FLC Ch 5 Math 100 Elementary Algebra Sec 5.1: The Greatest...

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FLC Ch 5 Page 1 of 13 Math 100 Elementary Algebra Sec 5.1: The Greatest Common Factor and Factor By Grouping (FBG) Ex 1 Factor. (Check work by multiplying.) a) b) 21 4 โˆ’ 14 3 + 35 2 96 2 2 โˆ’ 144 3 + 48 c) d) 6(3 + ) โˆ’ (3 + ) 3( โˆ’ 3) โˆ’ 12( โˆ’ 3) โˆ’ 6( โˆ’ 3) e) f) Factor out 1/3. 7 2 ( + 4) + + 4 2 3 2 (2 โˆ’ 1) โˆ’ 4 3 (2 โˆ’ 1) + 3(2 โˆ’ 1) g) h) 20 3 3 โˆ’ 18 3 4 + 22 4 4 6 โˆ’ 3 + 9 Ex 2 PP Find the area of the shaded region in factored form. is the radius of the larger circle and is the radius of the smaller circle. Ans: ( โˆ’ ) Recall: In the product , and are factors. Defn In an expression, any factor that is common to each term is called a common factor. The largest of all common factors is called the greatest common factor (GCF). Remark: The answer upon factoring is always a ___________________. When factoring, we ALWAYS start with the ___________________________ (unless itโ€™s 1).

Transcript of FLC Ch 5 Math 100 Elementary Algebra Sec 5.1: The Greatest...

FLC Ch 5

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Math 100 Elementary Algebra Sec 5.1: The Greatest Common Factor and Factor By Grouping (FBG)

Ex 1 Factor. (Check work by multiplying.) a) b)

21๐‘Ž4 โˆ’ 14๐‘Ž3 + 35๐‘Ž2 96๐‘ฅ2๐‘ฆ2 โˆ’ 144๐‘ฅ3๐‘ฆ + 48๐‘ฅ๐‘ฆ c) d)

6(3๐‘Ž + ๐‘) โˆ’ ๐‘ง(3๐‘Ž + ๐‘) 3๐‘(๐‘๐‘ โˆ’ 3๐‘Ž) โˆ’ 12(๐‘๐‘ โˆ’ 3๐‘Ž) โˆ’ 6๐‘(๐‘๐‘ โˆ’ 3๐‘Ž) e) f) Factor out 1/3.

7๐‘2(๐‘ + 4๐‘‘) + ๐‘ + 4๐‘‘ 2

3๐‘ฅ2(2๐‘ฅ โˆ’ 1) โˆ’

4

3๐‘ฅ(2๐‘ฅ โˆ’ 1) + 3(2๐‘ฅ โˆ’ 1)

g) h) 20๐‘Ž3๐‘3 โˆ’ 18๐‘Ž3๐‘4 + 22๐‘Ž4๐‘4 6๐‘ฅ โˆ’ 3๐‘ฅ๐‘ฆ + 9๐‘ฆ Ex 2 PP Find the area of the shaded region in factored form. ๐‘… is the radius of the larger circle and ๐‘Ÿ is the radius of the smaller circle. Ans: ๐…(๐‘น๐Ÿ โˆ’ ๐’“๐Ÿ)

Recall: In the product ๐‘‹๐‘Œ, ๐‘‹ and ๐‘Œ are factors. Defn In an expression, any factor that is common to each term is called a common factor. The largest of all common factors is called the greatest common factor (GCF). Remark: The answer upon factoring is always a ___________________.

When factoring, we ALWAYS start with the ___________________________ (unless itโ€™s 1).

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Practice Problems Factor. 1) 6๐‘ฅ๐‘ฆ โˆ’ 15๐‘ง + 21 2) 20๐‘ฅ2 โˆ’ 32๐‘ฅ๐‘ฆ + 12๐‘ฅ 3) 7(4๐‘ฅ โˆ’ 5) โˆ’ ๐‘(4๐‘ฅ โˆ’ 5)

