Question

(K3+2k2-82k-28)/(k+10)

Answer 1

So for this, we will be using synthetic division. To set it up, have the equation so that the divisor is -10 (since that is the solution of k + 10 = 0) and the dividend are the coefficients. Our equation will look as such:

(Note that synthetic division can only be used when the divisor is a 1st degree binomial)

• -10 | 1 + 2 - 82 - 28
• ---------------------------

Now firstly, drop the 1:

• -10 | 1 + 2 - 82 - 28
•       ↓
• -------------------------
•        1

Next, you are going to multiply -10 and 1, and then combine the product with 2.

• -10 | 1 + 2 - 82 - 28
•       ↓ - 10
• -------------------------
•        1 - 8

Next, multiply -10 and -8, then combine the product with -82:

• -10 | 1 + 2 - 82 - 28
•       ↓ -10 + 80
• -------------------------
•        1 - 8 - 2

Next, multiply -10 and -2, then combine the product with -28:

• -10 | 1 + 2 - 82 - 28
•       ↓ -10 + 80 + 20
• -------------------------
•        1 - 8 - 2 - 8

Now, since we know that the degree of the dividend is 3, this means that the degree of the quotient is 2. Using this, the first 3 terms are k^2, k, and the constant, or in this case k² - 8k - 2. Now what about the last coefficient -8? Well this is our remainder, and will be written as -8/(k + 10).

Putting it together, the quotient is $k^2-8k-2-\frac{8}{k+10}$