Algebra 2 worksheet on dividing polynomials using long or synthetic division, with eight practice problems.

Let’s solve problem 1 step by step using long division.

Problem:
Divide (k³ + 8k² + 10k + 21) by (k + 7)

Step 1: Set up the long division.
We write it like this:

________________
k + 7 ) k³ + 8k² + 10k + 21

Step 2: Divide the first term of the dividend by the first term of the divisor.
k³ ÷ k = k² → Write k² above the division bar.

k²
________________
k + 7 ) k³ + 8k² + 10k + 21

Step 3: Multiply k² by (k + 7):
k² × k = k³
k² × 7 = 7k²
Write that under the first two terms:

k²
________________
k + 7 ) k³ + 8k² + 10k + 21
k³ + 7k²
-----------

Step 4: Subtract:
(k³ + 8k²) - (k³ + 7k²) = k²
Bring down the next term (+10k):

k²
________________
k + 7 ) k³ + 8k² + 10k + 21
k³ + 7k²
-----------
k² + 10k

Step 5: Repeat — divide k² by k → k
Write +k above the division bar.

k² + k
________________
k + 7 ) k³ + 8k² + 10k + 21
k³ + 7k²
-----------
k² + 10k

Step 6: Multiply k by (k + 7):
k × k = k²
k × 7 = 7k
Write under current line:

k² + k
________________
k + 7 ) k³ + 8k² + 10k + 21
k³ + 7k²
-----------
k² + 10k
k² + 7k
---------

Step 7: Subtract:
(k² + 10k) - (k² + 7k) = 3k
Bring down +21:

k² + k
________________
k + 7 ) k³ + 8k² + 10k + 21
k³ + 7k²
-----------
k² + 10k
k² + 7k
---------
3k + 21

Step 8: Divide 3k by k → 3
Write +3 above the division bar.

k² + k + 3
________________
k + 7 ) k³ + 8k² + 10k + 21
k³ + 7k²
-----------
k² + 10k
k² + 7k
---------
3k + 21

Step 9: Multiply 3 by (k + 7):
3 × k = 3k
3 × 7 = 21
Write under:

k² + k + 3
________________
k + 7 ) k³ + 8k² + 10k + 21
k³ + 7k²
-----------
k² + 10k
k² + 7k
---------
3k + 21
3k + 21
--------
0

Step 10: Subtract — remainder is 0.

✔ Final Answer: k² + k + 3

Final Answer:
k² + k + 3