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Warm-Up Exercises
1. What is the degree of f (x) = 8x6 – 4x5 + 3x2 + 2?
ANSWER
6
2. Solve x2 – 2x + 3 = 0
ANSWER
_ i 2
1+
Warm-Up Exercises
3. The function P given by x4 + 3x3 – 30x2 – 6x = 56 model
the profit of a company. What are the real solution of
the function?
ANSWER
_ 2, 4
–7, +
Warm-Up1Exercises
EXAMPLE
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.)
Warm-Up1Exercises
EXAMPLE
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.)
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 1
1. How many solutions does the equation
x4 + 5x2 – 36 = 0 have?
ANSWER
4
Warm-Up
Exercises
GUIDED
PRACTICE
2.
for Example 1
How many zeros does the function
f (x) = x3 + 7x2 + 8x – 16 have?
ANSWER
3
Warm-Up2Exercises
EXAMPLE
Find the zeros of a polynomial function
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)
Warm-Up2Exercises
EXAMPLE
Find the zeros of a polynomial function
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 )
ANSWER
The zeros of f are – 1, – 1, 2, 2 + i 3 , and 2 – i 3.
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 2
Find all zeros of the polynomial function.
3.
f (x) = x3 + 7x2 + 15x + 9
SOLUTION
STEP 1 Find the rational zero of f. because f is a
polynomial function degree 3, it has 3 zero. The
possible rational zeros are –+ 1 , –+3, using
synthetic division, you can determine that 3 is
a zero reputed twice and –3 is also a zero
STEP 2 Write f (x) in factored form
Formula are (x +1)2 (x +3)
f(x) = (x +1) (x +3)2
The zeros of f are – 1 and – 3
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 2
4. f (x) = x5 – 2x4 + 8x2 – 13x + 6
SOLUTION
STEP 1 Find the rational zero of f. because f is a
polynomial function degree 5, it has 5 zero. The
possible rational zeros are –+ 1 , –+2, –+ 3 and –+ 6.
Using synthetic division, you can determine that
1 is a zero reputed twice and –3 is also a zero
STEP 2 Write f (x) in factored form dividing f(x)by its known
factor (x – 1),(x – 1)and (x+2) given a qualities x2 – 2x
+3 therefore
f (x) = (x – 1)2 (x+2) (x2 – 2x + 3)
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 2
STEP 3 Find the complex zero of f. use the quadratic
formula to factor the trinomial into linear factor
f (x) = (x –1)2 (x + 2) [x – (1 + i 2) [ (x – (1 – i 2)]
Zeros of f are 1, 1, – 2, 1 + i 2 , and 1 – i 2
Warm-Up3Exercises
EXAMPLE
Use zeros to write a polynomial function
Write a polynomial function f of least degree that has
rational coefficients, a leading coefficient of 1, and 3
and 2 + 5 as zeros.
SOLUTION
Because the coefficients are rational and 2 + 5 is a
zero, 2 – 5 must also be a zero by the irrational
conjugates theorem. Use the three zeros and the
factor theorem to write f (x) as a product of three
factors.
Warm-Up3Exercises
EXAMPLE
Use zeros to write a polynomial function
f (x) = (x – 3) [ x – (2 + √ 5 ) ] [ x – (2 – √ 5 ) ] Write f (x) in
factored form.
= (x – 3) [ (x – 2) – √ 5 ] [ (x – 2) +√ 5 ]
Regroup terms.
= (x – 3)[(x – 2)2 – 5]
Multiply.
= (x – 3)[(x2 – 4x + 4) – 5]
Expand binomial.
= (x – 3)(x2 – 4x – 1)
Simplify.
= x3 – 4x2 – x – 3x2 + 12x + 3
Multiply.
= x3 – 7x2 + 11x + 3
Combine like
terms.
Warm-Up3Exercises
EXAMPLE
Use zeros to write a polynomial function
CHECK
You can check this result by evaluating f (x) at each of
its three zeros.
f(3) = 33 – 7(3)2 + 11(3) + 3 = 27 – 63 + 33 + 3 = 0 
f(2 + √ 5 ) = (2 + √ 5 )3 – 7(2 + √ 5 )2 + 11( 2 + √ 5 ) + 3
= 38 + 17 √ 5 – 63 – 28 √ 5 + 22 + 11√ 5 + 3
=0
Since f (2 + √ 5 ) = 0, by the irrational conjugates
theorem f (2 – √ 5 ) = 0. 
