DO NOW
 Use
synthetic division to find a factor of
the following polynomial:
𝑥 3 − 14𝑥 2 + 51𝑥 − 54
Learning Target
Understand
how to use the rational
root theorem to find the roots (xintercepts) of polynomials.
Success Criteria
I
can multiply, and use synthetic
division.
Last Week
 Last
week we worked on division using
both long division and synthetic division
Today
 Today
we are working on using synthetic
division to factor polynomials and then
solve them for zeros.
Rational Root Theorem
 To
use synthetic division to factor, we need to find
what a values will give us a remainder of zero.

We could test numbers forever…
 To
make our lives easier, we are going to use the
rational root theorem to find all of the roots of
particular polynomials.
Not all polynomials are factorable, but the
Rational Root Theorem can help you find all
possible rational roots of a polynomial
equation.
Example: Rational Roots
 Find
all the possible rational roots of the
following polynomial: 2x3 - x2 + 2x+ 5 = 0
 Possible
roots are in the form of:
factor of the constant term .
factor of the leading coefficient
 Factors of constant term (p):
 ±1, ±5
 Factors of leading coefficient (q):
 ±1, ±2
 Possible Rational Roots (p/q):
 ±1, ±5, ±1/2, ±5/2,
Example 1:
What are the rational roots of x3 - 4x -15 = 0
Step 1: Use the Rational Root Theorem to identify all possible
rational roots.
Factors of –15: ±1, ±3, ±5, ±15
Step 2: Test the possible roots to find one that is actually a root. I like
to start with positive whole numbers.
Divisors (a’s for synthetic division)
Coefficients of polynomial
Use a synthetic division table to
organize your work.
p/q
1
0
-4
-15
1
1
1
-3
-18
3
1
3
5
0
5
1
5
21
90
Remainders
Example 1:
Step 4 Factor the polynomial.
The remainder when we used an a of 3 results in a
remainder of 0, so 3 is a root and the polynomial in
factored form is (x – 3)(x2 + 3x + 5).
x2 + 3x + 5 is not factorable so (x – 3)(x2 + 3x + 5) is our
factored form.
Example 2:
What are the rational roots of 15 x 3  32 x 2  3 x  2  0
Step 1: Use the Rational Root Theorem to identify all possible
rational roots.
Factors of 2 (p): ±1, ±2
Factors of 15 (q): ±1, ±3, ±5, ±15
Possible Rational Roots (p/q)
±1, ±2, ±1/3, ± 2/3, ±1/5, ±2/5, ±1/15, ±2/15
Example 2:
Step 2: Test the possible roots to find one that is actually a root. The
width must be positive, so try only positive rational roots.
Divisors (a’s for synthetic division)
Coefficients of polynomial
Use a synthetic division table to
organize your work.
p/q
15
-32
3
2
1
15
-1
-2
0
2
15
-17
-14
-12
Remainders
Example 2:
Step 4 Factor the polynomial.
The remainder when we used an a of 2 results in a
remainder of 0, so 2 is a root and the polynomial in
factored form is (x – 2)(15x2 –2 x – 1 ).
(x – 2)(15x2 – 2x – 1) = 0
(x – 2)(5x + 1)(3x - 1) = 0
x = 2, x = –1/5, or x = 1/3
Set the equation equal to 0.
Use ac method to
Factor 15x2 – x – 2.
Set each factor equal to 0, and
solve.
DO NOW:
What are the possible rational roots of
2x + x - 7x- 6 = 0
3
2
Step 1: Use the Rational Root Theorem to identify all possible
rational roots.
Factors of 6 (p): ±1, ±2, ±3, ±6,
Factors of 2 (q): ±1, ±2
Possible Rational Roots (p/q)
±1, ±2, ±3, ±6, ±1/2, ±3/2
Now find a root that works in the polynomial.
DO NOW:
Step 4 Factor the polynomial.
An a of 2 results in a remainder of 0, so 2 is a root and the
polynomial in factored form is (x – 2)(2x2 + 5x + 3 ).
