2.
Main ideas:
What is the numerical value of a polynomial for x = a?
What is a root of a polynomial?
What is a factor of a polynomial?
What is the relationship between roots and factors?
How can we determine the roots?
How can we factor a polynomial?
3.
Numerical value of a polynomial for x = a
If you evaluate P(x) for x = a, you get P(a).
We say that P(a) is the numerical value of
the polynomial P(x) when x = a
4.
Numerical value of a polynomial for x = a
Example:
Evaluate P(x)= -2x2+3x-5 for x = -3
P(-3)= -2(-3)2 + 3(-3) – 5 =
− 2 ⋅ 9 − 9 − 5 = −32
So, the numerical value of the polynomial P(x)
when x = -3 is:
P(-3) = -32
5.
Root of a polynomial P(x)
We say that the number “a” is a root of the
polynomial P(x) if the numerical value P(a)=0
In other words:
If, when you evaluate P(x) for x = a, you get
P(a)=0, then “a” is a root of P(x)
6.
Exercise:
Check that a = 2 and b = 1, are
both roots of the polynomial:
P(x) = x2 - 3x + 2
but c = -1 is not.
7.
Factors of a polynomial
We can write the
polynomial P(x) this way:
P(x) = x2 - 3x + 2
or this way:
P(x) = (x – 2)(x – 1)
The linear binomials (x-2) y (x-1) are the
simplest factors of P(x)
9.
There are several key ideas which will help us to find
roots:
The remainder theorem
The Ruffini’s rule
This useful property of polynomials:
“An integer root of a polynomial, is always a divisor of its constant
term”
10.
The remainder theorem :
"If a polynomial P(x) is divided by a binomial (x - a), then
the remainder is equal to the numerical value of the
polynomial when x = "a", and can be expressed as P(a)"
In other words, we can write:
P(x)
Remainder
x-a
C(x)
Remaider = P(a)
(numerical value when x = a)
11.
The remainder theorem :
"If a polynomial P(x) is divided by a binomial (x - a), then
the remainder is equal to the numerical value of the
polynomial when x = "a", and can be expressed as P(a)"
In other words, we can write:
P(x)
Remainder
x-a
C(x)
Remaider = P(a)
(numerical value when x = a)
So, if the remainder of the division is zero,
then P(a)=0 and therefore “a” is a root of P(x).
12.
The Ruffini’s rule :
This rule is a practical way for doing divisions like this:
(x2 – 3x + 2) : (x – 2)
13.
The Ruffini’s rule :
This rule is a practical way for doing divisions like this:
(x2 – 3x + 2) : (x – 2)
How?
14. We write this number here
(x2 – 3x + 2) : (x – 2)
1
2
-3
2
16.
We can see that the remainder of the division is zero.
As the remainder of P(x) : (x-2) is zero, “2” is a root of
P(x), and therefore, the linear binomial (x-2) is a factor
of the polynomial P(x)
Now, we will look for another root in
order to get a new factor, and so on…
1
1
2
2
2
-3
-2
-1
0
(x2 – 3x + 2) : (x – 2)
17. This rule makes it easier to find the values of "a".
“An integer root of a polynomial, is always a divisor of its constant term”
18. “An integer root of a polynomial, is always a divisor of its constant term”
Example:
Which integer numbers could be roots of the polynomial
P(x) = - 3x5 + 4x2 – 5x -3 ?
Answer: The divisors of the constant term
They are only +1, -1, +3 y -3
19. It is time to use Ruffini’s rule to make the
following divisions faster:
P(x):(x-1)
P(x):(x+1)
( x-(-1))
P(x):(x-3)
and
P(x):(x+3)
( x-(3))
20.
So, if we want to factor a polynomial, we need to
find its roots…
Roots
2, -1 y 3
1st Factor:
(x-2)
2nd Factor:
(x+1)
3th Factor:
(x-3)
21.
Exercise: Factor the polynomial
P(x) = x3 -4x2 + x + 6
Possible integer roots :
+1, -1, +2, -2, +3, -3, +6, -6
22. Possible integer root
Is it a root? Yes/No
Factor
+1
No
-1
Yes
(x+1)
+2
Yes
(x-2)
-2
No
+3
Yes
-3
No
+6
No
-6
No
(x-3)
23.
Now, we can write the polynomial:
P(x) = x3 -4x2 + x + 6
in its factored form
P(x) = x3 -4x2 + x + 6 = (x+1)(x-2)(x-3)
24.
In short, there are three strategies for finding roots…
If we “suspect” that a number “a” is a root of the
polynomial P(x), we can:
1. We can make the division P(x) : (x-a). If the
remainder is zero, then “a” is a root of P(x).
Ruffini’s rule
2.We can write the number “a” instead of the “x” and
make the calculations. If the numerical value, P(a)=0,
then “a” is a root of P(x).
Root of a polynomial: definition
3. We can solve the equation P(x)=0
Particularly if P(x)=0 is a quadratic equation
25. In a diagram form, we can say:
P(x)
Possible integer roots:
Divisors of the constant term:
a, b, c…
a is a root
(x – a) is a factor
b is not a root
26. P(x)
0
x-a
P(x)
C(x)
Possible integer roots:
Divisors of the constant term:
a, b, c…
When the
If the
remainder of
remainder of
the division is
division is
zero…
zero…
a is a root
(x – a) is a factor
b is not a root
27. P(x)
P(a) = 0
If the
numerical value
P(a) = 0
Possible integer roots:
Divisors of the constant term:
a, b, c…
a is a root
(x – a) is a factor
b is not a root
28. P(x)
If “a” is a
solution of the
equation P(x) = 0
Possible integer roots:
Divisors of the constant term:
a, b, c…
a is a root
(x – a) is a factor
b is not a root
29. Therefore, if we know every
root of the polynomial P(x), so
to speak,
a, b, c, etc…
The factors will be:
(x-a), (x-b), (x-c), etc…
