+6x^2+12x+10` and `x+2`

Solution:
`(x^3+6x^2+12x+10)/(x+2)` using remainder theorem

Remainder theorem : If an expression `f(x)` is divided by `(x-a)` then the remainder is `f(a)`.

To determine root divisor, we have to solve divisor equation `x+2=0`

`:.x=-2`

So `f(x)=x^3+6x^2+12x+10`

we have to find `f(-2)`

`f(-2)= (-2)^3+6 xx (-2)^2+12 xx (-2)+10`

`=-8+24-24+10`

`=2`

`:.` The remainder is `2`.

Method-2: solution using synthetic division method

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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5. polynomial Synthetic division
(Previous method)

1. Examples