Answer:
x^3 + 9 x^2 + 21 x + 9 = (x^2 + 6 x + 3)×(x + 3) + 0
Step-by-step explanation:
Set up the polynomial long division problem with a division bracket, putting the numerator inside and the denominator on the left:
x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9
To eliminate the leading term of the numerator, x^3, multiply x + 3 by x^2 to get x^3 + 3 x^2. Write x^2 on top of the division bracket and subtract x^3 + 3 x^2 from x^3 + 9 x^2 + 21 x + 9 to get 6 x^2 + 21 x + 9:
| | | x^2 | | | |
x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9
| -(x^3 | + | 3 x^2) | | | |
| | | 6 x^2 | + | 21 x | + | 9
To eliminate the leading term of the remainder of the previous step, 6 x^2, multiply x + 3 by 6 x to get 6 x^2 + 18 x. Write 6 x on top of the division bracket and subtract 6 x^2 + 18 x from 6 x^2 + 21 x + 9 to get 3 x + 9:
| | | x^2 | + | 6 x | |
x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9
| -(x^3 | + | 3 x^2) | | | |
| | | 6 x^2 | + | 21 x | + | 9
| | | -(6 x^2 | + | 18 x) | |
| | | | | 3 x | + | 9
To eliminate the leading term of the remainder of the previous step, 3 x, multiply x + 3 by 3 to get 3 x + 9. Write 3 on top of the division bracket and subtract 3 x + 9 from 3 x + 9 to get 0:
| | | x^2 | + | 6 x | + | 3
x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9
| -(x^3 | + | 3 x^2) | | | |
| | | 6 x^2 | + | 21 x | + | 9
| | | -(6 x^2 | + | 18 x) | |
| | | | | 3 x | + | 9
| | | | | -(3 x | + | 9)
| | | | | | | 0
The quotient of (x^3 + 9 x^2 + 21 x + 9)/(x + 3) is the sum of the terms on top of the division bracket. Since the final subtraction step resulted in zero, x + 3 exactly divides x^3 + 9 x^2 + 21 x + 9 and there is no remainder.
| | | x^2 | + | 6 x | + | 3 | (quotient)
x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9 |
| -(x^3 | + | 3 x^2) | | | | |
| | | 6 x^2 | + | 21 x | + | 9 |
| | | -(6 x^2 | + | 18 x) | | |
| | | | | 3 x | + | 9 |
| | | | | -(3 x | + | 9) |
| | | | | | | 0 | (remainder) invisible comma
(x^3 + 9 x^2 + 21 x + 9)/(x + 3) = (x^2 + 6 x + 3) + 0
Write the result in quotient and remainder form:
Answer: Set up the polynomial long division problem with a division bracket, putting the numerator inside and the denominator on the left:
x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9
To eliminate the leading term of the numerator, x^3, multiply x + 3 by x^2 to get x^3 + 3 x^2. Write x^2 on top of the division bracket and subtract x^3 + 3 x^2 from x^3 + 9 x^2 + 21 x + 9 to get 6 x^2 + 21 x + 9:
| | | x^2 | | | |
x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9
| -(x^3 | + | 3 x^2) | | | |
| | | 6 x^2 | + | 21 x | + | 9
To eliminate the leading term of the remainder of the previous step, 6 x^2, multiply x + 3 by 6 x to get 6 x^2 + 18 x. Write 6 x on top of the division bracket and subtract 6 x^2 + 18 x from 6 x^2 + 21 x + 9 to get 3 x + 9:
| | | x^2 | + | 6 x | |
x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9
| -(x^3 | + | 3 x^2) | | | |
| | | 6 x^2 | + | 21 x | + | 9
| | | -(6 x^2 | + | 18 x) | |
| | | | | 3 x | + | 9
To eliminate the leading term of the remainder of the previous step, 3 x, multiply x + 3 by 3 to get 3 x + 9. Write 3 on top of the division bracket and subtract 3 x + 9 from 3 x + 9 to get 0:
| | | x^2 | + | 6 x | + | 3
x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9
| -(x^3 | + | 3 x^2) | | | |
| | | 6 x^2 | + | 21 x | + | 9
| | | -(6 x^2 | + | 18 x) | |
| | | | | 3 x | + | 9
| | | | | -(3 x | + | 9)
| | | | | | | 0
The quotient of (x^3 + 9 x^2 + 21 x + 9)/(x + 3) is the sum of the terms on top of the division bracket. Since the final subtraction step resulted in zero, x + 3 exactly divides x^3 + 9 x^2 + 21 x + 9 and there is no remainder.
| | | x^2 | + | 6 x | + | 3 | (quotient)
x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9 |
| -(x^3 | + | 3 x^2) | | | | |
| | | 6 x^2 | + | 21 x | + | 9 |
| | | -(6 x^2 | + | 18 x) | | |
| | | | | 3 x | + | 9 |
| | | | | -(3 x | + | 9) |
| | | | | | | 0 | (remainder) invisible comma
(x^3 + 9 x^2 + 21 x + 9)/(x + 3) = (x^2 + 6 x + 3) + 0
Write the result in quotient and remainder form:
Answer: x^3 + 9 x^2 + 21 x + 9 = (x^2 + 6 x + 3)×(x + 3) + 0
