Finite Math Examples
x3-x2+7xx-5x3−x2+7xx−5
Step 1
Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of 00.
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 |
Step 2
Divide the highest order term in the dividend x3x3 by the highest order term in divisor xx.
x2x2 | |||||||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 |
Step 3
Multiply the new quotient term by the divisor.
x2x2 | |||||||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
+ | x3x3 | - | 5x25x2 |
Step 4
The expression needs to be subtracted from the dividend, so change all the signs in x3-5x2x3−5x2
x2x2 | |||||||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 |
Step 5
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2x2 | |||||||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 | ||||||||
+ | 4x24x2 |
Step 6
Pull the next terms from the original dividend down into the current dividend.
x2x2 | |||||||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 | ||||||||
+ | 4x24x2 | + | 7x7x |
Step 7
Divide the highest order term in the dividend 4x24x2 by the highest order term in divisor xx.
x2x2 | + | 4x4x | |||||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 | ||||||||
+ | 4x24x2 | + | 7x7x |
Step 8
Multiply the new quotient term by the divisor.
x2x2 | + | 4x4x | |||||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 | ||||||||
+ | 4x24x2 | + | 7x7x | ||||||||
+ | 4x24x2 | - | 20x20x |
Step 9
The expression needs to be subtracted from the dividend, so change all the signs in 4x2-20x4x2−20x
x2x2 | + | 4x4x | |||||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 | ||||||||
+ | 4x24x2 | + | 7x7x | ||||||||
- | 4x24x2 | + | 20x20x |
Step 10
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2x2 | + | 4x4x | |||||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 | ||||||||
+ | 4x24x2 | + | 7x7x | ||||||||
- | 4x24x2 | + | 20x20x | ||||||||
+ | 27x27x |
Step 11
Pull the next terms from the original dividend down into the current dividend.
x2x2 | + | 4x4x | |||||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 | ||||||||
+ | 4x24x2 | + | 7x7x | ||||||||
- | 4x24x2 | + | 20x20x | ||||||||
+ | 27x27x | + | 00 |
Step 12
Divide the highest order term in the dividend 27x27x by the highest order term in divisor xx.
x2x2 | + | 4x4x | + | 2727 | |||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 | ||||||||
+ | 4x24x2 | + | 7x7x | ||||||||
- | 4x24x2 | + | 20x20x | ||||||||
+ | 27x27x | + | 00 |
Step 13
Multiply the new quotient term by the divisor.
x2x2 | + | 4x4x | + | 2727 | |||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 | ||||||||
+ | 4x24x2 | + | 7x7x | ||||||||
- | 4x24x2 | + | 20x20x | ||||||||
+ | 27x27x | + | 00 | ||||||||
+ | 27x27x | - | 135135 |
Step 14
The expression needs to be subtracted from the dividend, so change all the signs in 27x-13527x−135
x2x2 | + | 4x4x | + | 2727 | |||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 | ||||||||
+ | 4x24x2 | + | 7x7x | ||||||||
- | 4x24x2 | + | 20x20x | ||||||||
+ | 27x27x | + | 00 | ||||||||
- | 27x27x | + | 135135 |
Step 15
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2x2 | + | 4x4x | + | 2727 | |||||||
xx | - | 55 | x3x3 | - | x2x2 | + | 7x7x | + | 00 | ||
- | x3x3 | + | 5x25x2 | ||||||||
+ | 4x24x2 | + | 7x7x | ||||||||
- | 4x24x2 | + | 20x20x | ||||||||
+ | 27x27x | + | 00 | ||||||||
- | 27x27x | + | 135135 | ||||||||
+ | 135135 |
Step 16
The final answer is the quotient plus the remainder over the divisor.
x2+4x+27+135x-5x2+4x+27+135x−5