4) 2๐‘ฅ(8๐‘ฆ + 3๐‘ง) โˆ’ 5๐‘ฆ(8๐‘ฆ + 3๐‘ง) 5) 4๐‘ฅ3(๐‘ฅ โˆ’ 1) โˆ’ (๐‘ฅ โˆ’ 1) Ex 3 Factor. a) ๐‘ฅ๐‘ฆ โˆ’ 4๐‘ฅ + 3๐‘ฆ โˆ’ 12 b) 10๐‘Ž๐‘ โˆ’ 3๐‘ + 5๐‘Ž๐‘ โˆ’ 6๐‘ c) ๐‘ฅ2 โˆ’ 2๐‘ฅ โˆ’ ๐‘ฅ๐‘ฆ + 2๐‘ฆ d) 15๐‘Ž3 โˆ’ 25๐‘Ž2๐‘ โˆ’ 18๐‘Ž๐‘2 + 30๐‘3 e) ๐‘Ž๐‘ฅ + ๐‘๐‘ฅ + ๐‘๐‘ฅ + ๐‘Ž๐‘ฆ + ๐‘๐‘ฆ + ๐‘๐‘ฆ Practice Problems Factor. 1) 18 + 3๐‘ฅ โˆ’ 6๐‘ฆ โˆ’ ๐‘ฅ๐‘ฆ 2) 15๐‘ฅ โˆ’ 9๐‘ฅ๐‘ + 20๐‘ค โˆ’ 12๐‘๐‘ค 3) ๐‘ฅ3 โˆ’ 5๐‘ฅ2 โˆ’ 3๐‘ฅ + 15 4) 7๐‘Ž + 21๐‘ + 2๐‘Ž๐‘ + 6๐‘2 5) 30๐‘Ž3 + 12๐‘Ž2๐‘ โˆ’ 25๐‘Ž๐‘2 โˆ’ 10๐‘3 Good Start?: (๐’™๐Ÿ‘ โˆ’ ๐Ÿ“๐’™๐Ÿ) โˆ’ (๐Ÿ‘๐’™ + ๐Ÿ๐Ÿ“)

Sec 5.2: Factoring Trinomials of the Form ๐’™๐Ÿ + ๐’ƒ๐’™ + ๐’„ (and 5.7)

We will โ€œdissectโ€ the FOIL method to factor trinomials. Consider different combinations of (๐‘ฅ 2)(๐‘ฅ 5). Observe numbers and signs. Ex 4 Factor. a) ๐‘ฅ2 + 12๐‘ฅ + 32 b) ๐‘ฆ2 โˆ’ 16๐‘ฆ + 60 c) ๐‘ฆ2 + 11๐‘ฆ โˆ’ 60

Factoring Trinomials of the Form ๐’™๐Ÿ + ๐’ƒ๐’™ + ๐’„ The factorization will have the form (๐‘ฅ + ๐‘š)(๐‘ฅ + ๐‘›) where ๐‘š๐‘› = ๐‘ and ๐‘š + ๐‘› = ๐‘.

Use FBG method when factoring a polynomial with 4 (or more) terms.

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d) ๐‘ฆ2 โˆ’ 11๐‘ฆ โˆ’ 60 e) 2๐‘ฅ2 + 2๐‘ฅ + 12 f) 6๐‘ฅ2 + 24๐‘ฅ + 18 How can we adjust?

g) 3๐‘ฅ2๐‘ฆ โˆ’ 6๐‘ฅ๐‘ฆ + 21๐‘ฅ๐‘ง h) ๐‘ฅ8 โˆ’ 2๐‘ฅ4 โˆ’ 15 i) ๐‘ฅ2 โˆ’2

3๐‘ฅ +

1

9

PP PP

j) ๐‘ฅ2 + 0.8๐‘ฅ + 0.15 k) If ๐‘ฅ โˆ’ 4 is a factor of ๐‘ฅ2 + ๐‘๐‘ฅ โˆ’ 20, what is the value of ๐‘?

Sec 5.7: Solving Quadratic Equations by Factoring

Exs Solve for ๐‘ฅ.

(๐‘ฅ โˆ’ 3)(๐‘ฅ + 2) = 0

1

2๐‘ฅ(2๐‘ฅ โˆ’ 1)(3๐‘ฅ + 4) = 0

Facts About Signs

Each represents some positive number.

๐‘ฅ2 + ๐‘ฅ + will factor as (๐‘ฅ +)(๐‘ฅ +) ๐‘ฅ2 โˆ’ ๐‘ฅ + will factor as (๐‘ฅ โˆ’)(๐‘ฅ โˆ’) ๐‘ฅ2 + ๐‘ฅ โˆ’ will factor as (๐‘ฅ +)(๐‘ฅ โˆ’) ๐‘ฅ2 โˆ’ ๐‘ฅ โˆ’ will factor as (๐‘ฅ +)(๐‘ฅ โˆ’)

Steps to Solve a Quadratic Equation by Factoring โ€“ Use ZFP

1) Make sure the equation is set to 0. 2) Factor, if possible, the quadratic expression. 3) Set each factor containing a variable equal to 0. 4) Solve the resulting equations to find each root. 5) Check each root.

