Examples
4x3+11x2−8x−10 , x+3
Step 1
Divide the higher order polynomial by the other polynomial in order to find the remainder.
4x3+11x2−8x−10x+3
Step 2
Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of 0.
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 |
Step 3
Divide the highest order term in the dividend 4x3 by the highest order term in divisor x.
4x2 | |||||||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 |
Step 4
Multiply the new quotient term by the divisor.
4x2 | |||||||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
+ | 4x3 | + | 12x2 |
Step 5
The expression needs to be subtracted from the dividend, so change all the signs in 4x3+12x2
4x2 | |||||||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 |
Step 6
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
4x2 | |||||||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 | ||||||||
- | x2 |
Step 7
Pull the next terms from the original dividend down into the current dividend.
4x2 | |||||||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 | ||||||||
- | x2 | - | 8x |
Step 8
Divide the highest order term in the dividend −x2 by the highest order term in divisor x.
4x2 | - | x | |||||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 | ||||||||
- | x2 | - | 8x |
Step 9
Multiply the new quotient term by the divisor.
4x2 | - | x | |||||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 | ||||||||
- | x2 | - | 8x | ||||||||
- | x2 | - | 3x |
Step 10
The expression needs to be subtracted from the dividend, so change all the signs in −x2−3x
4x2 | - | x | |||||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 | ||||||||
- | x2 | - | 8x | ||||||||
+ | x2 | + | 3x |
Step 11
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
4x2 | - | x | |||||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 | ||||||||
- | x2 | - | 8x | ||||||||
+ | x2 | + | 3x | ||||||||
- | 5x |
Step 12
Pull the next terms from the original dividend down into the current dividend.
4x2 | - | x | |||||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 | ||||||||
- | x2 | - | 8x | ||||||||
+ | x2 | + | 3x | ||||||||
- | 5x | - | 10 |
Step 13
Divide the highest order term in the dividend −5x by the highest order term in divisor x.
4x2 | - | x | - | 5 | |||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 | ||||||||
- | x2 | - | 8x | ||||||||
+ | x2 | + | 3x | ||||||||
- | 5x | - | 10 |
Step 14
Multiply the new quotient term by the divisor.
4x2 | - | x | - | 5 | |||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 | ||||||||
- | x2 | - | 8x | ||||||||
+ | x2 | + | 3x | ||||||||
- | 5x | - | 10 | ||||||||
- | 5x | - | 15 |
Step 15
The expression needs to be subtracted from the dividend, so change all the signs in −5x−15
4x2 | - | x | - | 5 | |||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 | ||||||||
- | x2 | - | 8x | ||||||||
+ | x2 | + | 3x | ||||||||
- | 5x | - | 10 | ||||||||
+ | 5x | + | 15 |
Step 16
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
4x2 | - | x | - | 5 | |||||||
x | + | 3 | 4x3 | + | 11x2 | - | 8x | - | 10 | ||
- | 4x3 | - | 12x2 | ||||||||
- | x2 | - | 8x | ||||||||
+ | x2 | + | 3x | ||||||||
- | 5x | - | 10 | ||||||||
+ | 5x | + | 15 | ||||||||
+ | 5 |
Step 17
The final answer is the quotient plus the remainder over the divisor.
4x2−x−5+5x+3
Step 18
The remainder is the part of the answer that is left after the division by x+3 is complete.
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