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 3
Write a polynomial function f of least degree that has
rational coefficients, a leading coefficient of 1, and
the given zeros.
5. – 1, 2, 4
Use the three zeros and the factor theorem to write f(x)
as a product of three factors.
SOLUTION
f (x) = (x + 1) (x – 2) ( x – 4)
= (x + 1) (x2 – 4x – 2x + 8)
= (x + 1) (x2 – 6x + 8)
= x3 – 6x2 + 8x + x2 – 6x + 8
= x3 – 5x2 + 2x + 8
Write f (x) in factored form.
Multiply.
Combine like terms.
Multiply.
Combine like terms.
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 3
6. 4, 1 + √ 5
Because the coefficients are rational and 1 + 5 is a zero,
1 – 5 must also be a zero by the irrational conjugates
theorem. Use the three zeros and the factor theorem to
write f (x) as a product of three factors
SOLUTION
f (x) = (x – 4) [ x – (1 + √ 5 ) ] [ x – (1 – √ 5 ) ]Write f (x) in factored
= (x – 4) [ (x – 1) – √ 5 ] [ (x – 1) +√ 5 ]
form.
Regroup terms.
= (x – 4)[(x – 1)2 – ( 5)2]
Multiply.
= (x – 4)[(x2 – 2x + 1) – 5]
Expand binomial.
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 3
= (x – 4)(x2 – 2x – 4)
Simplify.
= x3 – 2x2 – 4x – 4x2 + 8x + 16
Multiply.
= x3 – 6x2 + 4x +16
Combine like terms.
Warm-Up
Exercises
GUIDED
PRACTICE
7.
for Example 3
2, 2i, 4 – √ 6
Because the coefficients are rational and 2i is a zero, –2i
must also be a zero by the complex conjugates theorem.
4 + 6 is also a zero by the irrational conjugate theorem.
Use the five zeros and the factor theorem to write f(x) as a
product of five factors.
SOLUTION
= (x – 2) [ (x2 –(2i)2][x2–4)+√6][(x– 4) – √6 ]
Write f (x) in
factored form.
Regroup terms.
= (x – 2)[(x2 + 4)[(x– 4)2 – ( 6 )2]
Multiply.
= (x – 2)(x2 + 4)(x2 – 8x+16 – 6)
Expand binomial.
f (x) = (x–2) (x +2i)(x-2i)[(x –(4 –√6 )][x –(4+√6) ]
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 3
= (x – 2)(x2 + 4)(x2 – 8x + 10)
Simplify.
= (x–2) (x4– 8x2 +10x2 +4x2 –3x +40) Multiply.
= (x–2) (x4 – 8x3 +14x2 –32x + 40)
Combine like terms.
= x5– 8x4 +14x3 –32x2 +40x – 2x4
+16x3 –28x2 + 64x – 80
Multiply.
= x5–10x4 + 30x3 – 60x2 +10x – 80
Combine like terms.
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 3
8. 3, 3 – i
Because the coefficients are rational and 3 –i is a
zero, 3 + i must also be a zero by the complex
conjugates theorem. Use the three zeros and the
factor theorem to write f(x) as a product of three
factors
SOLUTION
= f(x) =(x – 3)[x – (3 – i)][x –(3 + i)]
= (x–3)[(x– 3)+i ][(x2 – 3) – i]
Write f (x) in factored
form.
Regroup terms.
= (x–3)[(x – 3)2 –i2)]
Multiply.
= (x– 3)[(x – 3)+ i][(x –3) –i]
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 3
= (x – 3)[(x – 3)2 – i2]=(x –3)(x2 – 6x + 9)
= (x–3)(x2 – 6x + 9)
Simplify.
= x3–6x2 + 9x – 3x2 +18x – 27
Multiply.
= x3 – 9x2 + 27x –27
Combine like terms.
