Venn diagram

1

Venn diagram

A

B

C

U

• The universal set U is usually represented by a rectangle.

• Inside this rectangle, subsets of the universal set are represented by

geometrical figures.

2

Venn diagrams help us identify some useful formulas in set operations.

2


• To represent (A ∪ B) ∩C:

• To represent (A ∩C) ∪ (B ∩C):

2



A

B

C


2



A

B

C

A

B

C


2



A

B

C

A

B

C


A

B

C

2



A

B

C

A

B

C


A

B

C

A

B

C

2



A

B

C

A

B

C


A

B

C

A

B

C

To prove (A ∪ B) ∩C = (A ∩C) ∪ (B ∩C) in a rigorous manner, should use

formal mathematical logic.

3

A few remarks about set notations

• If a set has finitely many elements, use the listing method to express the set

� write down all the elements

� enclosed the elements by braces

3





Example {1, 3, 5, 7, 9}

3





Example {1, 3, 5, 7, 9}

• Because intervals have infinitely many elements, we can’t use listing method

to express intervals. We introduce new notations:

3





Example {1, 3, 5, 7, 9}



� square bracket means endpoint included

� round bracket means endpoint excluded

3





Example {1, 3, 5, 7, 9}





Example [7, 11], (−2, 5]

3





Example {1, 3, 5, 7, 9}





Example [7, 11], (−2, 5]

Note The interval [7, 11] contains ALL numbers between 7 and 11 (including

integers, rational numbers, irrational numbers)

3





Example {1, 3, 5, 7, 9}





Example [7, 11], (−2, 5]

Note The interval [7, 11] contains ALL numbers between 7 and 11 (including

integers, rational numbers, irrational numbers)

For example, 10 ∈ [7, 11]

9.123 ∈ [7, 11]√

50 ∈ [7, 11]

4

Inequalities

To solve an inequality (or inequalities) in an unknown x means to find all real

numbers x such that the inequality is satisfied.

The set of all such x is called the solution set to the inequality.

4

Inequalities




Polynomial inequalities

anxn + an−1xn−1 + · · · + a1x + a0 < 0 ( or > 0, or ≤ 0, or ≥ 0)

where n ≥ 1 and an , 0.

4

Inequalities




Polynomial inequalities

anxn + an−1xn−1 + · · · + a1x + a0 < 0 ( or > 0, or ≤ 0, or ≥ 0)

where n ≥ 1 and an , 0.

(1) n = 1 Linear inequalities

(2) n = 2 Quadratic inequalities

(3) n ≥ 3 Higher degree inequalities

5

Example Find the solution set to the following compound inequality:

1 ≤ 3 − 2x ≤ 9

5


1 ≤ 3 − 2x ≤ 9

Solution The inequality means 1 ≤ 3 − 2x and 3 − 2x ≤ 9.

5


1 ≤ 3 − 2x ≤ 9


Solving separately: 1 ≤ 3 − 2x

5


1 ≤ 3 − 2x ≤ 9



2x ≤ 3 − 1

5


1 ≤ 3 − 2x ≤ 9



2x ≤ 3 − 1

x ≤ 1

5


1 ≤ 3 − 2x ≤ 9



2x ≤ 3 − 1

x ≤ 1

3 − 2x ≤ 9

5


1 ≤ 3 − 2x ≤ 9



2x ≤ 3 − 1

x ≤ 1

3 − 2x ≤ 9

3 − 9 ≤ 2x

5


1 ≤ 3 − 2x ≤ 9



2x ≤ 3 − 1

x ≤ 1

3 − 2x ≤ 9

3 − 9 ≤ 2x

−3 ≤ x

5


1 ≤ 3 − 2x ≤ 9



2x ≤ 3 − 1

x ≤ 1

3 − 2x ≤ 9

3 − 9 ≤ 2x

−3 ≤ x

Solution set = {x ∈ R : x ≤ 1 and − 3 ≤ x}

5


1 ≤ 3 − 2x ≤ 9



2x ≤ 3 − 1

x ≤ 1

3 − 2x ≤ 9

3 − 9 ≤ 2x

−3 ≤ x

Solution set = {x ∈ R : x ≤ 1 and − 3 ≤ x}= {x ∈ R : −3 ≤ x ≤ 1}

5


1 ≤ 3 − 2x ≤ 9



2x ≤ 3 − 1

x ≤ 1

3 − 2x ≤ 9

3 − 9 ≤ 2x

−3 ≤ x

Solution set = {x ∈ R : x ≤ 1 and − 3 ≤ x}= {x ∈ R : −3 ≤ x ≤ 1}= [−3, 1]

