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UNIT 1ALESSON 6

Linear, Quadratic and Polynomial Inequalities

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INTERVALS & INEQUALITIESInterval Notation

InequalityNotation

Graph

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Express the following intervals in terms of inequalities and graph the intervals

1,4

2

14

2x

[ 2,1) 2 1x

( 4, ) 4x

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REMEMBER:Linear inequalities are solved the same as equations

EXCEPT when the final step involves dividing by a NEGATIVE.

You must change the direction of the sign.

10 4

0 4

2 2

1

5 2

20 5

20 5

5 5

4 1

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Linear InequalitiesSolve the inequalities.State the answer in inequality form, interval form and graph.

5x + 7 > – 8

5x > – 15

x > – 3 inequality form

[– 3, ∞) interval form

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EXAMPLE 2

graph

(−𝟑 ,−𝟖)

𝒀 𝟐=𝟓 𝒙+𝟕

𝒀 𝟐=−𝟖

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Linear InequalitiesSolve the inequalities.State the answer in inequality form, interval form and graph.

3x + 1 < 7x – 7

– 4x < – 8

x > 2 inequality form

(2, ∞) interval form

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EXAMPLE 3

(𝟐 ,𝟕)

𝒀 𝟏=𝟑 𝒙+𝟏

𝒀 𝟐=𝟕 𝒙−𝟕graph

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Linear InequalitiesSolve the inequalities and graph. State the answer in inequality form and interval form.

8 – x > 5x + 2

– 6x > – 6

x < 1 inequality form

(–∞, 1) interval form

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EXAMPLE 4

−𝟔 𝒙−𝟔

<−𝟔−𝟔

(𝟏 ,𝟕)

𝒀 𝟏=𝟖−𝒙 𝒀 𝟐=𝟓 𝒙+𝟐

graph

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SOLVING POYNOMIAL INEQUALITIES In order to solve any polynomial equation or inequality you must FACTOR first!!!

EXAMPLE 5 : x2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

–4 2

x = –4 or x = 2

If x < – 4

( + 4)( – 2) is

positive

If x is between –4 and 2

( + 4)( – 2) isnegative

If x > 2

positive

( + 4)( – 2) is

+¿ +¿−

Let’s use our heads

−𝟏𝟎−𝟏𝟎−𝟓 −𝟓 𝟑𝟑 𝟔𝟔𝟎𝟎

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SOLVING POYNOMIAL INEQUALITIES

x2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

– 4 2

x = – 4 or x = 2

(x + 4)(x – 2) > 0 (x + 4)(x – 2) < 0

x2 + 2x – 8 > 0 x2 + 2x – 8 < 0

x < – 4 or x > 2 – 4 < x < 2

(−𝟒 ,𝟐)(−∞ ,−4 )∪ (2 ,∞ )

+¿ +¿−

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𝒀 𝟏=𝒙𝟐+𝟐𝒙 –𝟖

𝒙>𝟐𝒙<−𝟒

−𝟒<𝒙<𝟐

𝒙=𝟐𝒙=−𝟒

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EXAMPLE 6 Factor x3 + 2x2 – 5x – 6

-2 1 2 -5 -6 -2 -8 -6 1 4 3 0

23 + 2(2)2 – 5(2) – 6 = 8 + 8 – 10 – 6 = 0

(x – 2) is a factor

x3 + 2x2 – 5x – 6 = (x – 2)(1x2 + 4x + 3)

= (x – 2)(x + 3)(x + 1)

Test potential zeros ±1, ±2, ±3, ±6.

2 1 2 -5 -6 2 8 6 1 4 3 0

Subtraction Method Addition Method

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EXAMPLE 6 continued x3 + 2x2 – 5x – 6

x3 + 2x2 – 5x – 6 = (x – 2)(x + 3)(x + 1)

– 3 – 1 2

( – 2)( + 3)( + 1) is

− +¿

If x < – 3

pos

neg

If – 3 < x < – 1 ( – 2)( + 3)( + 1) is

( – 2)( + 3)( + 1) is

( – 2)( + 3)( + 1) is

− +¿

If – 1 < x < 2

If x > 2

neg

pos

−𝟐

−𝟒−𝟒 −𝟒

−𝟐−𝟐

𝟎 𝟎 𝟎

𝟓 𝟓 𝟓

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x3 + 2x2 – 5x – 6 = (x – 2)(x + 3)(x + 1)

–3 –1 2

neg pos neg pos

x3 + 2x2 – 5x – 6 < 0 x3 + 2x2 – 5x – 6 > 0

x < –3 or –1 < x < 2

–3 < x < –1 or x > 2

EXAMPLE 6 x3 + 2x2 – 5x – 6 EXAMPLE x3 + 2x2 – 5x – 6

(−∞ ,−𝟑 )∪ [−𝟏 ,𝟐 ] [−𝟑 ,𝟏 ]∪¿

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𝒙=−𝟑 𝒙=−𝟏 𝒙=𝟐

x < –3

–3 < x < –1

–1 < x < 2

x > 2 𝒙𝟑+𝟐𝒙𝟐−𝟓𝒙 −𝟔≥𝟎

𝒙𝟑+𝟐𝒙𝟐−𝟓𝒙 −𝟔≤𝟎