Finding the Roots of a
Polynomial
Classifying the Roots of a Polynomial
Describe the amount of roots and what number set
they belong to for each graph:
12Real
RealRepeated
Roots andRoot
1
24Real
Complex
RootsRoots
and 2
and
Repeated
2Roots
Complex
Real
Root
Rootsit
4
Real
because
Complex
because
Roots
it has
because
no
because
it has
onlythree
has 1
has
four
x-intercepts
it x-intercepts
only has 2 x-intercepts
but has 4
x-intercept
x-intercepts
thatand
“bounces
one
but has
“turns”
4 “turns”
off”“bounces
but has 4off”
“turns”
Degree = 4
(there are 4 changes of
directions)
An nth degree
Polynomial ALWAYS
has n roots
Here is an Interesting
Question.
Can an Odd-Degree Polynomial have
Zero Real Roots?
We have seen examples of even degree polynomials that
have zero real roots. Can an odd degree polynomial have
zero real roots?
NO!
Since odd degree polynomials have opposite end
behavior and are continuous (no gaps), they must
intersect the x-axis at least once.
Let’s try an example.
Example 1: Finding the Roots of a
Polynomial
Find the roots of y=x4 – 11x3 + 29x2 + 35x – 150.
Use the graph or table of the cubic to
find the roots.
The three roots that can be found from
the graph or table are x = -2, 3, and 5
because they are the x-intercepts.
Since the equation is degree 4, it must
have a total of 4 roots.
But the graph “bounces off” the
x-intercept 5. This means it is a double
root. Thus all of the roots have been
found.
Therefore the roots are:
y  x3  6 x 2  11x  12
x  2, 3, & 5
Let’s try another example.
Example 2: Finding the Roots of a
Polynomial
Find the roots of y=x4 – 3x3 – 2x2 + 4x.
Use the graph or table of the degree 4
equation to find all 4 roots.
The two EXACT roots that can be
found from the graph or table are x = 0
and 1. The other two x-intercepts are
irrational roots (not in a table).
To find the irrational roots, we need to
factor the polynomial.
x  x  3x  2 x  4 
Notice x is a GCF. We can factor it out:
3
2
Since x = 1 is a root, (x – 1) is a factor.
y  x  3x  2 x  4 x
4
3
2
Now use polynomial division to “factor
out” the (x – 1) of the cubic.
Example 2: Finding the Roots of a
Polynomial
Find the roots of y=x4 – 3x3 – 2x2 + 4x.
3
2
x
x

3
x
 2x  4
Current Factored form: 
Rewrite the polynomial:
2
-2x
x
3
x
2
-2x
-4x
–1
-x2
2x
4
x
x3
– 3x2 – 2x
-4
+4
x  x  1  x  2 x  4 
2
x=1
This quadratic can
x=0
not be factored
because the
x-intercepts are
Use the quadratic irrational.
formula to find the
roots.
In order to find the roots,
The first two factor’s
The third factor’s
find out when each factor is zeros are the roots we zeros are the missing
equal to 0.
found from the graph. two irrational roots.
Example 2: Finding the Roots of a
Polynomial
Find the roots of y=x4 – 3x3 – 2x2 + 4x.
2
x
x

1
x
Current Factored form: 
   2x  4
Find the roots of x2 – 2x – 4 with the quadratic formula:
x
2
 2 
2
 41 4 
21

2 20
2

2 2 5
2
x  1 5
Therefore, the four
roots are:
x  0, 1, 1  5, & 1  5
Let’s try another example.
Example 3: Finding the Roots of a
Polynomial
Find the roots of y=x3 + 6x2 + 11x + 12.
Use the graph or table of the cubic to
find the roots.
The only root that can be found from
the graph or table is x = -4 because it is
the only x-intercept.
Since the equation is a cubic, it must
have a total of 3 roots. Thus, this
equation must have 2 complex roots.
In order to find the two missing complex
roots, we must factor the cubic
equation.
y  x3  6 x 2  11x  12
Since x = -4 is a root, (x + 4) is a factor
of the cubic.
Now use polynomial division to “factor
out” the (x + 4) of the cubic.
Example 3: Finding the Roots of a
Polynomial
Find the roots of y=x3 + 6x2 + 11x + 12.
Rewrite the cubic:
x2
2x
3
2
3
x
x
2x
3x
+4
4x2
8x
12
x3 + 6x2 + 11x + 12
 x  4  x
x = -4
2
 2 x  3
This quadratic has
no real roots since
it does not have
x-intercepts.
Use the quadratic formula to
find the roots.
In order to find the roots, The first factor’s zero The second factor’s
find out when each factor is is the root we found zeros are the missing
equal to 0.
from the graph.
two complex roots.
Example 3: Finding the Roots of a
Polynomial
Find the roots of y=x3 + 6x2 + 11x + 12.
Find the roots of x2 + 2x + 3 with the quadratic formula:
x
2 
 2   41 3
21
2

2 8
2

2 2 2i
2
x  1  2i
Therefore, the
three roots are:
x  4,  1  2i, &  1  2i