Examples of division using Ruffini's Rule

1. RUFFINI'S RULE
We use Ruffini's rule when we want to divide a
polinomial P(x) by a binomial x-a

2. Let's divide
( x5−3 x3+x2−x+5):( x−2)
First, write only the coefficients;
if one term is missing, the coefficient is 0

3. Let's divide
( x5−3 x3+x2−x+5):( x−2)
First, write only the coefficients;
if one term is missing, the coefficient is 0
1 0 -3 1 -1 5

4. Let's divide
( x5−3 x3+x2−x+5):( x−2)
Then, the number a; in this case, a=2
1 0 -3 1 -1 5
2

5. Let's divide
( x5−3 x3+x2−x+5):( x−2)
The first term of the quotient will be:
x5 : x=x We only write the coefficient: 1
1 0 -3 1 -1 5
2
1

6. Let's divide
( x5−3 x3+x2−x+5):( x−2)
The terms of the quotient are to be multiplied
by the divisor and subtracted to the dividend
1 0 -3 1 -1 5
2
1
2
2

7. Let's divide
( x5−3 x3+x2−x+5):( x−2)
Don't worry about the minus sign when subtracting
because you already changed it writing 2 instead of -2
1 0 -3 1 -1 5
2
1
2
2

8. Let's divide
( x5−3 x3+x2−x+5):( x−2)
You already have the second term of the quotient.
Go on the same way multiplying 2 · 2.
1 0 -3 1 -1 5
2
1
2
2

9. Let's divide
( x5−3 x3+x2−x+5):( x−2)
Go on the same way.
1 0 -3 1 -1 5
2
1
2
2
4
1

10. Let's divide
( x5−3 x3+x2−x+5):( x−2)
Go on the same way.
1 0 -3 1 -1 5
2
1
2
2
4
1
2
3

11. Let's divide
( x5−3 x3+x2−x+5):( x−2)
Go on the same way.
1 0 -3 1 -1 5
2
1
2
2
4
1
2
3
6
5

12. Let's divide
( x5−3 x3+x2−x+5):( x−2)
Go on the same way.
1 0 -3 1 -1 5
2
1
2
2
4
1
2
3
6
5
10
15

13. Let's divide
( x5−3 x3+x2−x+5):( x−2)
This last term is the remainder.
1 0 -3 1 -1 5
2
1
2
2
4
1
2
3
6
5
10
15

14. Let's divide
( x5−3 x3+x2−x+5):( x−2)
This last term is the remainder.
1 0 -3 1 -1 5
2
1
2
2
4
1
2
3
6
5
10
15=R

15. Let's divide
( x5−3 x3+x2−x+5):( x−2)
Write the quotient; the last term is the constant
The term before it is the x term and so on
1 0 -3 1 -1 5
2
1
2
2
4
1
2
3
6
5
10
15=R

16. Let's divide
( x5−3 x3+x2−x+5):( x−2)
Q( x)=x4+2 x3+x2+3x+5
1 0 -3 1 -1 5
2
1
2
2
4
1
2
3
6
5
10
15=R

17. ● Another example; in this case, the number a
will be negative, so that the divisor will be in the
form x+a, being a a negative number

18. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)

19. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
First, write only the coefficients;
if one term is missing, the coefficient is 0

20. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
First, write only the coefficients;
if one term is missing, the coefficient is 0
2 1 0 4 -3 1

21. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
Now write a: remember a=-1
2 1 0 4 -3 1

22. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
Now write a: remember a=-1
2 1 0 4 -3 1
-1

23. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
And proceed to the division
2 1 0 4 -3 1
-1

24. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
And proceed to the division
2 1 0 4 -3 1
-1
2

25. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
And proceed to the division
2 1 0 4 -3 1
-1
2
-2
-1

26. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
And proceed to the division
2 1 0 4 -3 1
-1
2
-2
-1
1
1

27. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
And proceed to the division
2 1 0 4 -3 1
-1
2
-2
-1
1
1
-1
3

28. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
And proceed to the division
2 1 0 4 -3 1
-1
2
-2
-1
1
1
-1
3
-3
-6

29. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
And proceed to the division
2 1 0 4 -3 1
-1
2
-2
-1
1
1
-1
3
-3
-6
6
7

30. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
Write the remainder
2 1 0 4 -3 1
-1
2
-2
-1
1
1
-1
3
-3
-6
6
7

31. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
Write the remainder
2 1 0 4 -3 1
-1
2
-2
-1
1
1
-1
3
-3
-6
6
7=R

32. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
And the quotient:
2 1 0 4 -3 1
-1
2
-2
-1
1
1
-1
3
-3
-6
6
7=R

33. Let's divide
(2x5+ x4+4 x2−3x+1):( x+1)
And the quotient:
Q( x)=2 x4−x3+x2+3 x−6
2 1 0 4 -3 1
-1
2
-2
-1
1
1
-1
3
-3
-6
6
7=R