Transcript 4-5 & 6, Factor and Remainder Theorems revised

4-5, 4-6 Factor and
Remainder Theorems
If r is a real number that is a zero of a function then
f x   0
•r is an x intercept of the graph of the function
•x = r is a solution, or root, of the equation
•x - r is a factor of the polynomial
First: How many zeros are we expecting?
How many were there in each of our quadratic examples?
Fundamental Theorem of Algebra
Every polynomial of degree n≥1 has at least one complex zero.
Every polynomial of degree n≥1 has exactly n complex zeros, counting multiple
roots.
Complex zeros are real, imaginary or a combination
To find the zeros, solve the equation when
f x   0
Begin with quadratics.
Techniques for solving quadratic equations
•Factor, set factors = to zero, solve
• Complete the square
• Use quadratic formula
Reminder:
f(x)=x4 +27x
0=x4 + 27x
0= x(x3 + 27)
0=x(x3 + 33)
x(x + 3)(x2 – x  3 + 32)
Use the rule a3 + b3 = (a + b)
 (a2 – ab + b2).
0=x(x + 3)(x2 – 3x + 9)
x0
x3  0
x3
0  x  3x  9
2
f x   x  2 x  8
2
0  x  2x  8
2
x4
x  2
f x   x  2 x  8
2
15
10
fx = x2-2x-8
5
-20
-10
10
-5
-10
Roots are real and rational
20
f x   x  4 x  4
2
0  x  4x  4
2
x2
x2
The multiplicity of root r is the number of times it
occurs as a root.
-2 occurs twice as a root, so it has a multiplicity of 2
When a real root has even multiplicity, the graph of y =
P(x) touches the x-axis but does not cross it.
When a real root has odd multiplicity greater than 1,
the graph “bends” as it crosses the x-axis.
You cannot always determine the multiplicity of a
root from a graph. It is easiest to determine
multiplicity when the polynomial is in factored
form.
f x   x  4 x  4
2
10
5
x
-10
10
-5
2 is a double root, real, and rational
20
f ( x)  x  5 x  3
0  x  5x  3
2
2
 b  b  4ac
x
2a
2
5  13
x
2
5  13 4.3
x

