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Warm-Up Exercises
1. Use the quadratic formula to solve 2x2 – 3x – 1 = 0.
Round the nearest hundredth.
ANSWER
1.78, –0.28
2. Use synthetic substitution to evaluate
f (x) = x3 + x2 – 3x – 10 when x = 2.
ANSWER
–4
Warm-Up Exercises
3. A company’s income is modeled by the function
P = 22x2 – 571x. What is the value of P when x = 200?
ANSWER
765,800
Warm-Up1Exercises
EXAMPLE
Use polynomial long division
Divide f (x) = 3x4 – 5x3 + 4x – 6 by x2 – 3x + 5.
SOLUTION
Write polynomial division in the same format you
use when dividing numbers. Include a “0” as the
coefficient of x2 in the dividend. At each stage,
divide the term with the highest power in what is
left of the dividend by the first term of the divisor.
This gives the next term of the quotient.
Warm-Up1Exercises
EXAMPLE
Use polynomial long division
3x2 + 4x – 3
x2 – 3x + 5 )3x4 – 5x3 + 0x2 + 4x – 6
quotient
Multiply divisor by 3x4/x2 = 3x2
3x4 – 9x3 + 15x2
4x3 – 15x2 + 4x
Subtract.
Bring down next term.
4x3 – 12x2 + 20x
Multiply divisor by 4x3/x2 = 4x
– 3x2 – 16x – 6
–3x2 + 9x – 15
– 25x + 9
Subtract.
Bring down next term.
Multiply divisor by – 3x2/x2 = – 3
remainder
Warm-Up1Exercises
EXAMPLE
Use polynomial long division
ANSWER
3x4 – 5x3 + 4x – 6 = 3x2 + 4x – 3 + – 25x + 9
x2 – 3x + 5
x2 – 3x + 5
CHECK
You can check the result of a division problem by
multiplying the quotient by the divisor and adding the
remainder. The result should be the dividend.
(3x2 + 4x – 3)(x2 – 3x + 5) + (– 25x + 9)
= 3x2(x2 – 3x + 5) + 4x(x2 – 3x + 5) – 3(x2 – 3x + 5) – 25x + 9
= 3x4 – 9x3 + 15x2 + 4x3 – 12x2 + 20x – 3x2 + 9x – 15 – 25x + 9
= 3x4 – 5x3 + 4x – 6
Warm-Up2Exercises
EXAMPLE
Use polynomial long division with a linear divisor
Divide f (x) = x3 + 5x2 – 7x + 2 by x – 2.
x2 + 7x + 7
x – 2 ) x3 + 5x2 – 7x + 2
x3 – 2x2
7x2 – 7x
7x2 – 14x
7x + 2
7x – 14
16
ANSWER
quotient
Multiply divisor by x3/x = x2.
Subtract.
Multiply divisor by 7x2/x = 7x.
Subtract.
Multiply divisor by 7x/x = 7.
remainder
x3 + 5x2 – 7x +2
= x2 + 7x + 7 + 16
x–2
x–2
Warm-Up
YOU
TRY. .Exercises
.
for Examples 1 and 2
Divide using polynomial long division.
1.
(2x4 + x3 + x – 1)
(x2 + 2x – 1)
SOLUTION
Write polynomial division in the same format you
use when dividing numbers. Include a “0” as the
coefficient of x2 in the dividend. At each stage,
divide the term with the highest power in what is
left of the dividend by the first term of the divisor.
This gives the next term of the quotient.
Warm-Up
YOU
TRY. .Exercises
.
for Examples 1 and 2
2x2 – 3x + 8
x2 + 2x – 1 )2x4 + x3 + 0x2 + x – 1
Multiply divisor by 2x4/x2
= –2x2.
2x4 – 4x3 – 2x2
3x3 – 2x2 + x
–
3x3
–
6x2
quotient
+ 3x
Subtract. Bring down
next term.
Multiply divisor by –3x3/x2
= –3.
8x2 – 2x – 1
Subtract. Bring down
next term.
8x2 –16x – 8
Multiply divisor by 4x2/x2 =
8.
– 18x + 7
remainder
Warm-Up
YOU
TRY. .Exercises
.
ANSWER
for Examples 1 and 2
2x4 + 5x3 + x – 1
x2 + 2x – 1
= (2x2 – 3x + 8)+ –218x + 7
x + 2x – 1
Warm-Up
YOU
TRY. .Exercises
.
for Examples 1 and 2
2. (x3 – x2 + 4x – 10)  (x + 2)
SOLUTION
Write polynomial division in the same format you
use when dividing numbers. Include a “0” as the
coefficient of x2 in the dividend. At each stage,
divide the term with the highest power in what is
left of the dividend by the first term of the divisor.
