The document discusses polynomial functions and their roots. It begins by defining that the roots of a polynomial function are the values of x that make the function equal to 0. It then provides examples of finding the roots of linear and quadratic equations. Next, it introduces the Rational Root Theorem, which states that possible rational roots must be factors of the constant term and leading coefficient. Examples are given to demonstrate applying the theorem. The document concludes by using synthetic division to find all three roots of a cubic polynomial given one known root.

1. SUBTITLE

2. Let’s Remember:
The roots, zero’s or solutions of a polynomial
function is the value of x that will make the
polynomial function f(x) = 0

3. Linear Equation
1. x + 4 = -5
x = -5 – 4
x = -9
x + 4 = -5
-9 + 4 = -5
-5 = -5
2. 2x – 5 = 3
2x = 3 + 5
2x = 8
2x = 8 = 4
2 2
Quadratic Equation
1. x2 + x – 6
(x + 3)(x – 2) = 0
x + 3 = 0 and x – 2 = 0
x = -3 and x = 2
Checking:
x2 + x – 6 = 0
(-3)2 + (-3) – 6 = 0
9 – 3 – 6 = 0
0 = 0
x2 + x – 6 = 0
(2)2 + 2 – 6 = 0
4 + 2 – 6 = 0
0 = 0

4. Study the following:
Roots Factors Equation
1. x = 2 and x = 3 (x – 2)(x – 3) x2 – 3x + 6 = 0
2. x = -1 and x = 4 (x + 1) (x - 4) x2 – 3x – 4 = 0
3. x =
2
5
and x = - 2 (5x – 2)(x + 2) 5x2 + 8x – 4 = 0
4. x = 3, x= 1, and x = 2 (x -3)(x - 1)(x - 2) x3 – 6x2 + 11x – 6 = 0
5. -1, 1, and 2 (x + 1)(x – 1)(x – 2) x3 – 2x2 – x + 2 = 0

5. RATIONAL ROOT THEOREM
The Rational Root Theorem states that if the rational
number
𝑀
𝑁
in simplest form is a root of a polynomial whose
coefficients are integers, then the integer M is a factor of the
constant term, and the integer N is a factor of the leading
coefficient.

6. Example 1:
List all the possible rational zeros of the following polynomials.
a. 1x3 – 5x2 + 4x + 4 = 0
Let M be the integral factors of 4: ±1, ±2, ±4 = 1, 2, 4, -1, -2, -4
Let N be the integral factors of 1: ±1 = -1 . -1, 1 . 1
=-1, 1, -2, 2, -4, 4,
Since the leading coefficient is 1and the constant is 4, we form the ratio
𝑀
𝑁
=
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 4
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙𝑠 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 1
±1
±1
= ±1
±2
±1
= ±2
±4
±1
= ±4
Then, the possible rational roots are ±1, ±2, ±4.

7. Example 2.
What are the possible zeros of f(x) = x3 – 3x2 – 4x + 12.
Let M be the integral factors of 12: ±1, ±2, ±3, ±4, ±6, ±12
Let N be the integral factors of 1: ±1
𝑀
𝑁
=
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 12
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙𝑠 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 1
±1,±2, ±3, ±4, ±6, ±12
±1
= ±1, ±2, ±3, ±4, ±6, ±12
Therefore the possible zeros or roots of f(x) = x3 – 3x2 – 4x + 12 is the
set of ±1, ±2, ±3, ±4, ±6, ±12 .

8. Find the roots of f(x) = x3 – 3x2 – 4x + 12.
If x = 1 If x = 2
= x3 – 3x2 – 4x + 12
= (1)3 – 3(1)2 – 4(1) + 12
= 1 – 3(1) – 4 + 12
= 1 – 3 – 4 + 12
= -2 – 4 + 12
= 6
= x3 – 3x2 – 4x + 12
= (2)3 – 3(2)2 – 4(2) + 12
= 8 – 3(4) – 8 + 12
= 8 – 12 – 8 + 12
= -4 - 8 + 12
= 0

9. If x = -2 If x = 3
= x3 – 3x2 – 4x + 12
= (-2)3 – 3(-2)2 – 4(-2) + 12
= -8 – 3(4) + 8 + 12
= -8 – 12 + 8 + 12
= -20 + 8 + 12
= 12 + 12
= 0
= x3 – 3x2 – 4x + 12
= (3)3 – 3(3)2 – 4(3) + 12
= 27 – 3(9) – 12 + 12
=27 – 27 - 12 + 12
= 0 – 12 + 12
= -12 + 12
= 0
Therefore the zeros of the polynomials f(x) = x3 – 3x2 – 4x + 12 are -2, 2 and 3.
Since the roots or the zeros of polynomial are -2, 2 and 3, the factors are
(x + 2)(x - 2)(x - 3).

10. Depressed Equation
Use synthetic division to find the zeros of a polynomial.
Since -2 is a possible rational root of f(x) = x3 – 3x2 – 4x + 12, determine the
other roots using synthetic division.
-2 1 -3 -4 12
-2 10 -12
___________________________
1 -5 6 0
= (x2 – 5x + 6
Factor = (x – 3)(x – 2)
roots = 3 and 2
Zeros of a polynomial are -2, 2 and 3.

11. THANK
YOU