5


1 ≤ 3 − 2x ≤ 9



2x ≤ 3 − 1

x ≤ 1

3 − 2x ≤ 9

3 − 9 ≤ 2x

−3 ≤ x


Be careful 1 ≤ 3 − 2x

1 − 3 ≤ −2x

5


1 ≤ 3 − 2x ≤ 9



2x ≤ 3 − 1

x ≤ 1

3 − 2x ≤ 9

3 − 9 ≤ 2x

−3 ≤ x


Be careful 1 ≤ 3 − 2x

1 − 3 ≤ −2x−2−2

≥ x

6

Example Find the solution set to the following:

2x + 1 < 3 and 3x + 10 < 4

6


2x + 1 < 3 and 3x + 10 < 4

Solution Solve separately: 2x + 1 < 3

2x < 2

x < 1

6


2x + 1 < 3 and 3x + 10 < 4


2x < 2

x < 1

3x + 10 < 4

3x < −6

x < −2

6


2x + 1 < 3 and 3x + 10 < 4


2x < 2

x < 1

3x + 10 < 4

3x < −6

x < −2

Solution set = {x ∈ R : x < 1 and x < −2}

6


2x + 1 < 3 and 3x + 10 < 4


2x < 2

x < 1

3x + 10 < 4

3x < −6

x < −2

Solution set = {x ∈ R : x < 1 and x < −2}= {x ∈ R : x < −2}

6


2x + 1 < 3 and 3x + 10 < 4


2x < 2

x < 1

3x + 10 < 4

3x < −6

x < −2

Solution set = {x ∈ R : x < 1 and x < −2}= {x ∈ R : x < −2}= (−∞,−2)

7


2x + 1 > 9 and 3x + 4 < 10

7


2x + 1 > 9 and 3x + 4 < 10

Solution Solve separately: 2x + 1 > 9

2x > 8

x > 4

7


2x + 1 > 9 and 3x + 4 < 10


2x > 8

x > 4

3x + 4 < 10

3x < 6

x < 2

7


2x + 1 > 9 and 3x + 4 < 10


2x > 8

x > 4

3x + 4 < 10

3x < 6

x < 2

Solution set = {x ∈ R : x > 4 and x < 2}

7


2x + 1 > 9 and 3x + 4 < 10


2x > 8

x > 4

3x + 4 < 10

3x < 6

x < 2

Solution set = {x ∈ R : x > 4 and x < 2}= ∅

8

Wording

(1) Find the solution(s) to the inequality 2x − 1 > 0.

(2) Find the solution set to the inequality 2x − 1 > 0.

8

Wording



Answer

(1) Solutions x > 12

8

Wording



Answer


(2) Solution set{x ∈ R : x > 1

2

}=

(12,∞)

8

Wording



Answer


(2) Solution set{x ∈ R : x > 1

2

}=

(12,∞)

• Solve the inequality 2x − 1 > 0.

Can give solution or solution set.

9

Quadratic Inequalities ax2 + bx + c > 0

9


ax2 + bx + c ≥ 0

ax2 + bx + c < 0

ax2 + bx + c ≤ 0

9


ax2 + bx + c ≥ 0

ax2 + bx + c < 0

ax2 + bx + c ≤ 0

Properties of real numbers

(1) m > 0 and n > 0 =⇒ m · n > 0

9


ax2 + bx + c ≥ 0

ax2 + bx + c < 0

ax2 + bx + c ≤ 0


(1) m > 0 and n > 0 =⇒ m · n > 0

(2) m < 0 and n < 0 =⇒ m · n > 0

9


ax2 + bx + c ≥ 0

ax2 + bx + c < 0

ax2 + bx + c ≤ 0


(1) m > 0 and n > 0 =⇒ m · n > 0

(2) m < 0 and n < 0 =⇒ m · n > 0

(3) m > 0 and n < 0 =⇒ m · n < 0

9


ax2 + bx + c ≥ 0

ax2 + bx + c < 0

ax2 + bx + c ≤ 0


(1) m > 0 and n > 0 =⇒ m · n > 0

(2) m < 0 and n < 0 =⇒ m · n > 0

(3) m > 0 and n < 0 =⇒ m · n < 0

From these we get

(4) m · n > 0 ⇐⇒ (m > 0 and n > 0) or (m < 0 and n < 0)

