7.5.1 Zeros of Polynomial 7.5.1 Zeros of Polynomial FunctionsFunctions
7.5.1 Zeros of Polynomial 7.5.1 Zeros of Polynomial FunctionsFunctions
Objectives: •Use the Rational Root Theorem to find the zeros of a polynomial function
Rational Root Theorem
• p is a factor of the constant term of f(x)
Let f(x) be a polynomial function with integer coefficients in standard form. If (in lowest terms) is a root of f(x) = 0, then
pq
• q is a factor of the leading coefficient of f(x)
Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.
Step 1: Make an organized list of all possible roots.
factors of 2: 1, 2
factors of 8: 1, 2, 4, 8
1 1 1 1, , ,
1 2 4 8
2 2 2 2, , ,
1 2 4 8
List: ±1, ±½, ±¼, ± ⅛ , ±2
Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.
Step 2: Use substitution or synthetic division to test possible roots, until you find one that works.
List: ±1, ±½, ±¼, ± ⅛ , ±2
Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.
Step 3: Use factoring or quadratic formula to find the other two.
Found: ½ Resulting Equation: 8x2 + 14x – 4 = 0
Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.
Step 2: Use a graphing calculator to identify possible roots.
possible roots: -21
,4
1,2
1 1 1 1, , ,
1 2 4 8
2 2 2 2, , ,
1 2 4 8
Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.
Step 3: Use substitution or synthetic division to test all possible roots.
possible roots: -21
,4
1,2
8
-16
-6
121
-2
0
-2 8 10 -11 2
roots: -2
Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.
Step 3: Use substitution or synthetic division to test all possible roots.
possible roots: -21
,4
1,2
8 10 -11 2
roots -2
14
8
2
12
3
-8
-2
0
1,
4
Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.
Step 3: Use substitution or synthetic division to test all possible roots.
possible roots: -21
,4
1,2
8 10 -11 2
roots: -2
12
8
4
14
7
-4
-2
0
1,
41
,2
Example 2Find all of the zeros of Q(x) = x3 + 4x2 – 6x - 12.First, use the Rational Root Theorem and a graph of the polynomial function to determine some possibilities.Then use synthetic division to test your choices.
1
2
6
12
6
12
0
2 1 4 -6 -12
Since the remainder is 0, x – 2 is a factor of x3 + 4x2 – 6x - 12.
Example 2Find all of the zeros of Q(x) = x3 + 4x2 – 6x - 12.
Since the remainder is 0, x – 2 is a factor of x3 + 4x2 – 6x - 12.
x3 + 4x2 – 6x – 12 = 0
(x – 2)(x2 + 6x + 6) = 0x = 2
or
26 6 4(1)(6)
x2(1)
6 12x
2
6 2 3x
2
x 3 3
Homework
p.463 #11-21 Odd
7.5.2 Zeros of Polynomial 7.5.2 Zeros of Polynomial FunctionsFunctions
7.5.2 Zeros of Polynomial 7.5.2 Zeros of Polynomial FunctionsFunctions
Objectives: •Use the Complex Conjugate Root Theorem to find the zeros of a polynomial function•Use the Fundamental Theorem to write a polynomial function given sufficient information about its zeros
Example 1Find all of the zeros of P(x) = -4x3 + 2x2 – x + 3.First, use the Rational Root Theorem and a graph of the polynomial function to determine some possibilities.Then use synthetic division to test your choices.
-4
-4
-2
-2
-3
-3
0
1 -4 2 -1 3
Since the remainder is 0, x – 1 is a factor of -4x3 + 2x2 – x + 3.
-4x3 + 2x2 – x + 3 = 0
(x – 1)(-4x2 - 2x - 3) = 0
x = 1
or
2( 2) ( 2) 4( 4)( 3)x
2( 4)
2 44
x8
2 2 11
x8
1 i 11
x4
Example 1Find all of the zeros of P(x) = -4x3 + 2x2 – x + 3.
Since the remainder is 0, x – 1 is a factor of -4x3 + 2x2 – x + 3.
Complex Conjugate Root Theorem
If P is a polynomial function with real-number coefficients and a + bi (where b = 0) is a root of P(x) = 0, then a – bi is also a root of P(x) = 0.
Graph R(x) = 2x3 – x2 – 4x. How many real zeros does R have?
Graph S(x) = 2x3 – x2 – 4x + 3. How many real zeros does S have?
Graph T(x) = 2x3 – x2 – 4x + 6. How many real zeros does T have?
Homework
Page 464 #23-33 Odd
Example 2Write a polynomial function, P, given that P has a degree of 3, P(0) = 120, and its zeros are -3, 2, and 4.
x = -3 and
P(x) = a(x + 3)(x – 2)(x – 4)
Since the zeros are -3, 2, and 4, x = 4
x + 3= 0
x – 4= 0and
x = 2
x - 2 = 0
P(x) = a(x2 + x - 6)(x – 4)
P(x) = a(x3 - 3x2 – 10x + 24)
120 = a((0)3 - 3(0)2 – 10(0) + 24)120 = 24 a a = 5
P(x) = 5(x3 - 3x2 – 10x + 24)
P(x) = 5x3 - 15x2 – 50 x + 120
Example 3Write a polynomial function, P, given that P has a degree of 2, P(0) = 18, and its zeros are -3 + 3i and -3 – 3i.
x = -3 + 3i
and
x + 3 - 3i( )( ) = 0
x2 + 3x + 3xi x2 + 6x + 9 – 9(-1)
= 0
Since the zeros are -3 + 3i and -3 – 3i,
x = -3 - 3i
x – (-3 + 3i) = 0
x – (-3 - 3i) = 0
and
x + 3 - 3i = 0
x + 3 + 3i = 0
and
x + 3 + 3i
+ 3x + 9 + 9i
– 3xi – 9i – 9i2 = 0
x2 + 6x + 18 = 0
Homework
Page 464 #41-49 Odd