P.o.D. – Factor each of the
following:
1.) 𝑥 2 + 10𝑥 + 21
2.) 𝑡 2 − 3𝑡 − 4
3.) 𝑥 2 − 15𝑥 + 54
4.) 𝑝2 + 12𝑝 + 32
5.) 𝑥 2 − 100
1.)
2.)
3.)
4.)
(x+3)(x+7)
(t-4)(t+1)
(x-6)(x-9)
(p+8)(p+4)
5.) (x+10)(x-10)
11-4: The Factor Theorem
Learning Target(s): I can find
zeros of polynomial functions
by factoring; apply the zeroproduct theorem and the
factor theorem; graph
polynomial functions;
estimate zeros using graphs.
Zero(s) of a Polynomial:
- Places where the graph
crosses the x-axis
- Solutions to the equation
- Also known as a root
Zero Product Theorem:
For all a and b, ab=0 if and
only if a=0 or b=0.
EX: An open box has sides of
length x, (24-2x), and (24-2x).
Thus, its volume is given by
V(x)=x(24-2x)(24-2x). Find
the zeros of V.
The Zero-Product Theorem
says that in order for an
equation to equal zero, each of
its factors could equal zero.
X=0
24-2x=0 24-2x=0
24=2x
24=2x
12=x
12=x
Solutions: 0,12,12  12 is a
double root
Factor Theorem:
(x-r) is a factor of P(x) if and
only if P(r)=0. In other words,
r is a zero of P.
EX: Find the roots of 𝑃(𝑥 ) =
𝑥 4 − 𝑥 3 − 20𝑥 2 by factoring.
Begin by taking out a common
factor.
𝑃(𝑥 ) = 𝑥 2 (𝑥 2 − 𝑥 − 20)
Now factor the trinomial by
un-FOILing.
𝑃(𝑥 ) = 𝑥 2 (𝑥 − 5)(𝑥 + 4)
Use the Zero-Product
Property to set each factor
equal to zero and solve.
X+4=0
𝑥 2 = 0 x-5=0
X=-4
𝑥 = +0, −0 x=5
The zeros are x=0,0,5,-4.
O is a double root.
EX: Find the roots of 𝑃(𝑥 ) =
𝑥 4 − 14𝑥 2 + 45
Since this has 3 terms, we
should try to un-FOIL.
𝑃(𝑥 ) = (𝑥 2 − 9)(𝑥 2 − 5)
Notice that one of these
factors is the difference of
squares.
𝑃(𝑥 ) = (𝑥 − 3)(𝑥 + 3)(𝑥 2 − 5)
Apply the zero-product
property.
x-3=0
x=3
x+3=0
x=-3
𝑥2 − 5 = 0
𝑥2 = 5
𝑥 = ±√5
The roots are 3, -3, √5, −√5.
This is an example of the
Fundamental Theorem of
Algebra. The FTA states that
you will have the same
number of roots as the highest
exponent.
EX: A polynomial function p
with degree 4 and leading
coefficient 1 is graphed below.
Find the factors of p(x) and
use them to write a formula
for p(x).
This graph crosses the x-axis,
or has zeros, at -7, -4, 0, and 3.
Therefore, it has factors of
(x+7), (x+4), x, and (x-3).
Expand these factors to find
the polynomial.
𝑃(𝑥 ) = 𝑥(𝑥 + 7)(𝑥 + 4)(𝑥 − 3)
= (𝑥 2 + 7𝑥)(𝑥 + 4)(𝑥 − 3)
= (𝑥 3 + 4𝑥 2 + 7𝑥 2 + 28𝑥)(𝑥 − 3)
= (𝑥 3 + 11𝑥 2 + 28𝑥)(𝑥 − 3)
= 𝑥 4 + 11𝑥 3 + 28𝑥 2 − 3𝑥 3 − 33𝑥 2 − 84𝑥
= 𝑥 4 + 8𝑥 3 − 5𝑥 2 − 84𝑥
EX: Find the equation of a
polynomial whose zeros are -4,
7/2, and 5/3.
The factors are (x+4), (x-7/2),
and (x-5/3).
7
5
𝑃(𝑥 ) = (𝑥 + 4) (𝑥 − ) (𝑥 − )
2
3
Or 𝑃(𝑥 ) = (𝑥 + 4)(2𝑥 − 7)(3𝑥 − 5)
7
5
= (𝑥 − 𝑥 + 4𝑥 − 14) (𝑥 − )
2
3
1
5
2
= (𝑥 + 𝑥 − 14) (𝑥 − )
2
3
2
1 2
5 2 5
70
= 𝑥 + 𝑥 − 14𝑥 − 𝑥 − 𝑥 +
2
3
6
3
3
7 2 89
70
=𝑥 − 𝑥 − 𝑥+
6
6
3
Or 𝑃(𝑥 ) = 6𝑥 3 − 7𝑥 2 − 89𝑥 + 140
3
Upon completion of this
lesson, you should be able to:
1. Find zeros of a polynomial.
2. Find the roots of a
polynomial by factoring.
3. Use ZoomMath (not on
ACT).
For more information, visit
http://www.purplemath.com/modules/from
zero2.htm
HW Pg.756 2-12, 14-27
Quiz 11.1-11.4 tomorrow