1. 1. What is the degree of f (x) = 8x6 – 4x5 + 3x2 + 2?
ANSWER 6
2. Solve x2 – 2x + 3 = 0
1 i 2 + _
ANSWER

2. 3. The function P given by x4 + 3x3 – 30x2 – 6x = 56 model
3. The function P given by x4 + 3x3 – 30x2 – 6x = 56 model the profit of a company. What are the real solutions of the function?
ANSWER
-7.2, 4.5

4. 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.)

5. 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.)

6. YOU TRY
for Example 1
1. How many solutions does the equation
x4 + 5x2 – 36 = 0 have?
ANSWER 4

7. YOU TRY
for Example 1
2. How many zeros does the function
f (x) = x3 + 7x2 + 8x – 16 have?
ANSWER 3

8. EXAMPLE 2
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 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)

9. EXAMPLE 2
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 ) x – (2 – i )
The zeros of f are – 1, – 1, 2, 2 + i , and 2 – i 3.
ANSWER

10. YOU TRY
for Example 2
Find all zeros of the polynomial function.
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

11. EXAMPLE 3
Use zeros to write a polynomial function
SOLUTION
Because the coefficients are rational and is a zero, 2 – 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.

12. Use zeros to write a polynomial function
EXAMPLE 3
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.

13. EXAMPLE 3
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 – = 0 
f(2 + √ 5 ) = (2 + √ 5 )3 – 7(2 + √ 5 )2 + 11( 2 + √ 5 ) + 3
= √ 5 – 63 – 28 √ √ 5 + 3
= 0 
Since f (2 + √ 5 ) = 0, by the irrational conjugates theorem f (2 – √ 5 ) = 0. 

14. f (x) = (x – 4) [ x – (1 + √ 5 ) ] [ x – (1 – √ 5 ) ]
YOU TRY
for Example 3
Because the coefficients are rational and 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 form.
= (x – 4) [ (x – 1) – √ 5 ] [ (x – 1) +√ 5 ]
Regroup terms.
= (x – 4)[(x – 1)2 – ( 5)2]
Multiply.
= (x – 4)[(x2 – 2x + 1) – 5]
Expand binomial.

15. YOU TRY for Example 3 = (x – 4)(x2 – 2x – 4)
Simplify.
= x3 – 2x2 – 4x – 4x2 + 8x + 16
Multiply.
= x3 – 6x2 + 4x +16
Combine like terms.

16. YOU TRY
for Example 3
√ 6
Because the coefficients are rational and 2i is a zero, –2i must also be a zero by the complex conjugates theorem 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
f (x) = (x–2) (x +2i)(x-2i)[(x –(4 –√6 )][x –(4+√6) ]
Write f (x) in factored form.
= (x – 2) [ (x2 –(2i)2][x2–4)+√6][(x– 4) – √6 ]
Regroup terms.
= (x – 2)[(x2 + 4)[(x– 4)2 – ( 6 )2]
Multiply.
= (x – 2)(x2 + 4)(x2 – 8x+16 – 6)
Expand binomial.

17. YOU TRY 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 +104x – 80
Combine like terms.

18. EXAMPLE 4
Approximate real zeros of a polynomial model
s (x) = x3 – 0.225x x – 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

19. EXAMPLE 4
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 = x3 – 0.225x x – 26.0
Then, use a graphing calculator to approximate the real zeros of f (x) = x3 – 0.225x x – 26.0.
From the graph, there is one real zero: x ≈ 19.9.
The tachometer reading is about 1990 RPMs.
ANSWER

20. YOU TRY
for Examples 4 6.
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

21. YOU TRY
7. The profit P for printing envelopes is modeled
by P = x – 0.001x3 – 0.06x x, 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

22. KEEP GOING
1. Find all the zeros of f(x) = x4 – x2 – 20.
+ √5, + 2i
ANSWER
2. Write a polynomial function of least degree
that has rational coefficients, a leading
coefficient of 1, and – 3 and 1 – 7i
ANSWER
x3 + x x