(x – 2)(2x2 + 5x + 3) = 0
(x – 2)(2x + 3)(x + 1) = 0
x = 2, x = –3/2, or x = -1
Set the equation equal to 0.
Use ac method to
Factor 15x2 – x – 2.
Set each factor equal to 0, and
solve.
Example 1:
Rational Roots with higher degrees
 Find
all the possible rational roots of the
following polynomial: x 4  x 3  5 x 2  x  6  0
 Possible
roots are in the form of:
factor of the constant term .
factor of the leading coefficient
 Factors of constant term (p):
 ±1, ±2, ±3, ±6
 Factors of leading coefficient (q):
 ±1
 Possible Rational Roots (p/q):
 ±1, ±2, ±3, ±6,
Example 2:
Rational Roots with higher degrees
 Find
all the possible rational roots of the
following polynomial: 2 x 4  3 x 3  17 x 2  27 x  9  0
 Possible
roots are in the form of:
factor of the constant term .
factor of the leading coefficient
 Factors of constant term (p):
 ±1, ±2, ±3, ±6
 Factors of leading coefficient (q):
 ±1
 Possible Rational Roots (p/q):
 ±1, ±2, ±3, ±6,
Irrational Root Theorem
(conjugate root theorem)
Polynomial equations may also have irrational
roots.
Example 1:
A
quartic polynomial P(x) has the roots -2i
and 𝟑
Find two additional roots:
The irrational root theorem states that all
irrational or unreal roots come in conjugate
pairs.
The pair to – 2i is + 2i
The pair to 𝟑 is − 𝟑
Example 2:
A
quartic polynomial P(x) has the roots 𝟐
and 𝟏 + 𝒊
Find the missing roots:
 The
irrational root theorem states that all
irrational roots come in conjugate pairs.
 Since
the polynomial is quartic (degree of
4), we are missing 2 roots.
The pair to 𝟏 + 𝒊 is 𝟏 − 𝒊
The pair to 𝟎 + 𝟐 is 𝟎 − 𝟐
Example 3: Using Roots to
Write Polynomials
 What
is a third degree polynomial function with
rational coefficients whose roots are 5 and -3i?
Step 1: Write out the polynomial in factored form.
Step 2: Multiply the complex conjugates
Step 3: Write the polynomial function in standard
form
Example 4: Using Roots to
Write Polynomials
 What
is a third degree polynomial function with
rational coefficients whose roots are -1 and i?
Step 1: Write out the polynomial in factored form.
Step 2: Multiply the complex conjugates
Step 3: Write the polynomial function in standard
form
Example:
The design of a box specifies that its length is 4 inches
greater than its width. The height is 1 inch less than the
width. The volume of the box is 12 cubic inches. What is the
width of the box?
Step 1 Write an equation to model the volume of the box.
Let x represent the width in inches. Then the length is x + 4,
and the height is x – 1.
x(x + 4)(x – 1) = 12
x3 + 3x2 – 4x = 12
x3 + 3x2 – 4x – 12 = 0
V = lwh.
Multiply the left side.
Set the equation equal to 0.
Example 3:
Step 2 Use the Rational Root Theorem to identify all possible rational
roots.
Factors of –12: ±1, ±2, ±3, ±4, ±6, ±12
Step 3 Test the possible roots to find one that is actually a root. The
width must be positive, so try only positive rational roots.
Use a synthetic division table to organize
your work. The first row represents the
coefficients of the polynomial. The first
column represents the divisors and the
last column represents the remainders.
Test divisors to identify at least one root.
Example 3:
Step 4 Factor the polynomial. The synthetic remainder of
2 results in a remainder of 0, so 2 is a root and the
polynomial in factored form is (x – 2)(x2 + 5x + 6).
(x – 2)(x2 + 5x + 6) = 0
Set the equation equal to 0.
(x – 2)(x + 2)(x + 3) = 0
Factor x2 + 5x + 6.
x = 2, x = –2, or x = –3
Set each factor equal to 0, and
solve.
The width must be positive, so the width should be 2
inches.