The document discusses the Remainder Theorem and using it to find the value of a polynomial P(x) at a given point c. It provides examples of using synthetic division and the Factor Theorem to factor polynomials and determine if (x-c) is a factor based on whether c is a root. Exercises provide additional problems applying these concepts to find remainders, factor polynomials, and solve for constants.

1. Remainder Theorem: Let P(x) be a polynomial, the
remainder of P(x) ÷ (x – c) is r = P(c).
Example A. Find P(1/2) given P(x) = 4x4 – 2x3 + 6x – 1
using synthetic division.
We want the remainder of P(x) / (x – ½)
Set up the synthetic division
4 –2 0 6 –11/2
4
2
0 0
0
6
0
2
3
Since the remainder r is the same as P(1/2),
therefore P(1/2) = 2.
Properties of Division and Roots
This gives a way to find the value of P(c) via synthetic
division.
the remainder r

2. Properties of Division and Roots
The Factor Theorem says that knowing
“x = c is a root of a polynomial P(x), i.e. P(c) = 0”
is same as knowing that
“(x – c) is a factor of P(x), i.e. P(x) = Q(x)(x – c)”.
Example B. Use the Factor Theorem to show that
(x – 1) is a factor P(x) = 3x100 – 5x50 + 9x30 – 13x20 + 6.
We check to see if x = 1 is a root of P(x).
P(1) = 3 – 5 + 9 – 13 + 6 = 0.
Hence x = 1 is a root of P(x).
Therefore (x – 1) is a factor of P(x).

3. Properties of Division and Roots
Example E. Factor x4 – 2x3 – 3x2 + 2x + 2 completely
into real factors.
By observation that 1 – 2 – 3 + 2 + 2 = 0, we see that
(x – 1) is a factor.
At the same time, 1 – 3 + 2 = – 2 + 2 = 0, we see
that (x + 1) is a factor. Use synthetic division twice
1 –2 –3 2 21
1
1
–1
–1
–4
–4
–2
–2
0–1
1
–1
–2
2
–2
2
0
So x4 – 2x3 – 3x2 + 2x + 2
= (x – 1)(x +1)(x2 – 2x – 2)

4. Properties of Division and Roots
(x2 – 2x – 2) is irreducible, its roots are 1 ± 3 by
the quadratic formula.
So x4 – 2x3 – 3x2 + 2x + 2 factors completely as
(x – 1)(x +1)(x – (1 + 3))(x – (1 – 3)).

5. Remainder Theorem: Let P(x) be a polynomial,
the remainder of P(x) ÷ (x – c) is r = P(c).
Exercise A. Find the remainder r of P(x) ÷ (x – c)
by evaluating P(c).
Properties of Division and Roots
1. x – 1
–2x + 3
4.
x2 – 9
7.
x + 3
x2 – 2x + 3
2. x + 1
3x + 2 3.
x – 2
3x + 1
8.
x – 3
2x2 – 2x + 1 9.
x + 2
–2x2 + 4x + 1
5. x + 2
x2 + 4
6. x – 3
x2 + 9
10.
x3 – 2x + 3 11.
x – 3
2x3 – 2x + 1 12.
x + 2
–2x4 + 4x + 1
x – 3
x – 2

6. Remainder Theorem: Let P(x) be a polynomial,
the remainder of P(x) ÷ (x – c) is r = P(c).
Properties of Division and Roots
B. Find P(c) by finding the remainder of P(x) ÷ (x – c).
1. Find P(1/2) given P(x) = 2x3 – 5x2 + 6x – 1
2. Find P(1/3) given P(x) = 3x3 – 4x2 + 7x – 1
3. Find P(1/4) given P(x) = 4x4 – 5x3 + 2x – 1
4. Find P(1/5) given P(x) = 5x4 + 4x3 + x – 1
5. Find P(1/6) given P(x) = 12x5 – 2x4 + 60x – 1
6. Find P(1/7) given P(x) = 7x20 – x19 + x – 1
7. Find P(1/8) given P(x) = 64x200 – x198 + x – 1
8. Find P(1/10) given
P(x) = 20x200 – 2x199 – 100x19 + 10x18 – 1

7. C. Use the Factor Theorem to show that
(x – c) is a factor of P(x) by checking that c is a root.
Properties of Division and Roots
1. (x – 1) is a factor of P(x) = 2x3 – 5x2 + 4x – 1
2. (x – 1) is a factor of P(x) = 9x3 – 7x2 + 6x – 8
3. (x + 1) is a factor of P(x) = 100x3 + 99x2 – 98x – 97
4. (x + 1) is a factor of
P(x) = 10x71 – 9x70 – 9x61 + 10x60 + 98x7 + 98
5. Check that (x + 2) is a factor of P(x) = x3 – 7x – 6,
then factor P(x) completely.
7. Check that (x + 3) and (x + 4) are factors of
P(x) = x4 + x3 –16x2 – 4x + 48 then factor it completely.
6. Check that (x + 3) is a factor of
P(x) = x3 + 3x2 – 4x –12 then factor P(x) completely.

8. Properties of Division and Roots
10. Create a 4th degree polynomial in the expanded
with (x + 1) as a factor.
11. Create a 5th degree polynomial in the expanded
with (x + 1) as a factor.
8. Create a 4th degree polynomial in the expanded
with (x – 1) as a factor.
9. Create a 5th degree polynomial in the expanded
with (x – 1) as a factor.
12. Solve for k if (x + 1) is a factor of 2x3 – 5x2 + 4x + k.
13. Solve for k if (x + 2) is a factor of 2x3 – kx2 + 4x + k.
14. Solve for k if (x + 2) is a factor of kx3 – x2 + 3kx + k.

9. (Answers to odd problems) Exercise A.
1. 1
7. 18
3. 7
9. −15
5. 8
11. 49
Exercise B.
1. 1 3. −9/16 5. 9 7. 0
Exercise C.
5. 𝑃(𝑥) = (𝑥 + 2)(𝑥 + 1)(𝑥 − 3)
7. 𝑃(𝑥) = (𝑥 − 3)(𝑥 + 4)(𝑥 − 2)(𝑥 + 2)
11. 𝑃 𝑥 = 𝑥5 + 𝑥 = 𝑥4(𝑥 + 1)
9. 𝑃 𝑥 = 𝑥5 − 𝑥 = 𝑥4(𝑥 − 1)
13. 𝑘 = 8
Properties of Division and Roots