9


ax2 + bx + c ≥ 0

ax2 + bx + c < 0

ax2 + bx + c ≤ 0


(1) m > 0 and n > 0 =⇒ m · n > 0

(2) m < 0 and n < 0 =⇒ m · n > 0

(3) m > 0 and n < 0 =⇒ m · n < 0

From these we get

(4) m · n > 0 ⇐⇒ (m > 0 and n > 0) or (m < 0 and n < 0)

(5) m · n < 0 ⇐⇒ (m > 0 and n < 0) or (m < 0 and n > 0)

10

Example Find the solution set to the inequality

x2 + 2x − 15 > 0

10


x2 + 2x − 15 > 0

Solution

Method 1 Factorize the quadratic polynomial

10


x2 + 2x − 15 > 0

Solution


x2 + 2x − 15 > 0

(x + 5)(x − 3) > 0

10


x2 + 2x − 15 > 0

Solution


x2 + 2x − 15 > 0

(x + 5)(x − 3) > 0

Apply Rule (4)

(x + 5 > 0 and x − 3 > 0) or (x + 5 < 0 and x − 3 < 0)

10


x2 + 2x − 15 > 0

Solution


x2 + 2x − 15 > 0

(x + 5)(x − 3) > 0

Apply Rule (4)

(x + 5 > 0 and x − 3 > 0) or (x + 5 < 0 and x − 3 < 0)

(x > −5 and x > 3) or (x < −5 and x < 3)

10


x2 + 2x − 15 > 0

Solution


x2 + 2x − 15 > 0

(x + 5)(x − 3) > 0

Apply Rule (4)

(x + 5 > 0 and x − 3 > 0) or (x + 5 < 0 and x − 3 < 0)

(x > −5 and x > 3) or (x < −5 and x < 3)

x > 3 or x < −5

10


x2 + 2x − 15 > 0

Solution


x2 + 2x − 15 > 0

(x + 5)(x − 3) > 0

Apply Rule (4)

(x + 5 > 0 and x − 3 > 0) or (x + 5 < 0 and x − 3 < 0)

(x > −5 and x > 3) or (x < −5 and x < 3)

x > 3 or x < −5

Solution set = {x ∈ R : x < −5 or x > 3}

10


x2 + 2x − 15 > 0

Solution


x2 + 2x − 15 > 0

(x + 5)(x − 3) > 0

Apply Rule (4)

(x + 5 > 0 and x − 3 > 0) or (x + 5 < 0 and x − 3 < 0)

(x > −5 and x > 3) or (x < −5 and x < 3)

x > 3 or x < −5

Solution set = {x ∈ R : x < −5 or x > 3}= (−∞,−5) ∪ (3,∞)

11


x2 + 2x − 15 > 0

Method 2 Graphical method

11


x2 + 2x − 15 > 0


• Graph of y = x2 + 2x − 15

11


x2 + 2x − 15 > 0


• Graph of y = x2 + 2x − 15

-5 3

11


x2 + 2x − 15 > 0


• Graph of y = x2 + 2x − 15

-5 3

• To solve the inequality x2 + 2x − 15 > 0 means

to find all (real numbers) x such that y > 0

11


x2 + 2x − 15 > 0


• Graph of y = x2 + 2x − 15

-5 3

• To solve the inequality x2 + 2x − 15 > 0 means

to find all (real numbers) x such that y > 0

• Solution set: (−∞,−5) ∪ (3,∞)

12


x2 + 2x − 15 > 0Method 3

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5

x − 3

(x + 5)(x − 3)

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 0

x − 3

(x + 5)(x − 3)

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 0 + + +

x − 3

(x + 5)(x − 3)

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3

(x + 5)(x − 3)

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 0

(x + 5)(x − 3)

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 0 +

(x + 5)(x − 3)

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3)

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) +

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) + 0

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) + 0 −

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) + 0 − 0

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) + 0 − 0 +

12


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

(x + 5)(x − 3) + 0 − 0 +

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) + 0 − 0 +

Solution set: (−∞,−5) ∪ (3,∞)

13


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) + 0 − 0 +

Steps

13


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) + 0 − 0 +

Steps

• Factorize left-side x2 + 2x − 15 = (x + 5)(x − 3)

13


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) + 0 − 0 +

Steps


• Zeros of left-side −5 and 3

13


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) + 0 − 0 +

Steps



• Divide real number line into three parts: (−∞,−5), (−5, 3), (3,∞)

13


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) + 0 − 0 +

Steps




• On each of these intervals, determine the sign of (x + 5) and (x − 3),

13


x2 + 2x − 15 > 0Method 3

x < −5 x = −5 −5 < x < 3 x = 3 x > 3

x + 5 − 0 + + +

x − 3 − − − 0 +

(x + 5)(x − 3) + 0 − 0 +

Steps




• On each of these intervals, determine the sign of (x + 5) and (x − 3),

hence the sign of (x + 5)(x − 3)

