EXAMPLE 5 Use the result to write f (x) as a product of two factors. Then factor completely. f (x) =...
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Transcript of EXAMPLE 5 Use the result to write f (x) as a product of two factors. Then factor completely. f (x) =...
EXAMPLE 5
Use the result to write f (x) as a product of two factors. Then factor completely.
f (x) = x3 – 2x2 – 23x + 60
The zeros are 3, – 5, and 4.
Standardized Test Practice
The correct answer is A. ANSWER
= (x – 3)(x + 5)(x – 4)
= (x – 3)(x2 + x – 20)
EXAMPLE 1 List possible rational zeros
List the possible rational zeros of f using the rational zero theorem.
a. f (x) = x3 + 2x2 – 11x + 12
Factors of the constant term: + 1, + 2, + 3, + 4, + 6, + 12
Factors of the leading coefficient: + 1
Possible rational zeros: + , + , + , + , + , +
11
21
31
41
61
121
Simplified list of possible zeros: + 1, + 2, + 3, + 4, + 6, + 12
EXAMPLE 1 List possible rational zeros
b. f (x) = 4x4 – x3 – 3x2 + 9x – 10
Factors of the constant term: + 1, + 2, + 5, + 10
Factors of the leading coefficient: + 1, + 2, + 4
+ , + , + , + , + , + , + , + , + , + , +
Possible rational zeros:11
21
51
101
12
22
52
102
14
24
54
+ 104
Simplified list of possible zeros: + 1, + 2, + 5, + 10, + , + , +
52
14
125
4+
EXAMPLE 2
Find all real zeros of f (x) = x3 – 8x2 +11x + 20.
SOLUTION
List the possible rational zeros. The leading coefficient is 1 and the constant term is 20. So, the possible rational zeros are:
x = + , + , + , + , + , +51
41
21
11
101
201
STEP 1
Find zeros when the leading coefficient is 1
EXAMPLE 2
STEP 2
Find zeros when the leading coefficient is 1
1 1 – 8 11 20Test x =1:
1 – 7 41 – 7 4 24
Test x = –1:
–1 1 –8 11 20
1 – 9 20 0 –1 9 20
1 is not a zero.↑
–1 is a zero↑
Test these zeros using synthetic division.
EXAMPLE 2
Because –1 is a zero of f, you can write f (x) = (x + 1)(x2 – 9x + 20).
STEP 3
f (x) = (x + 1) (x2 – 9x + 20)
Factor the trinomial in f (x) and use the factor theorem.
The zeros of f are –1, 4, and 5.
ANSWER
= (x + 1)(x – 4)(x – 5)
Find zeros when the leading coefficient is 1
EXAMPLE 1 Find the number of solutions or zeros
a. How many solutions does the equation x3 + 5x2 + 4x + 20 = 0 have?
SOLUTION
Because x3 + 5x2 + 4x + 20 = 0 is a polynomial equation of degree 3,it has three solutions. (The solutions are – 5, – 2i, and 2i.)
EXAMPLE 1 Find the number of solutions or zeros
b. How many zeros does the function f (x) = x4 – 8x3 + 18x2 – 27 have?
SOLUTION
Because f (x) = x4 – 8x3 + 18x2 – 27 is a polynomial function of degree 4, it has four zeros. (The zeros are – 1, 3, 3, and 3.)
EXAMPLE 2
Find all zeros of f (x) = x5 – 4x4 + 4x3 + 10x2 – 13x – 14.
SOLUTION
STEP 1 Find the rational zeros of f. Because f is a polynomial function of degree 5, it has 5 zeros. The possible rational zeros are + 1, + 2, + 7, and + 14. Using synthetic division, you can determine that – 1 is a zero repeated twice and 2 is also a zero.
STEP 2 Write f (x) in factored form. Dividing f (x) by its known factors x + 1, x + 1, and x – 2 gives a quotient of x2 – 4x + 7. Therefore:
f (x) = (x + 1)2(x – 2)(x2 – 4x + 7)
Find the zeros of a polynomial function
EXAMPLE 2
STEP 3 Find the complex zeros of f . Use the quadratic formula to factor the trinomial into linear factors.
f(x) = (x + 1)2(x – 2) x – (2 + i 3 ) x – (2 – i 3 )
The zeros of f are – 1, – 1, 2, 2 + i 3 , and 2 – i 3.
ANSWER
Find the zeros of a polynomial function
EXAMPLE 4 Use Descartes’ rule of signs
Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for f (x) = x6 – 2x5 + 3x4 – 10x3 – 6x2 – 8x – 8.
f (x) = x6 – 2x5 + 3x4 – 10x3 – 6x2 – 8x – 8.
The coefficients in f (x) have 3 sign changes, so f has 3 or 1 positive real zero(s).
f (– x) = (– x)6 – 2(– x)5 + 3(– x)4 – 10(– x)3 – 6(– x)2 – 8(– x) – 8
= x6 + 2x5 + 3x4 + 10x3 – 6x2 + 8x – 8
SOLUTION
EXAMPLE 4 Use Descartes’ rule of signs
The coefficients in f (– x) have 3 sign changes, so f has 3 or 1 negative real zero(s) .
The possible numbers of zeros for f are summarized in the table below.
EXAMPLE 5 Approximate real zeros
Approximate the real zeros of f (x) = x6 – 2x5 + 3x4 – 10x3 – 6x2 – 8x – 8.
SOLUTION
Use the zero (or root) feature of a graphing calculator, as shown below.
From these screens, you can see that the zeros are x ≈ – 0.73 and x ≈ 2.73.
ANSWER
EXAMPLE 6 Approximate real zeros of a polynomial model
s (x) = 0.00547x3 – 0.225x2 + 3.62x – 11.0
What is the tachometer reading when the boat travels 15 miles per hour?
A tachometer measures the speed (in revolutions per minute, or RPMs) at which an engine shaft rotates. For a certain boat, the speed x of the engine shaft (in 100s of RPMs) and the speed s of the boat (in miles per hour) are modeled by
TACHOMETER
EXAMPLE 6 Approximate real zeros of a polynomial model
Substitute 15 for s(x) in the given function. You can rewrite the resulting equation as:
0 = 0.00547x3 – 0.225x2 + 3.62x – 26.0
Then, use a graphing calculator to approximate the real zeros of f (x) = 0.00547x3 – 0.225x2 + 3.62x – 26.0.
SOLUTION
From the graph, there is one real zero: x ≈ 19.9.
The tachometer reading is about 1990 RPMs.ANSWER