1. Remainder
Theorem
Prepared by:
TeresitaM. Joson
BSEd IV – Math

2. The Remainder Theorem
-is a useful mathematical
theorem that can be used to
factorize polynomials of any
degree in a neat and fast
manner.

3. The Remainder Theorem states
that when you divide a
polynomial P(x) by any factor
(x - a); which is not necessarily a
factor of the polynomial; you'll
obtain a new smaller polynomial
and a remainder, and this
remainder is the value
of P(x) at x = a

4. Remainder Theorem operates on
the fact that a polynomial is
completely divisible once by its
factor to obtain a smaller
polynomial and a remainder of
zero. This provides an easy way
to test whether a value a is a
root of the polynomial P(x).

5. There are two(2) ways to solve for
the remainder:
 Long division/Evaluation
 Synthetic division

6. Using Evaluation:
f (x) = x6 + 5x5 + 5x4 + 5x3 + 2x2 – 10x – 8
x-2
In using evaluation method, we first find the
value of X, we will use the divisor x-2.
x-2 =0 transposition method
x = 2 + 0
x = 2 you have the value of x.

7. Next step:
Substitute the vale of x in the equation
F(x) = x6 + 5x5 + 5x4 + 5x3 + 2x2 – 10x – 8
= (2)6 + 5(2)5 + 5(2)4 + 5(2)3 + 2(2)2 – 10(2) – 8
= 64 + 5 (32) + 5 (16) + 5 (8) + 2(4) – 20 – 8
= 64 + 160 + 80 + 40 + 8 – 20 – 8
= 352 – 28
= 324 Remainder

8. Synthetic Division
Rules:
 see to it that the exponents are
arranged in descending order
 get the numerical coefficients

9. Using Synthetic Division
Using the same equation:
f (x) = x6 + 5x5 + 5x4 + 5x3 + 2x2 – 10x – 8
x-2
We have the value of x = 2

10. Arrange the value of numerical coefficient:
Steps:
1. Bring down the first value.
2. Multiply 1 to your divisor/x.
3. Put it down the second value.
4. Then add.
5. Repeat step 2 until you get the last value.
x=2 1 5 5 5 2 -10 -8
Bring
Down 2 14 38 86 176 332
1 7 19 43 88 166 324 remainder

11. The quotient now in this equation
f (x) = x6 + 5x5 + 5x4 + 5x3 + 2x2 – 10x – 8
x-2
• You have to less 1 to your first exponent.
f (x) = x5 + 7x4 + 19x3 + 43x2 + 88x + 166
remainder is 324
x-2

12. Differences:
Evaluation method:
You can not determine the quotient.
Synthetic Division:
You can get the final quotient of the
equation.

13. Boardwork!
f (x) = x3 – 4x2 + x – 6 / x + 4
Using evaluation:
x + 4 = 0 x = -4
Substitute:
f (x) = (-4)3 – 4(-4)2 + (-4)– 6
= -64 – 4(16) – 4 – 6
= -64 – 64 – 4 – 6
= -138 remainder

14. Using Synthetic Division:
f (x) = x3 – 4x2 + x – 6 / x + 4
-4  1 -4 1 -6
-4 32 -132
1 -8 33 -138
We have come up with the same remainder.
Now quotient is:
f (x) = x2 - 8x + 33 -138
x + 4

15. Now! Try this!
Find the remainder using evaluation and
Synthetic Division:
1. - 2a5 + 10a4 + a3 + 2a2 – 10a – 3 / a +3
2. 2x4 - 15x3 + 11x2 – 10x – 12 / x – 5
3. x3 + 5x2 – 7x + 2 / x + 2
4. 2x3 - 3x2 – x + 4 / x – 3
5. 5x6 + 2x4 + 5x3 – 5x – 12 / x + 2

16. Sometimes the
questions are
complicated
yet the answers are
simple…
Thank You!!!