zeros of polynomials 2.5. properties of polynomial equations a polynomial of degree n, has n roots...
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Zeros of Polynomials
2.5
Properties of Polynomial Equations
A polynomial of degree n, has n roots (counting multiple roots separately)
If a + bi is a root, then a – bi is also a root. Complex roots occur in conjugate pairs!
Descartes’ Rule of Signs
State the possible number of positive real zeros, negative real zeros, and imaginary zeros of p(x) = –x6 + 4x3 – 2x2 – x – 1.
Since p(x) has degree 6, it has 6 zeros. However, some of them may be imaginary. Use Descartes’ Rule of Signs to determine the number and type of real zeros. Count the number of changes in sign for the coefficients of p(x).
Find Numbers of Positive and Negative Zeros
p(x) = –x6 + 4x3 – 2x2 – x –1
yes– to +
yes+ to –
no– to –
no– to –2 or 0 positive real
zeros
Since there are two sign changes, there are 2 or 0 positive real zeros. Find p(–x) and count the number of sign changes for its coefficients.
Since there are two sign changes, there are 2 or 0 negative real zeros. Make a chart of possible combinations.
Find Numbers of Positive and Negative Zeros
p(–x) = –(–x)6 + 4(–x)3 – 2(–x)2 – (–x) –1
no– to –
no– to –
yes– to +
yes+ to –
Answer:
Find Numbers of Positive and Negative Zeros
Find all of the zeros of f(x) = x3 – x2 + 2x + 4.
Since f(x) has degree of 3, the function has three zeros. To determine the possible number and type of real zeros, examine the number of sign changes in f(x) and f(–x).
yes yes no
no no yes
f(x) = x3 – x2 + 2x + 4
f(–x) = –x3 – x2 – 2x + 4
2 or 0 positive real zeros
1 negative real zero
Rational Zero Theorem
n
0
a of Factors
a of Factors ZerosPossible
Identify Possible Zeros
A. List all of the possible rational zeros of f(x) = 3x4 – x3 + 4.
Answer:
B. List all of the possible rational zeros of f(x) = x3 + 3x + 24. <YU TRY>
A.
B.
C.
D.
Use the Factor Theorem
Show that x – 3 is a factor of x3 + 4x2 – 15x – 18. Then find the remaining factors of the polynomial.
The binomial x – 3 is a factor of the polynomial if 3 is a zero of the related polynomial function. Use the factor theorem and synthetic division.
1 7 6 0
1 4 –15 –18
3 21 18
Use the Factor Theorem
Since the remainder is 0, (x – 3) is a factor of the polynomial. The polynomial x3 + 4x2 – 15x –18 can be factored as (x – 3)(x2 + 7x + 6). The polynomial x2 + 7x + 6 is the depressed polynomial. Check to see if this polynomial can be factored.
x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial.
Answer: So, x3 + 4x2 – 15x – 18 = (x – 3)(x + 6)(x + 1).
Example
24860 Solve 24 xx x
24 12, 8, 6, 4, 3, 2, 1, : ZerosPossible
0 12- 2- 2 1
24- 4- 4 2
24 8- 6- 0 1 2
0 6 4 1
12 8 2
12- 2- 2 1 2
2234 2642486 xxxxxx
Repeated zero at x = 2
Upper and Lower Bounds
Suppose f(x) is divided by x – c using synthetic division
If c > 0 and each number in the last row is either positive or zero, then c is an upper bound for real zeros
If c < 0 and each number in the last row are alternatively positive or negative (zero counts as both), then c is a lower bound
13302260 Solve 234 xxxx
0 13- 17 5- 1
13- 17 5- 1
13 30- 22 6- 1 1
The degree is 4, so there are 4
roots!
Possible roots are ±1, ±13
Use Synthetic Division to
find the roots0 13 4- 1
13 4- 1
13- 17 5- 1 1
1342 xxUse Quadratic Formula to find
remaining solutions
ii
322
64
2
364
)1(2
)13)(1(4164
ix 32 1, ,1 Multiplicity
of 2
f(x) = x3 – x2 + 2x + 4.
There are 3 zeros, we know there are 2 or 0
positive and 1 negative
Possible roots are ±1, ±2, ±4
Use Synthetic Division to
find the roots
-4, -2, -1, 1, 2, 4
6 2 0 1
2 0 1
4 2 1- 1 1
12 4 1 1
8 2 2
4 2 1- 1 2
We need either 2 or 0 positive, so 4 cannot be a zero
0 4 2- 1
4- 2 1-
4 2 1- 1 1
-1 works, so it is our 1 negative zero. The
other two have to be imaginary.
Use Quadratic Formula to find
remaining solutions 31
2
322
2
122
)1(2
)4)(1(442i
i
1,31 ix
151620 Solve 24 xxx
There are 4 zeros, We know there are 1 positive
and 3 or 1 negative15)(16)(2)()( 24 xxxxf
Possible roots are ±1, ±3, ±5, ±15
-15, -5, -3 -1, 1, 3, 5, 15
480 99 23 5 1
495 115 25 5
15- 16- 2- 0 1 5
All positive so 5 is an upper bound
0 5 7 3 1
15 21 9 3
15- 16- 2- 0 1 3
3 is a solution. This is our 1 positive zero
Now try negatives
16- 7 0 1
21- 0 3-
5 7 3 1 3Alternates between
positive and negative so -3 is a lower bound0| 5 2 1
5- 2- 1-
5 7 3 1 1-1 is our 1 negative. Use the quadratic to find the
remaining 2 zeros
ii
212
42
2
162
)1(2
)5)(1(442
3,1,21 ix
Linear Factorization Theorem
Find a fourth degree polynomial with real coefficients that has 2, -2 and i as zeros and such that f(3) = -150
ixixxxaxf n 22)( Foil
14)( 22 xxaxf nFoil
43)( 24 xxaxf nSubstitute f(3) = -150
4)3(3)3(150 24 na Solve for a
na 3 Substitute back in equation
1293433)( 2424 xxxxxf
Linear Factorization Theorem
An nth – degree polynomial can be expressed as the product of a nonzero constant and n linear factors
283factor Completely 24 xx
rationals over the factored 47 22 xx
reals over the factored 477 2 xxx
factored completely 2i-x277 ixxx