This gives the next term of the quotient.
Warm-Up
YOU
TRY. .Exercises
.
for Examples 1 and 2
x2 – 3x + 10
x + 2 )x3 – x2 + 4x – 10
quotient
Multiply divisor by x3/x =
x2.
x3 + 2x2
–3x2 + 4x
Subtract. Bring down
next term.
–
Multiply divisor by –3x2/x
= –3x.
3x2
– 6x
10x – 1
Subtract. Bring down
next term.
10x + 20
Multiply divisor by 10x/x =
10.
– 30
remainder
Warm-Up
YOU
TRY. .Exercises
.
ANSWER
for Examples 1 and 2
x3 – x2 +4x – 10
x+2
= (x2 – 3x +10)+
– 30
x+2
Warm-Up3Exercises
EXAMPLE
Use synthetic division
Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic
division.
SOLUTION
–3
ANSWER
1
–8
5
–6
15
– 21
2 –5
7
– 16
2
2x3 + x2 – 8x + 5
16
= 2x2 – 5x + 7 –
x+3
x+3
Warm-Up4Exercises
EXAMPLE
Factor a polynomial
Factor f (x) = 3x3 – 4x2 – 28x – 16 completely given that
x + 2 is a factor.
SOLUTION
Because x + 2 is a factor of f (x), you know that
f (– 2) = 0. Use synthetic division to find the other
factors.
– 2 3 – 4 – 28 – 16
–6
20
16
3 – 10
–8
0
Warm-Up4Exercises
EXAMPLE
Factor a polynomial
Use the result to write f (x) as a product of two
factors and then factor completely.
f (x) = 3x3 – 4x2 – 28x – 16
Write original polynomial.
= (x + 2)(3x2 – 10x – 8)
Write as a product of two
factors.
= (x + 2)(3x + 2)(x – 4)
Factor trinomial.
Warm-Up
YOU
TRY. .Exercises
.
for Examples 3 and 4
Divide using synthetic division.
3.
(x3 + 4x2 – x – 1)  (x + 3)
SOLUTION
(x3 + 4x2 – x – 1)  (x + 3)
–3
1
3
ANSWER
–1
4
–1
–3
–3
12
1
–4
11
x3 + 4 x2 – x – 1
11
= x2 + x – 4 +
x+3
x+3
Warm-Up
YOU
TRY. .Exercises
.
4.
for Examples 3 and 4
(4x3 + x2 – 3x + 7)  (x – 1)
SOLUTION
(4x3 + x2 – 3x + 7)  (x – 1)
1
4
4
ANSWER
1
–3
7
4
5
2
2
9
5
4x3 + x2 – 3x + 1
9
2
= 4x + 5x + 2 +
x–1
x–1
Warm-Up
YOU
TRY. .Exercises
.
for Examples 3 and 4
Factor the polynomial completely given that x – 4 is a
factor.
5.
f (x) = x3 – 6x2 + 5x + 12
SOLUTION
Because x – 4 is a factor of f (x), you know that
f (4) = 0. Use synthetic division to find the other
factors.
4
1 –6
5
12
4
–8
–12
1 – 2
–3
0
Warm-Up
YOU
TRY. .Exercises
.
for Examples 3 and 4
Use the result to write f (x) as a product of two
factors and then factor completely.
f (x) = x3 – 6x2 + 5x + 12
Write original polynomial.
= (x – 4)(x2 – 2x – 3)
Write as a product of two
factors.
= (x – 4)(x –3)(x + 1)
Factor trinomial.
Warm-Up
YOU
TRY. .Exercises
.
6.
for Examples 3 and 4
f (x) = x3 – x2 – 22x + 40
SOLUTION
Because x – 4 is a factor of f (x), you know that
f (4) = 0. Use synthetic division to find the other
factors.
4
1
-1
– 22 40
4
1
3
12 –40
– 10
0
Warm-Up
YOU
TRY. .Exercises
.
for Examples 3 and 4
Use the result to write f (x) as a product of two
factors and then factor completely.
f (x) = x3 – x2 – 22x + 40
= (x – 4)(x2 + 3x – 10)
= (x – 4)(x –2)(x +5)
Write original polynomial.
Write as a product of two
factors.
Factor trinomial.
Warm-Up5Exercises
EXAMPLE
Standardized Test Practice
SOLUTION
Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic
division.