Defn A quadratic equation is an equation of the form ๐‘Ž๐‘ฅ2 + ๐‘๐‘ฅ + ๐‘ = 0, where ๐‘Ž, ๐‘, and ๐‘ are real numbers and ๐‘Ž โ‰  0. ๐‘Ž๐‘ฅ2 + ๐‘๐‘ฅ + ๐‘ = 0 is the standard form of a quadratic equation.

Zero Factor Property

If ๐‘Ž โˆ™ ๐‘ = 0, then ๐‘Ž = 0 or ๐‘ = 0 (or both)

Last Sign of any Poly um ame signs ifference ifferent signs

S

D

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Ex 5 Solve. a) ๐‘ฅ2 โˆ’ 6๐‘ฅ โˆ’ 7 = 0 b) 2๐‘ฅ2 โˆ’ 12๐‘ฅ = 54 c) ๐‘ฅ(๐‘ฅ + 1) = 110

๐Ÿ๐’™(๐’™ โˆ’ ๐Ÿ”) = ๐Ÿ“๐Ÿ’

Ex 6 Find the area of the shaded region of the figure below in factored form. The dimensions of the smaller rectangle are ๐‘ฅ ร— (๐‘ฅ + 2). Practice Problems Factor. 1) ๐‘ฅ2๐‘ฆ + 14๐‘ฅ๐‘ฆ + 48๐‘ฆ 2) 2๐‘ฅ2 โˆ’ 12๐‘ฅ โˆ’ 54

Sec 5.3: Factoring Trinomials of the Form ๐’‚๐’™๐Ÿ + ๐’ƒ๐’™ + ๐’„ AND

Sec 5.4: The Difference of Two Squares and Perfect Square Trinomials Ex 7 Factor using the trial-and-error method.

a) 3๐‘ฅ2 + 11๐‘ฅ + 10 b) 12๐‘ฅ2 + 5๐‘ฅ โˆ’ 3 c) 4๐‘ฅ2 โˆ’ 7๐‘ฅ โˆ’ 15 What multiplies to 10 and adds to 11?

Ex 8 ____________ 4๐‘ฅ2 โˆ’ 7๐‘ฅ โˆ’ 15 = 0 Ex 9 Factor. a) 3๐‘ฅ2 โˆ’ 13๐‘ฅ โˆ’ 10 b) 12๐‘ฅ2 + 7๐‘ฅ โˆ’ 12

12

10

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c) 36๐‘ฅ2 โˆ’ 1 d) 25๐‘Ž2 โˆ’ 64๐‘2 e) 25๐‘ฅ2 + 10๐‘ฅ + 1 ๐Ÿ‘๐Ÿ”๐’™๐Ÿ + ๐ŸŽ๐’™ โˆ’ ๐Ÿ

f) 25๐‘ฅ2 โˆ’ 10๐‘ฅ + 1 g) 9 โˆ’ 100๐‘ฆ2 h) 36๐‘ฅ2 + 1 i) If 2๐‘ฅ โˆ’ 5 is a factor of 6๐‘ฅ2 + ๐‘๐‘ฅ + 10, what is the value of ๐‘? Ex 10 #68 At the beginning of every football game, the referee flips a coin to see who will kick off. The equation that gives the height (in feet) of the coin tossed in the air is โ„Ž = 6 + 29๐‘ก โˆ’ 16๐‘ก2. a) Factor the equation. b) Use the factored form of the equation to find the height of the quarter after 0 seconds, 1 second, and 2 seconds.

Factoring the Difference of Two Squares

๐‘Ž2 โˆ’ ๐‘2 = (๐‘Ž + ๐‘)(๐‘Ž โˆ’ ๐‘) Note: ๐‘Ž2 + ๐‘2 is ________.

Perfect Square Trinomials

๐‘Ž2 + 2๐‘Ž๐‘ + ๐‘2 = (๐‘Ž + ๐‘)2 ๐‘Ž2 โˆ’ 2๐‘Ž๐‘ + ๐‘2 = (๐‘Ž โˆ’ ๐‘)2

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Ex 11 Factor or solve. 12๐‘ฅ2 + ๐‘ฅ โˆ’ 6 12๐‘ฅ2 = โˆ’๐‘ฅ + 6 ๐‘ฅ2 โˆ’ ๐‘ฅ + 5 = 0 Can this appear on exam 2?

In-Class Problems/Quiz: How Am I Doing?