Warm-Up4Exercises
EXAMPLE
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.
SOLUTION
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
Warm-Up4Exercises
EXAMPLE
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.
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 4
Determine the possible numbers of positive real
zeros, negative real zeros, and imaginary zeros for
the function.
9. f (x) = x3 + 2x – 11
SOLUTION
f (x) = x3 + 2x – 11
The coefficients in f (x) have 1 sign changes, so f
has 1 positive real zero(s).
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 4
f (– x) = (– x)3 + 2(– x) – 11
= – x3 – 2x – 11
The coefficients in f (– x) have no sign changes.
The possible numbers of zeros for f are summarized in
the table below.
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 4
10. g(x) = 2x4 – 8x3 + 6x2 – 3x + 1
SOLUTION
f (x) = 2x4 – 8x3 + 6x2 – 3x + 1
The coefficients in f (x) have 4 sign changes, so f
has 4 positive real zero(s).
f (– x) = 2(– x)4 – 8(– x)3 + 6(– x)2 + 1
= 2x4 + 8x + 6x2 + 1
The coefficients in f (– x) have no sign changes.
Warm-Up
Exercises
GUIDED
PRACTICE
for Example 4
The possible numbers of zeros for f are summarized in
the table below.
Warm-Up5Exercises
EXAMPLE
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.
ANSWER
From these screens, you can see that the
zeros are x ≈ – 0.73 and x ≈ 2.73.
Warm-Up6Exercises
EXAMPLE
Approximate real zeros of a polynomial model
TACHOMETER
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
s (x) = 0.00547x3 – 0.225x2 + 3.62x – 11.0
What is the tachometer reading when the boat travels
15 miles per hour?
Warm-Up6Exercises
EXAMPLE
Approximate real zeros of a polynomial model
SOLUTION
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.
From the graph, there is one real zero: x ≈ 19.9.
ANSWER
The tachometer reading is about 1990 RPMs.
Warm-Up
Exercises
GUIDED
PRACTICE
11.
for Examples 5 and 6
Approximate the real zeros of
f (x) = 3x5 + 2x4 – 8x3 + 4x2 – x – 1.
ANSWER
The zeros are x ≈ – 2.2, x ≈ – 0.3, and x ≈ 1.1.
Warm-Up
Exercises
GUIDED
PRACTICE
for Examples 5 and 6
12. What If? In Example 6, what is the tachometer
reading when the boat travels 20 miles per hour?
SOLUTION
Substitute 20 for s(x) in the given function. You can
rewrite the resulting equation as:
0 = 0.00547x3 – 0.225x2 + 3.62x – 31.0
Then, use a graphing calculator to approximate the
real zeros of f (x) = 0.00547x3 – 0.225x2 + 3.62x – 31.0.
From the graph, there is one real zero: x ≈ 23.1.
Warm-Up
Exercises
GUIDED
PRACTICE
ANSWER
for Examples 5 and 6
The tachometer reading is about 2310 RPMs.
Warm-Up
Exercises
Daily
Homework
Quiz
1. Find all the zeros of f(x) = x4 – x2 – 20.
ANSWER
+ √5, + 2i
2. Write a polynomial function of least degree
that has rational coefficients, a leading
coefficient of 1, and – 3 and 1 – 7i
ANSWER
x3 + x2 + 44x + 150.
Warm-Up
Exercises
Daily
Homework
Quiz
3. Determine the possible numbers of positive
real zeros, negative real zeros, and imaginary
zeros for f(x) = 2x5 – 3x4 – 5x3 + 10x2 + 3x – 5.
ANSWER
3 positive, 2 negative, 0 imaginary;
3 positive, 0 negative, 2 imaginary;
1 positive, 2 negative, 2 imaginary;
1 positive, 0 negative, 4 imaginary
Warm-Up
Exercises
Daily
Homework
Quiz
4. The profit P for printing envelopes is modeled
by P = x – 0.001x3 – 0.06x2 + 30.5x, where x is
the number of envelopes printed in thousands.
What is the least number of envelopes that
can be printed for a profit of $1500?
ANSWER
about 70,000 envelopes