3
1
1
–2
– 23
60
3
3
– 60
1
– 20
0
Warm-Up5Exercises
EXAMPLE
Standardized Test Practice
Use the result to write f (x) as a product of two
factors. Then factor completely.
f (x) = x3 – 2x2 – 23x + 60
= (x – 3)(x2 + x – 20)
= (x – 3)(x + 5)(x – 4)
The zeros are 3, – 5, and 4.
ANSWER The correct answer is A.
Warm-Up6Exercises
EXAMPLE
Use a polynomial model
BUSINESS
The profit P (in millions
of dollars) for a shoe
manufacturer can be
modeled by
P = – 21x3 + 46x
where x is the number
of shoes produced
(in millions). The company
now produces 1 million shoes and makes a profit of
$25,000,000, but would like to cut back production.
What lesser number of shoes could the company
produce and still make the same profit?
Warm-Up6Exercises
EXAMPLE
Use a polynomial model
SOLUTION
Substitute 25 for P in P = – 21x3 + 46x.
25 = – 21x3 + 46x
0 = 21x3 – 46x + 25
Write in standard form.
You know that x = 1 is one solution of the equation.
This implies that x – 1 is a factor of 21x3 – 46x + 25.
Use synthetic division to find the other factors.
1
21
0
21
21
– 46
25
21 –25
21 – 25
0
Warm-Up6Exercises
EXAMPLE
Use a polynomial model
So, (x – 1)(21x2 + 21x – 25) = 0. Use the quadratic formula
to find that x  0.7 is the other positive solution.
ANSWER
The company could still make the same profit
producing about 700,000 shoes.
Warm-Up
YOU
TRY. .Exercises
.
for Examples 5 and 6
Find the other zeros of f given that f (– 2) = 0.
7.
f (x) = x3 + 2x2 – 9x – 18
SOLUTION
Because f (– 2 ) = 0, x + 2 is a factor of f (x). Use
synthetic division.
–2 1
1
2
– 9
– 18
–2
0
18
0
–9
0
Warm-Up
YOU
TRY. .Exercises
.
for Examples 5 and 6
Use the result to write f (x) as a product of two
factors. Then factor completely.
f (x) = x3 + 2x2 – 9x – 18
= (x + 2)(x2 – 92)
= (x + 2)(x + 3)(x – 3)
The zeros are 3, – 3, and – 2.
Warm-Up
YOU
TRY. .Exercises
.
8.
for Examples 5 and 6
f (x) = x3 + 8x2 + 5x – 14
SOLUTION
Because f (– 2 ) = 0, x + 2 is a factor of f (x). Use
synthetic division.
–2 1
1
8
5
– 14
–2
–12
14
6
–7
0
Warm-Up
YOU
TRY. .Exercises
.
for Examples 5 and 6
Use the result to write f (x) as a product of two
factors. Then factor completely.
f (x) = x3 + 8x2 + 5x – 14
= (x + 2)(x2 + 6x – 7 )
= (x + 2)(x + 7)(x – 1)
The zeros are 1, – 7, and – 2.
Warm-Up
YOU
TRY. .Exercises
.
9.
for Examples 5 and 6
What if? In Example 6, how does the answer
change if the profit for the shoe manufacturer is
modeled by P = – 15x3 + 40x?
SOLUTION
25 = – 15x3 + 40x
0 = 15x3 – 40x + 25
Substitute 25 for P in P = – 15x3 + 40x.
Write in standard form.
You know that x = 1 is one solution of the equation.
This implies that x – 1 is a factor of 15x3 – 40x + 25.
Use synthetic division to find the other factors.
1
15
0 – 40 25
15
15 –25
15 15 – 25
0
Warm-Up
YOU
TRY. .Exercises
.
for Examples 5 and 6
So, (x – 1)(15x2 + 15x – 25) = 0. Use the quadratic formula
to find that x  0.9 is the other positive solution.
ANSWER
The company could still make the same profit
producing about 900,000 shoes.
Warm-Up
Exercises
KEEP
GOING
1. Divide 6x4 – x3 – x2 + 11x – 18 by 2x2 + x – 3.
ANSWER
3x2
–3
– 2x + 5 + 2x2 + x – 3
2. Use synthetic division to divide f(x) = x3 – 3x2 – 5x – 25
by x – 5.
ANSWER
x2 + 2x + 5
Warm-Up
Exercises
KEEP
GOING
3. Find the other zero if one zero of f(x) = x3 – x2 – 17x
– 15 is x = – 1.
ANSWER
-3, -1 and 5
4. One of the costs to print a novel can be modeled
by C = x3 – 10x2 + 28x, where x is the number of
novels printed in thousands. The company now
prints 5000 novels at a cost of $15,000. What
other numbers of novels would cost about the
same amount?
ANSWER
About 4300 or about 700