Algebra Examples
x3-5x+6x3−5x+6 , x+2x+2
Step 1
Divide the higher order polynomial by the other polynomial in order to find the remainder.
x3-5x+6x+2x3−5x+6x+2
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 | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 |
Step 3
Divide the highest order term in the dividend x3x3 by the highest order term in divisor xx.
x2x2 | |||||||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 |
Step 4
Multiply the new quotient term by the divisor.
x2x2 | |||||||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
+ | x3x3 | + | 2x22x2 |
Step 5
The expression needs to be subtracted from the dividend, so change all the signs in x3+2x2x3+2x2
x2x2 | |||||||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
- | x3x3 | - | 2x22x2 |
Step 6
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2x2 | |||||||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
- | x3x3 | - | 2x22x2 | ||||||||
- | 2x22x2 |
Step 7
Pull the next terms from the original dividend down into the current dividend.
x2x2 | |||||||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
- | x3x3 | - | 2x22x2 | ||||||||
- | 2x22x2 | - | 5x5x |
Step 8
Divide the highest order term in the dividend -2x2−2x2 by the highest order term in divisor xx.
x2x2 | - | 2x2x | |||||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
- | x3x3 | - | 2x22x2 | ||||||||
- | 2x22x2 | - | 5x5x |
Step 9
Multiply the new quotient term by the divisor.
x2x2 | - | 2x2x | |||||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
- | x3x3 | - | 2x22x2 | ||||||||
- | 2x22x2 | - | 5x5x | ||||||||
- | 2x22x2 | - | 4x4x |
Step 10
The expression needs to be subtracted from the dividend, so change all the signs in -2x2-4x−2x2−4x
x2x2 | - | 2x2x | |||||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
- | x3x3 | - | 2x22x2 | ||||||||
- | 2x22x2 | - | 5x5x | ||||||||
+ | 2x22x2 | + | 4x4x |
Step 11
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2x2 | - | 2x2x | |||||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
- | x3x3 | - | 2x22x2 | ||||||||
- | 2x22x2 | - | 5x5x | ||||||||
+ | 2x22x2 | + | 4x4x | ||||||||
- | xx |
Step 12
Pull the next terms from the original dividend down into the current dividend.
x2x2 | - | 2x2x | |||||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
- | x3x3 | - | 2x22x2 | ||||||||
- | 2x22x2 | - | 5x5x | ||||||||
+ | 2x22x2 | + | 4x4x | ||||||||
- | xx | + | 66 |
Step 13
Divide the highest order term in the dividend -x−x by the highest order term in divisor xx.
x2x2 | - | 2x2x | - | 11 | |||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
- | x3x3 | - | 2x22x2 | ||||||||
- | 2x22x2 | - | 5x5x | ||||||||
+ | 2x22x2 | + | 4x4x | ||||||||
- | xx | + | 66 |
Step 14
Multiply the new quotient term by the divisor.
x2x2 | - | 2x2x | - | 11 | |||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
- | x3x3 | - | 2x22x2 | ||||||||
- | 2x22x2 | - | 5x5x | ||||||||
+ | 2x22x2 | + | 4x4x | ||||||||
- | xx | + | 66 | ||||||||
- | xx | - | 22 |
Step 15
The expression needs to be subtracted from the dividend, so change all the signs in -x-2−x−2
x2x2 | - | 2x2x | - | 11 | |||||||
xx | + | 22 | x3x3 | + | 0x20x2 | - | 5x5x | + | 66 | ||
- | x3x3 | - | 2x2 | ||||||||
- | 2x2 | - | 5x | ||||||||
+ | 2x2 | + | 4x | ||||||||
- | x | + | 6 | ||||||||
+ | x | + | 2 |
Step 16
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2 | - | 2x | - | 1 | |||||||
x | + | 2 | x3 | + | 0x2 | - | 5x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | 2x2 | - | 5x | ||||||||
+ | 2x2 | + | 4x | ||||||||
- | x | + | 6 | ||||||||
+ | x | + | 2 | ||||||||
+ | 8 |
Step 17
The final answer is the quotient plus the remainder over the divisor.
x2-2x-1+8x+2
Step 18
The remainder is the part of the answer that is left after the division by x+2 is complete.
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