2
5  13
x

2
.7
f ( x)  x  5 x  3
2
4
2
5
-2
Roots are real but irrational
10
In all of the previous 3 examples, the roots were real, they were all xintercepts, places where the graph crossed the x axis.
Now consider the function
f x   x  2 x  5
2
f x   x  2 x  5
2
 b  b  4ac
x
2a
2   16
2  i 16
2
2
21  2i 
1 2i
2
2
Roots are imaginary.
2  4i
2
f x   x  2 x  5
2
12
10
8
6
4
2
-10
-5
5
10
-2
Roots are imaginary, graph does NOT cross the x axis.
What about other polynomials, not quadratic?
f(x)=4x4 + 108x
f(x)x4 + 25 = 26x2
f(x)= 2x6 – 10x5 – 12x4 = 0
Where to start?
Synthetic division is a shorthand method of
dividing a polynomial by a linear binomial by
using only the coefficients. For synthetic division
to work, the polynomial must be written in
standard form, using 0 and a coefficient for any
missing terms, and the divisor must be in the
form (x – a).
Divide using synthetic division.
(3x4 – x3 + 5x – 1) ÷ (x + 2)
Step 1 Find a.
a = –2
For (x + 2), a = –2.
Step 2 Write the coefficients and a in the synthetic
division format.
–2 3 – 1
0
5 –1
Use 0 for the coefficient
of x2.
Step 3 Bring down the first coefficient. Then
multiply and add for each column.
–2 3 –1
0
5 –1
–6 14 –28 46
3 –7 14 –23 45
Draw a box around the
remainder, 45.
Step 4 Write the quotient.
3x3
–
7x2
45
+ 14x – 23 +
x+2
Write the remainder over
the divisor.
You can use synthetic division to evaluate polynomials.
This process is called synthetic substitution. The
process of synthetic substitution is exactly the same as
the process of synthetic division, but the final answer is
interpreted differently, as described by the Remainder
Theorem.
Use synthetic substitution to evaluate the
polynomial for the given value.
P(x) = 2x3 + 5x2 – x + 7 for x = 2.
2
2
5 –1
2
4
9
7
18 34
17 41
P(2) = 41
Check Substitute 2 for x in P(x) = 2x3 + 5x2 – x + 7.
P(2) = 2(2)3 + 5(2)2 – (2) + 7
P(2) = 41 
Use synthetic substitution to evaluate the
polynomial for the given value.
1
P(x) = 6x4 – 25x3 – 3x + 5 for x = –
.
3
1
6 –25 0 –3 5
–
3
–2 9 –3 2
6 –27
P( 1 ) = 7
3
9 –6
7
The Remainder Theorem states that if a
polynomial is divided by (x – a), the remainder
is the value of the function at a.
Remember that the value of a function is the y value at any
given x value. If a is a root (or zero) of a function, then the
value of the function at a is zero.
So, if a is a zero, then (x – a) is a factor of P(x), then P(a) = 0.
Determine whether the given binomial is a factor
of the polynomial P(x).
A. (x + 1); (x2 – 3x + 1)
Find P(–1) by synthetic
substitution.
–1
1 –3 1
1
–1
–4
4
5
B. (x + 2);
(3x4 + 6x3 – 5x – 10)
Find P(–2) by synthetic
substitution.
–2 3
6
0 –5 –10
–6 0 0
3 0 0 –5
10
0
5-Minute Check Lesson 4-4A
Not all polynomials are factorable, but the Rational Root
Theorem can help you find all possible rational roots of
a polynomial equation.
Lesson Overview 4-4A
5-Minute Check Lesson 4-5A
Descartes’ Rule of Signs
Suppose P(x) is a polynomial whose terms are
arranged in descending powers of the variable.
Then the number of positive real zeros is the same as
the number of changes in sign of the coefficients of the
terms or is less than this by an even number.
f ( x )  2 x 5  3x 4  6 x 3  6 x 2  8 x  3
no
yes
yes
yes
yes
There are 4 changes. Therefore there are 4, 2, or 0
positive real roots.
Descartes’ Rule of Signs
The rule can also be applied to find the number of negative
real roots. First find f(-x) and count the sign changes.
f (  x )  2 x  3x  6 x  6 x  8x  3
5
4
yes
no
3
no
2
no
no
The number of negative real zeros is the same as the
number of changes in sign of the coefficients of the terms
or is less than this by an even number.
There is 1 change. Therefore there is 1 negative real
root.
You can use a combination of the Fundamental
Theorem of Algebra, the Rational Root Theorem,
Descartes’ Rule, Graphing, and Synthetic Substitution
to determine all roots for any given polynomial.
Identify all the real roots of 2x3 – 9x2 + 2 = 0.
By the Fundamental Theorem of Algebra, we know
that there are 3 complex roots.
The Rational Root Theorem identifies possible
rational roots. p = 2 and q = 2
±1, ±2 = ±1, ±2, ± 1 .
±1, ±2
2
Lesson Overview 4-4A
5-Minute Check Lesson 4-5A
You can use a combination of the Fundamental
Theorem of Algebra, the Rational Root Theorem,
Descartes’ Rule, Graphing, and Synthetic Substitution
to determine all roots for any given polynomial.
Identify all the real roots of 2x3 – 9x2 + 2 = 0.
By the Fundamental Theorem of Algebra, we know
that there are 3 complex roots.
The Rational Root Theorem identifies possible
rational roots. p = 2 and q = 2
±1, ±2 = ±1, ±2, ± 1 .
±1, ±2
2
Applying Descartes’ Rule of Signs
2x3 – 9x2 + 2 = 0.
-2x3 – 9x2 + 2 = 0.
There are 2 or 0 positive real roots
There is 1 negative real root.
Graph y = 2x3 – 9x2 + 2 to find the x-intercepts.
gx = 2x3-9x2+2
3
2
1
-4
-2
2
-1
-2
-3
4
6
Test the possible rational root 1 .
2
1
2
2 –9
0 2
1 –4 –2
2 –8 –4 0
Test 1 . The remainder is
2 1
0, so (x – ) is a factor.
2
remainder
constant
coefficient of x
coefficient of x2
2 x  8x  4  0
2
Is the Depressed Equation.
1
The polynomial factors into (x –
)(2x2 – 8x – 4).
2
Solve 2x2 – 8x – 4 = 0 to find the remaining roots.
2(x2 – 4x – 2) = 0
x
4±
16+8
2 
2
6
Factor out the GCF, 2
Use the quadratic formula to
identify the irrational roots.
The fully factored equation is

1
2 x –
2





 x – 2 –
x
–
2
+
6




6  = 0
1
The roots are
, 2  6 , and 2  6 .
2
Identify all the real roots of
f ( x )  x  7 x  13x  3x  18
4
3
2
By the Fundamental Theorem of Algebra, we know
that there are 4 roots. They may be real, imaginary or
complex.
The Rational Root Theorem identifies possible
rational roots. p = 18 and q = 1
P : 1, 2 , 3, 6 , 9 , 18
Q : 1
Applying Descartes’ Rule of Signs
3 or 1 positive real roots
1 negative real root
Graph the function to get a starting place.
Use synthetic substitution
and the factor theorem to
test for roots
-1
1
1
2
1
1
-7
13
3
-18
-1
8
-21
18
-8
21
-18
0
-8
21
-18
2
-12
18
-6
9
0
We are left with the depressed equation
x  6x  9  0
2
This quadratic factors easily into
 x  3 x  3  0
Therefore 3 is a double root.
Finally the roots are -1, 2 and 3 (double).
The factors are
 x  3 x  3 x  1 x  2
The Conjugate Pairs Theorem says that irrational
roots and imaginary roots come in conjugate
pairs. For example, if you know that 1 +
is a
root of x3 – x2 – 3x – 1 = 0, then you know that
1–
is also a root.
If you know that -3i is a root, then 3i is also a root.
If you know that 4+2i is a root, then 4-2i is also a root.
Identify the roots of each equation. State the
multiplicity of each root.
0 and 2 each with
1. 5x4 – 20x3 + 20x2 = 0
multiplicity 2
2. x3 – 12x2 + 48x – 64 = 0
4 with multiplicity 3
3. Identify all the real roots of x3 + 5x2 – 3x – 3 = 0.
1, 3 + 6, 3  6
4. x3 + 9 = x2 + 9x
–3, 3, 1