Lesson 3.4 Rational Root Test and Zeros of Polynomials

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Transcript Lesson 3.4 Rational Root Test and Zeros of Polynomials

Lesson 3.4 – Zeros of Polynomial Functions
Rational Zero Theorem a0 x n  a1 x n 1  ...  an 1 x  an  0
Represent a polynomial equation of degree n . If a rational
number p , where p and q have no common factors, is a
q
root of the equation, then p is a factor of the constant term
and q is a factor of the leading coefficient.
Ex. 1
List all possible roots of
Then determine the rational roots.
6 x3  11x 2  3x  2  0
List possible values of p:
1, 2
List possible values of q:
1, 2, 3, 6
Possible rational roots:
p
q
1 1 1 2
1, 2,  ,  ,  , 
2 3 6 3
You Try:
List all possible rational zeros of
f(x) = x3 + 2x2 – 5x – 6
Possible values of p: 1, 2, 3, 6
Possible values of q: 1
Possible rational roots(p/q): 1, 2, 3, 6
Finding Zeros of a Polynomial
Function
Now, use synthetic division to test and find
the roots/factors. The last number must be
a zero to show the root is a factor. Degree
is 3, so there should be 3 solutions.
6 x  11x  3x  2  0
3
2
Possible rational roots:
1
1
1
2
1, 2,  ,  ,  , 
2
3
6
3
Checking with Synthetic Division
1
6 11 -3 -2
6 17 14
6 17 14 12
Now let’s try -2.
-2 6 11 -3 -2
-12 2 2
6 -1 -1 0
1 is not a zero
because the
remainder does
not equal 0!!
-2 is a zero!!!
Finding the Zero (cont.)
Take -2 and write it as a factor which is x+ 2 and
take your answer from synthetic division and put
it into a polynomial 6x2 – x -1.
Now factor 6x2 – x -1
(2x – 1 )(3x + 1)
Now put all the factors together
(x+2)(2x-1)(3x+1).
Put factors equal to zero to find the zeros.
X= -2, ½, -1/3 (3 real rational solutions)
The process:
Don’t forget:
Step 1:Find your p’s and q’s and list all possible roots.
Step 2:Number of roots/zeros is based on highest
degree.
Use synthetic division to find your first root. If that does
not work, USE YOUR CALCULATOR!!! Remember your
multiplicity ideas as well. If the polynomial crosses the
x axis, the multiplicity is odd. If the polynomial touches
and turns around, it is even.
Step3: After finding a root, factor the rest on your own.
If not factorable, use the quadratic formula.
Step 4: Then, solve for the rest of the roots. Roots can be
real or imaginary. If the roots are imaginary, then they
occur in conjugate pairs!
To set up factors (in parenthesis) just change their signs.
You Try!!
Find all zeros of f(x) = x3 + 7x2 + 11x – 3
Step 1 – Find possible rational roots.
p: 1, 3 q: 1
possible rational roots:
1, 3
Use synthetic division to find one rational root or by the
calculator. By using the calculator, find one zero. Show
on the calculator to class.
Hint: You will need to use the quadratic formula
One root is 3 from calculator. Now find the other roots.
How many should there be?
3
Answer:
The solution set is {-3,-2 - √5, -2+√5}
Your solutions can be imaginary or real. If
your solution is imaginary, it will be written
as a complex conjugate. If it is real, it
could be rational (nice numbers) or
irrational (not nice numbers).
You Try Again:
Solve: x4 + 6x3 + 22x2 – 30x + 13
Use Calculator to find two zeros.
Answer: {1,2-3i,2+3i}
Zeros of Polynomial Functions
Complex
Numbers
(a+bi)
Imaginary
Numbers
REAL number
system
(+bi)
Rational
Numbers
Irrational
Numbers
General shapes of graphs with a positive leading coefficient.
Degree 1
1 zero
Degree 2
Degree 3
2 zeros
3 zeros
Degree 5
Degree 4
4 zeros
Remember, zeros are just x-intercepts.
5 zeros
Finding a Polynomial Function with Given Zeros
EXAMPLE 1: Find a 3rd degree polynomial function f(x) with real
coefficients that has -3 and i as zeros and such that f(1) =8.
f(x)= an (x-c1)(x-c2)(x-c3)
Now substitute in the zeros with what you know.
Do not forget about the conjugate pairs.
f(x) = an(x+3)(x+i)(x-i)
Multiply the polynomial out.
)
f(x)= an(x3 + 3x2 + x + 3
f(1)= an[(1)3 + 3(1)2 + 1 +3] = 8an
f(1) =8
8 = 8an
an = 1
Polynomial Equation is f(x) = (x+3) (x2+1) or x3 + 3x2 +x + 3
YOU TRY: Find a 3rd degree polynomial function f(x)
with real coefficients that has 4 and 2i as zeros and
such that f(-1) =-50.
Answer: f(x) = 2x3 – 8x2 +8x -32
Summary:
Describe how to find the possible rational
zeros of a polynomial function.