Answer:
(a + 7)(a² + 15a + 59) + 0 = a³ + 15a² + 59a + 12
Explanation:
To solve this problem, we can use the long division method.
First, we need to divide the first term of the dividend, which is a³, by the first term of the divisor, which is a. This gives us a quotient of a² with a remainder of 0. We write this as follows:
a³ + 15a² + 59a + 12
a + 7 |
—————
a² | 0
Next, we multiply the divisor by the quotient and subtract it from the dividend, giving us a new dividend of 15a² + 59a + 12. We write this as follows:
a³ + 15a² + 59a + 12
a + 7 |
—————
a² | 0
- |
a² | 15a² + 59a + 12
Next, we divide the first term of the new dividend, which is 15a², by the first term of the divisor, which is a. This gives us a quotient of 15a with a remainder of 0. We write this as follows:
a³ + 15a² + 59a + 12
a + 7 |
—————
a² | 0
- |
a² | 15a² + 59a + 12
|
| 15a | 0
Next, we multiply the divisor by the quotient and subtract it from the dividend, giving us a new dividend of 59a + 12. We write this as follows:
a³ + 15a² + 59a + 12
a + 7 |
—————
a² | 0
- |
a² | 15a² + 59a + 12
|
| 15a | 0
- |
15a | 59a + 12
Finally, we divide the first term of the new dividend, which is 59a, by the first term of the divisor, which is a. This gives us a quotient of 59 with a remainder of 0. We write this as follows:
a³ + 15a² + 59a + 12
a + 7 |
—————
a² | 0
- |
a² | 15a² + 59a + 12
|
| 15a | 0
- |
15a | 59a + 12
|
| 59 | 0
Since the remainder is 0, this means that the expression (a³ + 15a² + 59a + 12) is divisible by (a + 7). Therefore, the quotient is:
a³ + 15a² + 59a + 12
a + 7
—————
a² + 15a + 59
We can check our answer by multiplying the divisor by the quotient and adding the remainder (which is 0), and verifying that the result is equal to the original dividend
(a + 7)(a² + 15a + 59) + 0 = a³ + 15a² + 59a + 12