Examples
2x3-3x2+4x-12x3−3x2+4x−1 , x+1x+1
Step 1
Divide the higher order polynomial by the other polynomial in order to find the remainder.
2x3-3x2+4x-1x+12x3−3x2+4x−1x+1
Step 2
Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of 00.
xx | + | 11 | 2x32x3 | - | 3x23x2 | + | 4x4x | - | 11 |
Step 3
Divide the highest order term in the dividend 2x32x3 by the highest order term in divisor xx.
2x22x2 | |||||||||||
xx | + | 11 | 2x32x3 | - | 3x23x2 | + | 4x4x | - | 11 |
Step 4
Multiply the new quotient term by the divisor.
2x22x2 | |||||||||||
xx | + | 11 | 2x32x3 | - | 3x23x2 | + | 4x4x | - | 11 | ||
+ | 2x32x3 | + | 2x22x2 |
Step 5
The expression needs to be subtracted from the dividend, so change all the signs in 2x3+2x22x3+2x2
2x22x2 | |||||||||||
xx | + | 11 | 2x32x3 | - | 3x23x2 | + | 4x4x | - | 11 | ||
- | 2x32x3 | - | 2x22x2 |
Step 6
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
2x22x2 | |||||||||||
xx | + | 11 | 2x32x3 | - | 3x23x2 | + | 4x4x | - | 11 | ||
- | 2x32x3 | - | 2x22x2 | ||||||||
- | 5x25x2 |
Step 7
Pull the next terms from the original dividend down into the current dividend.
2x22x2 | |||||||||||
xx | + | 1 | 2x3 | - | 3x2 | + | 4x | - | 1 | ||
- | 2x3 | - | 2x2 | ||||||||
- | 5x2 | + | 4x |
Step 8
Divide the highest order term in the dividend -5x2 by the highest order term in divisor x.
2x2 | - | 5x | |||||||||
x | + | 1 | 2x3 | - | 3x2 | + | 4x | - | 1 | ||
- | 2x3 | - | 2x2 | ||||||||
- | 5x2 | + | 4x |
Step 9
Multiply the new quotient term by the divisor.
2x2 | - | 5x | |||||||||
x | + | 1 | 2x3 | - | 3x2 | + | 4x | - | 1 | ||
- | 2x3 | - | 2x2 | ||||||||
- | 5x2 | + | 4x | ||||||||
- | 5x2 | - | 5x |
Step 10
The expression needs to be subtracted from the dividend, so change all the signs in -5x2-5x
2x2 | - | 5x | |||||||||
x | + | 1 | 2x3 | - | 3x2 | + | 4x | - | 1 | ||
- | 2x3 | - | 2x2 | ||||||||
- | 5x2 | + | 4x | ||||||||
+ | 5x2 | + | 5x |
Step 11
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
2x2 | - | 5x | |||||||||
x | + | 1 | 2x3 | - | 3x2 | + | 4x | - | 1 | ||
- | 2x3 | - | 2x2 | ||||||||
- | 5x2 | + | 4x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 9x |
Step 12
Pull the next terms from the original dividend down into the current dividend.
2x2 | - | 5x | |||||||||
x | + | 1 | 2x3 | - | 3x2 | + | 4x | - | 1 | ||
- | 2x3 | - | 2x2 | ||||||||
- | 5x2 | + | 4x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 9x | - | 1 |
Step 13
Divide the highest order term in the dividend 9x by the highest order term in divisor x.
2x2 | - | 5x | + | 9 | |||||||
x | + | 1 | 2x3 | - | 3x2 | + | 4x | - | 1 | ||
- | 2x3 | - | 2x2 | ||||||||
- | 5x2 | + | 4x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 9x | - | 1 |
Step 14
Multiply the new quotient term by the divisor.
2x2 | - | 5x | + | 9 | |||||||
x | + | 1 | 2x3 | - | 3x2 | + | 4x | - | 1 | ||
- | 2x3 | - | 2x2 | ||||||||
- | 5x2 | + | 4x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 9x | - | 1 | ||||||||
+ | 9x | + | 9 |
Step 15
The expression needs to be subtracted from the dividend, so change all the signs in 9x+9
2x2 | - | 5x | + | 9 | |||||||
x | + | 1 | 2x3 | - | 3x2 | + | 4x | - | 1 | ||
- | 2x3 | - | 2x2 | ||||||||
- | 5x2 | + | 4x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 9x | - | 1 | ||||||||
- | 9x | - | 9 |
Step 16
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
2x2 | - | 5x | + | 9 | |||||||
x | + | 1 | 2x3 | - | 3x2 | + | 4x | - | 1 | ||
- | 2x3 | - | 2x2 | ||||||||
- | 5x2 | + | 4x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 9x | - | 1 | ||||||||
- | 9x | - | 9 | ||||||||
- | 10 |
Step 17
The final answer is the quotient plus the remainder over the divisor.
2x2-5x+9-10x+1
Step 18
The remainder is the part of the answer that is left after the division by x+1 is complete.
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