14

Polynomial inequalities (degree ≥ 3)

anxn + an−1xn−1 + · · · a1x + a0 > 0

14


anxn + an−1xn−1 + · · · a1x + a0 > 0

≥ 0

< 0

≤ 0

14


anxn + an−1xn−1 + · · · a1x + a0 > 0

≥ 0

< 0

≤ 0Method 1 2n−1 cases

For example, n = 3: a · b · c > 0

4 cases: + + + + − − − + − − − +

14


anxn + an−1xn−1 + · · · a1x + a0 > 0

≥ 0

< 0



4 cases: + + + + − − − + − − − +

n = 4: a · b · c · d > 0

8 cases: + + ++ + + −− − + +− − − ++

+ − +− − + −+ + − −+ − − −−

14


anxn + an−1xn−1 + · · · a1x + a0 > 0

≥ 0

< 0



4 cases: + + + + − − − + − − − +

n = 4: a · b · c · d > 0

8 cases: + + ++ + + −− − + +− − − ++

+ − +− − + −+ + − −+ − − −−Method 2 Need graphs polynomials of degrees ≥ 3

14


anxn + an−1xn−1 + · · · a1x + a0 > 0

≥ 0

< 0



4 cases: + + + + − − − + − − − +

n = 4: a · b · c · d > 0

8 cases: + + ++ + + −− − + +− − − ++


Method 3 By table

14


anxn + an−1xn−1 + · · · a1x + a0 > 0

≥ 0

< 0



4 cases: + + + + − − − + − − − +

n = 4: a · b · c · d > 0

8 cases: + + ++ + + −− − + +− − − ++


Method 3 By table

ALL three methods need to factorize L.S.

15

Example Solve the following inequality x3 + 3x2 − 4x − 12 ≤ 0

15


Solution Denote p(x) = x3 + 3x2 − 4x − 12.

• Factorize p(x)

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0

By Factor Theorem, (x − 2) is a factor of p(x).

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0


� Long division or compare coeffficients p(x) = (x − 2)(x2 + 5x + 6)

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

−3 −2 2

x − 2

x + 2

x + 3

p(x)

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2

x + 2

x + 3

p(x)

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 0

x + 2 0

x + 3 0

p(x) 0 0 0

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 0 +

x + 2 0

x + 3 0

p(x) 0 0 0

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 − − − − − 0 +

x + 2 0

x + 3 0

p(x) 0 0 0

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 − − − − − 0 +

x + 2 0 + + +

x + 3 0

p(x) 0 0 0

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 − − − − − 0 +

x + 2 − − − 0 + + +

x + 3 0

p(x) 0 0 0

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 − − − − − 0 +

x + 2 − − − 0 + + +

x + 3 0 + + + + +

p(x) 0 0 0

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 − − − − − 0 +

x + 2 − − − 0 + + +

x + 3 − 0 + + + + +

p(x) 0 0 0

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 − − − − − 0 +

x + 2 − − − 0 + + +

x + 3 − 0 + + + + +

p(x) − 0 0 0

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 − − − − − 0 +

x + 2 − − − 0 + + +

x + 3 − 0 + + + + +

p(x) − 0 + 0 0

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 − − − − − 0 +

x + 2 − − − 0 + + +

x + 3 − 0 + + + + +

p(x) − 0 + 0 − 0

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 − − − − − 0 +

x + 2 − − − 0 + + +

x + 3 − 0 + + + + +

p(x) − 0 + 0 − 0 +

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 − − − − − 0 +

x + 2 − − − 0 + + +

x + 3 − 0 + + + + +

p(x) − 0 + 0 − 0 +

• Solution: x < −3 or x = −3 or x = −2 or −2 < x < 2 or x = 2, that is,

15



• Factorize p(x)

� p(2) = 23 + 3(22) − 4(2) − 12 = 0



= (x − 2)(x + 2)(x + 3)•

x < −3 −3 −3 < x < −2 −2 −2 < x < 2 2 2 < x

x − 2 − − − − − 0 +

x + 2 − − − 0 + + +

x + 3 − 0 + + + + +

p(x) − 0 + 0 − 0 +

• Solution: x < −3 or x = −3 or x = −2 or −2 < x < 2 or x = 2, that is,

x ≤ −3 or − 2 ≤ x ≤ 2.

Venn diagram

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Transcript of Venn diagram