Polynomial and
Rational Functions
Adv. Math +
1) Is -2 a root of x4 – 3x2 – 2x + 4 = 0?
2) Write a polynomial of least degree
with roots -4 and 2i.
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Quadratic Equations
Adv. Math +
! Solve
quadratic equations
! Use
the discriminant to
describe the roots of
quadratic equations
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A quadratic equation is one that can
be written in the form
ax2 + bx + c = 0,
where a, b and c are real numbers
and a ≠ 0.
! The form ax2 + bx + c = 0 is
referred to as standard form.
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If a · b = 0, what do you
know about a or b?
Why is that?
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If a and b are real numbers and
ab = 0, then a = 0 or b = 0.
If the product of two numbers is zero, then
one of the factors must be zero.
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To Solve Quadratic Equations by Factoring
Step 1: Write the equation in standard form.
Step 2: Factor the quadratic completely.
Step 3: Set each factor equal to 0.
Step 4: Solve the resulting equations.
Step 5: C
heck each solution in the original
equation.
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Solve: x2 – 5x = 24
First write the quadratic equation in standard form.
x2 – 5x – 24 = 0
Now factor.
x2 – 5x – 24 = 0
(x – 8)(x + 3) = 0
Set each factor equal to 0.
x – 8 = 0 or x + 3 = 0
which simplifies to
x = 8 or x = –3
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Check both possible answers in the
original equation.
82 – 5(8) = 64 – 40 = 24
True
(–3)2 – 5(–3) = 9 + 15) = 24 True
So the solutions are x = 8 or –3. Adv. Math +
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2. x2 + 3x – 18 = 0
3. 2p2 - 17p + 45 = 3p – 5
(x + 6)(x – 3) = 0
2p2 – 20p + 50 = 0
x + 6 = 0 x – 3 = 0
2(p2 – 10p + 25) = 0
x=–6
x = 3
2(p – 5)(p – 5) = 0
2 = 0 p – 5 = 0 p – 5 = 0
p=5
Adv. Math +
p = 5
11
4. 2w2 - 10w = 23w – w2
5. 3x – 6 = x2 – 10
3w2 – 33w = 0
0 = x2 – 3x – 4
3w(w – 11) = 0
0 = (x – 4)(x + 1)
3w = 0 w – 11 = 0
x – 4 = 0 x + 1 = 0
w=0
Adv. Math +
w = 11
x=4
x = –1 12
6. 6x2 – 17x = 3
7. 32x3 – 50x = 0
6x2 – 17x - 3 = 0
2x(16x2 – 25) = 0
(6x + 1)(x – 3) = 0
2x(4x – 5)(4x + 5) = 0
6x + 1 = 0 x – 3 = 0
2x = 0 4x – 5 = 0 4x + 5 = 0
x = - 1/6
x = 3
x=0
x = 5/4
x = -5/4
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Square Root Property
If x2 = a for a ≥ 0, then
x = a or x = − a
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Isolate the squared term
2.  Take the square root of both
sides
1. 
!  if you introduce a √ into the
problem, then YOU MUST put a
± into the answer
3. 
Simplify, if possible
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1. 2x2 + 1 = 17
2x2 = 16
2.
(x + 5)2 = 7
3( )(x + 5)2 = 7 ∙ 3
x2 = 8
(x + 5)2 = 21
√x2 = √8
√(x + 5)2 = √21
x = ± √4 ∙ √2
x + 5 = ± √21
x = ± 2√2
Adv. Math +
x = –5 ± √21 16
3. 2p2 – 15 = 5p2 + 12 4. -6(u + 5)2 = 120
-3p2 = 27
(u + 5)2 = -20
p2 = –9
√(u + 5)2 = √-20
√p2 = √-9
u + 5 = ± 2i√5
p = ± 3i
u = -5 ± 2i√5
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5. 3x2 + 10 = -26
3x2
x2
√x2
6. 2(x – 4)2 = 50 = -36
(x – 4)2 = 25
= -12
√(x – 4)2 = √25
= √-12
x = ± 2i√3
x – 4 = ± 5
x = 4 ± 5
x = 4 + 5 x = 4 – 5
x=9
Adv. Math +
x = -1
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! 
Factoring is the easiest method to solve a
quadratic, however some trinomials don't factor. ! 
When you cannot factor a trinomial, you can
“complete the square” so that you can solve the
trinomial by using square roots. ! 
