Warm-Up Exercises 1. What is the degree of f (x) = 8x 6 – 4x 5 + 3x 2 + 2 ? 2. Solve x 2 – 2x + 3 = 0 ANSWER 6 1 i 2 + _

Warm-Up Exercises

1. What is the degree of f (x) = 8x6 – 4x5 + 3x2 + 2?

2. Solve x2 – 2x + 3 = 0

ANSWER 6

ANSWER 1 i 2+_

Warm-Up Exercises

3. The function P given by x4 + 3x3 – 30x2 – 6x = 56 modelthe profit of a company. What are the real solution of the function?

ANSWER –7, 2, 4+_

Warm-Up ExercisesEXAMPLE 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.)

Warm-Up ExercisesEXAMPLE 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.)

Warm-Up ExercisesGUIDED PRACTICE for Example 1

1. How many solutions does the equation x4 + 5x2 – 36 = 0 have?

ANSWER 4


2. How many zeros does the function f (x) = x3 + 7x2 + 8x – 16 have?

ANSWER 3

Warm-Up ExercisesEXAMPLE 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

Warm-Up ExercisesEXAMPLE 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


Find all zeros of the polynomial function.

3. f (x) = x3 + 7x2 + 15x + 9

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

SOLUTION


4. f (x) = x5 – 2x4 + 8x2 – 13x + 6

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 Using synthetic division, you can determine that 1 is a zero reputed twice and –3 is also a zero

+– +– +–+– 6.

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)

SOLUTION


Find the complex zero of f. use the quadratic formula to factor the trinomial into linear factor

STEP 3

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-Up ExercisesEXAMPLE 3 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.


f (x) = (x – 3) [ x – (2 + √ 5 ) ] [ x – (2 – √ 5 ) ] Write f (x) in factored form.

Regroup terms.= (x – 3) [ (x – 2) – √ 5 ] [ (x – 2) +√ 5 ]

= (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 liketerms.


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.

You can check this result by evaluating f (x) at each of its three zeros.

CHECK

Warm-Up Exercises

= (x + 1) (x2 – 4x – 2x + 8)

f (x) = (x + 1) (x – 2) ( x – 4)

GUIDED PRACTICE

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

for Example 3

Write f (x) in factored form.

= (x + 1) (x2 – 6x + 8)

Multiply.= x3 – 6x2 + 8x + x2 – 6x + 8= x3 – 5x2 + 2x + 8

Multiply.

Combine like terms.

Combine like terms.

Use the three zeros and the factor theorem to write f(x) as a product of three factors.SOLUTION


6. 4, 1 + √ 5

f (x) = (x – 4) [ x – (1 + √ 5 ) ] [ x – (1 – √ 5 ) ]Write f (x) in factored form.Regroup terms.= (x – 4) [ (x – 1) – √ 5 ] [ (x – 1) +√ 5 ]

Multiply.

= (x – 4)[(x2 – 2x + 1) – 5] Expand binomial.

= (x – 4)[(x – 1)2 – ( 5)2]

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


= (x – 4)(x2 – 2x – 4) Simplify.

= x3 – 2x2 – 4x – 4x2 + 8x + 16 Multiply.

= x3 – 6x2 + 4x +16 Combine like terms.

Warm-Up ExercisesGUIDED PRACTICE

7. 2, 2i, 4 – √ 6

for Example 3

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.

f (x) = (x–2) (x +2i)(x-2i)[(x –(4 –√6 )][x –(4+√6) ] Write f (x) in factored form.Regroup terms.= (x – 2) [ (x2 –(2i)2][x2–4)+√6][(x– 4) – √6 ]

Multiply.

= (x – 2)(x2 + 4)(x2 – 8x+16 – 6) Expand binomial.

= (x – 2)[(x2 + 4)[(x– 4)2 – ( 6 )2]

SOLUTION


= (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

= x5–10x4 + 30x3 – 60x2 +10x – 80 Combine like terms.

Multiply.

Warm-Up ExercisesGUIDED PRACTICE

8. 3, 3 – i

for Example 3

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

= f(x) =(x – 3)[x – (3 – i)][x –(3 + i)]

= (x–3)[(x– 3)+i ][(x2 – 3) – i] Regroup terms.

= (x–3)[(x – 3)2 –i2)]

= (x– 3)[(x – 3)+ i][(x –3) –i]

Multiply.

Write f (x) in factored form.

SOLUTION


= (x – 3)[(x – 3)2 – i2]=(x –3)(x2 – 6x + 9)

Simplify.= (x–3)(x2 – 6x + 9)

= x3–6x2 + 9x – 3x2 +18x – 27

Combine like terms.= x3 – 9x2 + 27x –27

Multiply.

Warm-Up ExercisesEXAMPLE 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

Warm-Up ExercisesEXAMPLE 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.


Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.

9. f (x) = x3 + 2x – 11

The coefficients in f (x) have 1 sign changes, so f has 1 positive real zero(s).

SOLUTION

f (x) = x3 + 2x – 11


The coefficients in f (– x) have no sign changes.


f (– x) = (– x)3 + 2(– x) – 11

= – x3 – 2x – 11


10. g(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

SOLUTION

f (x) = 2x4 – 8x3 + 6x2 – 3x + 1

The coefficients in f (– x) have no sign changes.



Warm-Up ExercisesEXAMPLE 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

Warm-Up ExercisesEXAMPLE 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

Warm-Up ExercisesEXAMPLE 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

Warm-Up ExercisesGUIDED PRACTICE for Examples 5 and 6

11.

Approximate the real zeros of

f (x) = 3x5 + 2x4 – 8x3 + 4x2 – x – 1.

The zeros are x ≈ – 2.2, x ≈ – 0.3, and x ≈ 1.1.ANSWER


12. What If? In Example 6, what is the tachometer reading when the boat travels 20 miles per hour?

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.

SOLUTION

From the graph, there is one real zero: x ≈ 23.1.


The tachometer reading is about 2310 RPMs.ANSWER

Warm-Up ExercisesDaily Homework Quiz

1. Find all the zeros of f(x) = x4 – x2 – 20.

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.

+ √5, + 2i

ANSWER

Warm-Up ExercisesDaily 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.

3 positive, 2 negative, 0 imaginary; 3 positive, 0 negative, 2 imaginary; 1 positive, 2 negative, 2 imaginary;1 positive, 0 negative, 4 imaginary

ANSWER

Warm-Up Exercises

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?

Daily Homework Quiz

ANSWER

about 70,000 envelopes

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Warm-Up Exercises 1. What is the degree of f (x) = 8x 6 – 4x 5 + 3x 2 + 2 ? 2. Solve x 2 – 2x + 3 = 0 ANSWER 6 1 i 2 + _