In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree, a generalized version of the familiar arithmetic technique called long division. It can be done easily by hand, because it separates an otherwise complex division problem into smaller ones.
In algebra, the synthetic division is a method for manually performing Euclidean division of polynomials, with less writing and fewer calculations than the long division. It is mostly taught for division by linear monic polynomials, but the method can be generalized to division by any polynomial.
References:
https://en.wikipedia.org/wiki/Polynomial_long_division
https://en.wikipedia.org/wiki/Synthetic_division
2. Jahid as a Math Tutor
Sir Jahid Dalidig is a math tutor in a tutorial center..
He is preparing a test for his tutees for upcoming test, he want to
creat a worksheet for them to practice their skill in dividing
polynomials. So he will prepare a 10-item worksheet which includes
five items for long division and five items for synthetic division. He
will going to submit this worksheet to his coordinator, Mr. Lance
Cobrado with the corresponding answer key that shows the detailed
solution. Mr. Lance Cobrado will evaluate his work based on the
correctness and appropriateness of the given items and excercises,
and the accuracy of solutions and answers.
3. Yes sir, I am! And I will
submit it to your desk
sir.
Mr. Dalidig, are you
preparing the test for your
tutees?
9. Step 4
Subtract the dividend from the obtained
result, bring down the next term and
repeat the process from step 1 to 4:
y-3
-3y2
-3y3
+16y2 +3y -10
-3y3
-
-
16. Step 4
Subtract the dividend from the obtained
result, bring down the next term and
repeat the process from step 1 to 4:
y-9
-4y
-4y2
+20y-18
--4y2
-
17. As you can see that
there is a remaining
term, it is the
remainder. So, in
step 5 you will..
y-9
-4y
-4y2
+20y-18
--4y2
-
+36y
-16y -18
-16
-16y +144
-162
18. Step 5
Arrange your final
answer it should be
like this...
y-9
-4y
-4y2
+20y-18
--4y2
-
+36y
-16y -18
-16
-16y +144
-162
-4y-16 + -162
y-9
30. Step 4
Subtract the dividend from the obtained
result, bring down the next term and
repeat the process from step 1 to 4:
n-7 3
+8n
-4n
-4n
2
+19n
--4n
3
-
+8
2
31. As you can see that
there is a remaining
term, it is the
remainder. So, in
step 5 you will..
n-7 3
+8n
-4n
-4n
--4n3
-20n2
-28n2
-19n+8
+19n+8
-20n2
- +140n
-159n+8
- -159n+1113
-1105
-20n -159
2
32. Step 5
Arrange your final
answer it should be
like this...
-4n -20n-159-
-1105
y-9
n-7 3
+8n
-4n
-4n
--4n3
-20n2
-28n2
-19n+8
+19n+8
-20n2
- +140n
-159n+8
- -159n+1113
-1105
-20n -159
2
2
34. Step 1
Divide the leading term of the
dividend by the leading term of
the divisor:
-3n
3
n
= -3n
2
35. Step 2
Write down the calculated result
in the upper part of the table:
n-3 3
+0n -17
-3n
-3n2
2
-10n
36. Step 3
Multiply it by the divisor:
n-3 3
+0n -17
-3n
-3n2
2
-10n
-3n
3
-
37. Step 4
Subtract the dividend from the obtained
result, bring down the next term and
repeat the process from step 1 to 4:
n-3 3
+0n -17
-3n
-3n2
2
-10n
-3n
3
-
-
38. As you can see that
there is a remaining
term, it is the
remainder. So, in
step 5 you will..
n-3 3
+0n -17
-3n
-3n
2
-10n
-3n
3
-
-
+9n
+9n
2
+9n
2
-10n
-37
-27n
+9n
2
-37n-17
-37n
-
-111
-94
39. Step 5
Arrange your final
answer it should be
like this...
-3n +9n-37-
-94
n-3
n-3 3
+0n -17
-3n
-3n
2
-10n
-3n
3
-
-
+9n
+9n
2
+9n
2
-10n
-37
-27n
+9n
2
-37n-17
-37n
-
-111
-94
40. Now, we will proceed for
five items left whic is the
synthetic division..
43. Take the constant term of the divisor
with the opposite sign and write it to
the left.
Write the coefficients of the dividend
to the right.
To do this:
1.
2.
45. Step 2
Multiply the entry in the left part of the table by the last
entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend,
and write down the sum.
7 1 -3 -21
1 (-3)
7 1
+7 =
=7
4
46. Step 3
Multiply the entry in the left part of the table by the last
entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend,
and write down the sum.
7 1 -3 -21
1 + =
=
7
4
4
7
(-21) 28 7
28
47. All the coefficients except the last one are the coefficients of
the quotient, the last coefficient is the remainder. Thus, the
quotient is n+4, and the remainder is 7.
7 1 -3 -21
1
7
4 7
28
50. Take the constant term of the divisor
with the opposite sign and write it to
the left.
Write the coefficients of the dividend
to the right.
To do this:
1.
2.
52. Step 2
Multiply the entry in the left part of the table by the last
entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend,
and write down the sum.
1 1 -7 10
1 (-7)
1 1
+1 =
=1
6
53. Step 3
Multiply the entry in the left part of the table by the last
entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend,
and write down the sum.
1 1 -7 10
1 + =
=
1
-6
(-6)
1
-21 (-6) 4
-6
54. All the coefficients except the last one are the coefficients of
the quotient, the last coefficient is the remainder. Thus, the
quotient is k−6, and the remainder is 4.
1 1 -7 10
1
1
-6 4
-6
58. Step 2
Multiply the entry in the left part of the table by the last
entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend,
and write down the sum.
-5 1 10 18
1 10
(-5) 1
+(-5)=
=-5
5
59. Step 3
Multiply the entry in the left part of the table by the last
entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend,
and write down the sum.
-5 1 10 18
1 + =
=
-5
5
5
(-5)
18 (-25) -7
25
60. All the coefficients except the last one are the coefficients of
the quotient, the last coefficient is the remainder. Thus, the
quotient is n+5, and the remainder is −7.
-5 1 10 18
1
-5
5 -7
-25
64. Step 2
Multiply the entry in the left part of the table by the last
entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend,
and write down the sum.
6 1 -1 -29
1 (-1)
6 1
+6 =
=6
5
65. Step 3
Multiply the entry in the left part of the table by the last
entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend,
and write down the sum.
6 1 -1 -29
1 + =
=
6
5
5
6
(-29) 30 1
30
66. All the coefficients except the last one are the coefficients of
the quotient, the last coefficient is the remainder. Thus, the
quotient is n+5, and the remainder is 1.
6 1 -1
30
1
6
5 1
-29
70. Step 2
Multiply the entry in the left part of the table by the last
entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend,
and write down the sum.
8 1 -7 -11
1 (-7)
8 1
+8 =
=8
1
71. Step 3
Multiply the entry in the left part of the table by the last
entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend,
and write down the sum.
8 1 -7 -11
1 + =
=
8
1
1
8
(-11) 8 -3
8
72. All the coefficients except the last one are the coefficients of
the quotient, the last coefficient is the remainder. Thus, the
quotient is m+1, and the remainder is −3.
8 1 -7
8
1
8
1 -3
-11
74. Awesome,
good job Mr. Dalidig!
ALMOST ALL OF THE STUDENTS GOT HIGH
SCORES IN TEST, THANKS TO MR. DALIDIG
WORKSHEETS TO TEST THE SKILLS OF THE
STUDENTS!