A perfect square trinomial is one that can be
factored into (a+b)2 or (a-b)2
"  a2 + 2ab + b2 = (a+b)2
"  a2 - 2ab + b2 = (a-b)2
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1. x2 – 12x + c
3. x2 – 8x + c
5. x2 – 4x + c
x2 – 12x + 36 x2 – 8x + 16
x2 – 4x + 4
(x – 6)2
(x – 4)2
(x – 2)2
2. x2 + 16x + c
4. x2 + 10x + c
x2 + 16x + 64
x2 + 10x + 25
x2 + 9x + 81/4
(x + 8)2
(x + 5)2
(x + 9/2)2
Adv. Math +
6. x2 + 9x + c
What pattern are you seeing? The c must be (b/2)2
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1. Check to see if the trinomial is factorable. 2. Put in the form ax2 + bx = c. 3. Divide all terms by a. 4. Add (b/2)2 to both sides. 5. Factor the left side. **(remember it is a perfect square)**
6. Solve by square roots.
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1. x2 – 6x + 11 = 0
2. x2 + 8x = 9 x2 – 6x = -11
x2 + 8x + 16 = 9 + 16
x2 – 6x + 9 = -11 + 9
(x + 4)2 = 25
x2 – 6x + 9 = -2
√(x + 4)2 = √25
(x – 3)2 = -2
x + 4 = ± 5
√(x – 3)2 = √-2
x = - 4 ± 5
x – 3 = ± i√2
x = 3 ± i√2
x = -4 + 5
x = - 9
x=1
x = -4 – 5 Adv. Math +
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3. 2x2 – 12x + 14 = 0
4. 2x2 + 20x – 6 = 0
2x2 – 12x = -14
2x2 + 20x = 6
x2 – 6x = -7
x2 + 10x = 3
x2 – 6x + 9 = -7 + 9
x2 + 10x + 25 = 3 + 25
(x – 3)2 = 2
(x + 5)2 = 28
√(x – 3)2 = √2
√(x + 5)2 = √28
x – 3 = ± √2
x + 5 = ± 2√7
x = 3 ± √2
x = -5 ± 2√7
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5. 3x2 – 12x – 15 = 0
6. 5x2 – 10x + 15 = 0
3x2 – 12x = 15
5x2 – 10x = -15
x2 – 4x = 5
x2 – 2x = -3
x2 – 4x + 4 = 5 + 4
x2 – 2x + 1 = -3 + 1
(x – 2)2 = 9
(x – 1)2 = -2
√(x – 2)2 = √9
√(x – 1)2 = √-2
x – 2 = ± 3
x – 1 = ± √-2
x = 2 ± 3
x = 1 ± i√2
x=5
Adv. Math +
x = -1
24
Consider the quadratic equation
ax2 + bx + c = 0
When is factoring the appropriate method to use to
solve?
when it is factorable
When is square roots the appropriate method to use to
solve?
when b = 0
When is completing the square the appropriate method to
use to solve?
when a = 1 and b is even
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1) Solve 15x2 – 4x = 3 by factoring.
2) Solve (3x + 2)2 + 7 = 0 by square roots.
3) Solve 2x2 – 8x – 6 = 0 by completing the square.
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If
2
ax + bx + c = 0
then,
2
−b ± b − 4ac
x=
2a
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What is the Quadratic Formula?
x equals
the opposite of b
Plus or minus the square root
of b squared minus four ac
all over two a
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Solve using the QF
3x2 + 7x + 2 = 0
a = 3 b = 7 c = 2
x = -7 ± 5
6
x = -7 + 5 x = -7 – 5
6
6
x = -2
6
x = -7 ± √72 – 4(3)(2)
2(3)
x = -7 ± √25
6
Adv. Math +
If there is not
a radical left in your solution you
will always simplify!
x = -1/3
x = -12
6
x = -2
29
Solve using the QF x = 4 ± √4 • √14
2x2 – 4x = 5
4
2x2 – 4x – 5 = 0
a = 2 b = -4 c = -5 x = 4 ± 2√14 x=4±
√(-4)2
2(2)
x = 4 ± √56
4
Adv. Math +
– 4(2)(-5)
4
Exact Answers
You can not reduce unless you reduce all three terms by a constant!
x = 2 ± √14
2
Approximate Answers x ≈ 2.87, x ≈ -.87
30
Solve using the QF
x2 + 7 = -5x
x2 + 5x + 7 = 0
a = 1 b = 5 c = 7
x = -5 ± √(5)2 – 4(1)(7)
2(1)
x = -5 ± √ -3 2
Adv. Math +
x = -5 ± √-1 • √3
2
x = -5 ± i√3 2
Exact Answers. Imaginary Solutions
do not have approximate Solutions.
