The document discusses finding the zeros or roots of polynomial functions using the Rational Root Theorem. It provides three examples of applying the Rational Root Theorem and synthetic division to determine the actual roots of given polynomial functions. The key steps are to find the possible rational roots using the Rational Root Theorem, then use synthetic division or factoring to determine which possible roots are the actual roots that make the polynomial function equal to zero.

1. SUBTITLE

2.  Example 1. Determine the possible zeros or roots of the
polynomial function f(x) = 2x3 – x2 – 13x – 6.
Find the factors of -6 (represented by m) = {±1, ±2, ±3, ±6}
Find the factors of 2 (represented by (n) = {±1, ±2}
By using the Rational Theorem,
𝑀
𝑁
=
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 −6
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙𝑠 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 2
=
{±1,±2,±3,±6}
{±1,±2}
To find the zeros or roots of a polynomial
function f(x), we use the Rational Theorem

3. 𝑀
𝑁
=
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 −6
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙𝑠 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 2
=
{±1,±2,±3,±6}
{±1,±2}
𝑀
𝑁
=
{±1,±2,±3,±6}
±1
,
{±1,±2,±3,±6}
±2
𝑀
𝑁
= { ± 1, ±2, ±3, ±6} , ±
1
2
, ±1, ±
3
2
, ±𝟑
𝑀
𝑁
= ±
1
2
, ±1, ±
3
2
, ±2, ±3, ±6 simplified
Therefore, the possible roots or zeros of f(x) = 2x3 – x2 – 13x – 6
is the set
𝑴
𝑵
= ±
𝟏
𝟐
, ±𝟏, ±
𝟑
𝟐
, ±𝟐, ±𝟑, ±𝟔 .

4. Steps Solution
1. Consider the possible zeros or
roots of f(x) = 2x3 – x3 – 13x – 6.
𝑀
𝑁
= ±
1
2
, ±1, ±
3
2
, ±2, ±3, ±6
2. By trial and error method,
determine the roots or zeros
using synthetic division
(depressed equation)
−𝟐 2 -1 -13 -6
-4 10 6
2 -5 -3 0
= (2x2 – 5x – 3)(quotient)
Example 2. Find the zeros or roots of f(x) = 2x3 – x2 – 13x – 6.

5. Steps Solution
3. Since -2 is one of the roots of
f(x) = 2x3 – x3 – 13x – 6,
we can write it as (x + 2).
Factor (2x2 – 5x – 3) using the
quadratic formula.
a= 2, b = -5, c = -3
x =
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
x =
−(−5)± (−5)2−4(2)(−3)
2(2)
x =
5± 25+24
4
x =
5± 49
4
=
5±7
4
x =
5+7
4
=
12
4
= 3
x =
5 −7
4
=
−2
4
= -
𝟏
𝟐

6. Therefore, from the set
𝑀
𝑁
= ±
1
2
, ±1, ±
3
2
, ±2, ±3, ±6 of
possible zeros or roots of f(x) = 2x3 – x3 – 13x – 6, the actual
zeros or roots are -2, -
𝟏
𝟐
, and 3.

7. Solution
1. f(x) = 2x3 + 4x2 – 2x – 4
M = ±1, ±2, ±4
N = ±1, ±2
𝑀
𝑁
=
±1,±2,±4
±1
,
±1,±2,±4
±2
𝑀
𝑁
= ±1, ±
1
2
, ±2, ±4
Depressed equation:
(Using synthetic division)
𝟏 2 4 -2 -4
2 6 4
2 6 4 0
= (2x2 + 6x + 4)
Example 3. Find the zeros of a polynomial
f(x) = 2x3 + 4x2 – 2x – 4.

8. Solution
= (2x2 + 6x + 4)
Using factoring by groupings
in solving Quadratic Equation
a = 2, b = 6, c = 4
Multiply(a)(c) = (2)(4) = 8
Find the factors of 8 that gives
the sum of b (which is 6) = 4 . 2
= 2x2 + 6x + 4
= 2x2 + 4x + 2x + 4 = 0
(2x2 + 4x ) + (2x + 4) = 0
2x(x + 2) + 2(x + 2) = 0
(x + 2)(2x + 2) = 0
x + 2 = 0 and 2x + 2 = 0
x = -2 and 2x = 0 – 2
2x = -2
x = -1
Therefore the zeros of the
polynomial are 1, -1 and -2.