The document discusses the Remainder Theorem, which provides a way to factorize polynomials by dividing them by factors and obtaining a remainder. There are two methods for finding the remainder: long division/evaluation and synthetic division. Evaluation involves substituting the factor value into the polynomial, while synthetic division arranges the coefficients and repeatedly multiplies and adds down the line. The document provides examples of using both methods and notes that synthetic division allows determining the full quotient polynomial.

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!!!