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Solve using the QF x = -10 ± 14
3x2 = 8 – 10x
6
3x2 + 10x – 8 = 0
a = 3 b = 10 c = -8 x = -10 6+ 14 x = -10 6– 14
x = -10 ±
√(10)2
2(3)
x = -10 ± √196
6
Adv. Math +
– 4(3)(-8)
x=4
6
If there is not
a radical left in your solution you
will always simplify!
x = 2/3
x = -24
6
x = -4
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Solve using the QF x = 16 ± √-256 • √3
4x2 + 64 = 16x
8
4x2 – 16x + 64 = 0
a = 4 b = -16 c = 64 x = 16 ± 16i√3 x = 16 ±
√(-16)2
2(4)
– 4(4)(64)
x = 16 ± √-768 8
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You can not reduce unless you reduce all three terms by a constant!
x = 2 ± 2i√3
Exact Answers. Imaginary Solutions
do not have approximate Solutions.
33
Solve using the QF x = 8 ± √4 • √31
3x2 – 5 = 8x
6
3x2 – 8x – 5 = 0
a = 3 b = -8 c = -5 x = 8 ± 2√31 x=8±
√(-8)2
2(3)
x = 8 ± √124
6
Adv. Math +
– 4(3)(-5)
6
Exact Answers
You can not reduce unless you reduce all three terms by a constant!
x = 4 ± √31
3
Approximate Answers x ≈ 3.19, x ≈ -.52
34
!  Discriminant
# b2 – 4ac
!  In
the quadratic formula, the
expression underneath the
radical that describes the
nature of the roots is called the
discriminant.
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Discriminant
Desc. of Roots
b2 – 4ac > 0 2 real roots
b2 – 4ac = 0 1 real root
b2 – 4ac < 0 2 Imaginary
Roots
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Graph
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Use the discriminant to fully describe the roots.
4y2 + 3y – 1 = 0
a=4
Discriminant:
b=3
c = -1
b2 – 4ac
(3)2 – 4(4)(-1)
= 25
25 > 0, so there are:
2 Real Roots #
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Use the discriminant to fully describe the roots.
2x2 – 4x = -5
2x2 – 4x + 5 = 0
a = 2 b = -4 c = 5
Discriminant:
b2 – 4ac
(-4)2 – 4(2)(5)
= -24
-24 < 0, so there are:
2 Imaginary Roots #
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Use the discriminant to fully describe the roots.
3x2 = 25
3x2 – 25 = 0
a = 3 b = 0 c = -25
Discriminant:
b2 – 4ac
(0)2 – 4(3)(-25)
= 300
300 > 0, so there are:
2 Real Roots #
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Use the discriminant to fully describe the roots.
x2 = 8x – 16
x2 – 8x + 16 = 0
a = 1 b = -8 c = 16
Discriminant:
b2 – 4ac
(-8)2 – 4(1)(16)
= 0
0 = 0, so there are:
1 Real Root #
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General Strategy for Problem Solving
1. U
NDERSTAND the problem.
•  Read and reread the problem
•  Choose a variable to represent the
unknown
•  Construct a drawing, whenever possible
2. MODEL the problem with an equation.
3. SOLVE the equation.
4. INTERPRET the result.
•  Check proposed solution in original
problem.
•  State your conclusion.
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At a local university, students often leave the sidewalk
and cut across the lawn to save walking distance. Given
the diagram below of a favorite place to cut across the
lawn, approximate to the nearest foot how many feet of
walking distance a student saves by cutting across the
lawn instead of walking on the sidewalk. Use the
Pythagorean
Theorem!
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Use the Pythagorean Theorem!
x2 + (x + 20)2 = 502
x2 + x2 + 40x + 400 = 2500
2x2 + 40x – 2100 = 0
x2 + 20x – 1050 = 0
a = 1, b = 20, c = -1050
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Use the Quadratic Formula!
a = 1, b = 20, c = -1050
−20 ± (20) 2 − 4(1)(−1050)
x=
2(1)
x = −20 ± 400 + 4200
2
x = −20 ± 4600 = −20 ± 100 ⋅ 46
2
2
−20
±10
46
x=
= −10 + 5 46
2
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−10 + 5 46 ≈ 24 feet
x + (x + 20) ≈ 24 + (24 + 20) = 68 feet
A person saves about 68 – 50, or
18 feet of walking
distance by cutting
across the lawn.
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What is the Quadratic Formula?
x equals
the opposite of b
Plus or minus the square root
of b squared minus four ac
all over two a
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p. 218
3, 4, 5, 9, 10, 11, 12, 14,
16, 23, 24, 25, 28-33 all,
36, 37
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