1) 3๐‘Ž2 โˆ’ 10๐‘Ž โˆ’ 8 2) 10๐‘ฅ2 + ๐‘ฅ โˆ’ 2 3) 3๐‘ฅ2 โˆ’ 23๐‘ฅ + 14 4) 4๐‘ฅ2 โˆ’ 11๐‘ฅ โˆ’ 3 5) 12๐‘ฅ2 โˆ’ 24๐‘ฅ + 9 6) 20๐‘ฅ2 โˆ’ 38๐‘ฅ + 12 7) 6๐‘ฅ2 + 17๐‘ฅ๐‘ฆ + 12๐‘ฆ2 8) 14๐‘ฅ3 โˆ’ 20๐‘ฅ2 โˆ’ 16๐‘ฅ 9) 24๐‘ฅ2 โˆ’ 98๐‘ฅ โˆ’ 45 (Quiz EC)

Ex 12 Factor. a) 36๐‘ฅ2 + 60๐‘ฅ๐‘ฆ + 25๐‘ฆ2 b) 121๐‘ฆ2 โˆ’ 49 c) 50๐‘Ž2 โˆ’ 160๐‘Ž๐‘ + 128๐‘2 d) 5๐‘ฅ2 + 40๐‘ฅ + 80 e) 2๐‘ฅ2 โˆ’ 32๐‘ฅ + 110 f) ๐‘ฆ2๐‘ง โˆ’ 12๐‘ฆ๐‘ง + 36๐‘ง g) ๐‘ฅ2 + 16 h) ๐‘ฅ2 โˆ’ 16 i) 2๐‘ฅ4 โˆ’ 32

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Practice Problems Factor. 16๐‘ฅ2 โˆ’ 36๐‘ฆ2 49๐‘ฅ2 โˆ’ 28๐‘ฅ + 4 100๐‘ฅ2 โˆ’ 9 18๐‘ฆ2 โˆ’ 50๐‘ฅ2 25๐‘ฅ2 + 20๐‘ฅ + 4 25๐‘ฅ2 โˆ’ 20๐‘ฅ + 4 49๐‘ฅ2 โˆ’ 28๐‘ฅ๐‘ฆ + 4๐‘ฆ2 3๐‘ฅ2 โˆ’ 75 72๐‘ฅ2 โˆ’ 192๐‘ฅ + 128 144๐‘ฅ2 ยฑ ________ + 81๐‘ฆ2

Sec 5.6: Factoring: A General Review AND Sec 5.7: Solving Quadratic Equations by Factoring Refer to โ€œFactoring Polynomials Guideโ€. Indicate number of terms for each type.

Factor completely or solve for the roots of each quadratic equation. If the polynomial is not

factorable, you must state that itโ€™s prime. Check answers! How? What will each answer look like?

14) 3๐‘Ž๐‘ฅ + 9๐‘๐‘ฅ โˆ’ 12๐‘Ž๐‘ค โˆ’ 36๐‘๐‘ค 15) ๐‘ฅ3 + 2๐‘ฅ2๐‘ฆ โˆ’ 15๐‘ฅ๐‘ฆ2 16) 8 + 7๐‘ฅ โˆ’ ๐‘ฅ2 Do DO Do 17) 4๐‘ฅ2 + 2๐‘ฅ = 0 18) ๐‘ฅ2 + ๐‘ฅ โˆ’ 42 19) 7๐‘ฅ2 โˆ’ 252 Do PP Do 20) ๐‘ฅ2 + 36 21) ๐‘ฅ2 โˆ’ 7๐‘ฅ โˆ’ 14 22) 4๐‘ฅ2 โˆ’ 4๐‘ฅ โˆ’ 80 = 0 Do Do Do 23) 8๐‘ฅ3 โˆ’ 22๐‘ฅ2 + 5๐‘ฅ = 0 24) 100๐‘ฅ2 + 25 25) 10๐‘ฅ2 + ๐‘ฅ โˆ’ 2 Do Do Do

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26) 25๐‘ฅ2 + 16๐‘ฆ2 27) 64๐‘ฅ2 + 48๐‘ฅ = โˆ’9 28) ๐‘ฅ3 โˆ’ 5๐‘ฅ2 โˆ’ 4๐‘ฅ + 20 Do Do

29) 2๐‘ฅ2 โˆ’ 10๐‘ฅ โˆ’ 14 30) 3๐‘ฅ2 โˆ’ 33๐‘ฅ + 54 31) 4๐‘ฅ4 โˆ’ 11๐‘ฅ2 โˆ’ 3 Do 32) 18๐‘ฅ2 โˆ’ 69๐‘ฅ + 60 33) 2๐‘ฅ2 + ๐‘ฅ + 6 34) 12๐‘ฅ2 + 11๐‘ฅ๐‘ฆ โˆ’ 5๐‘ฆ2 Do Do 35) 4๐‘ฅ2 โˆ’ 13๐‘ฅ โˆ’ 12 36) ๐‘ฅ2 + 7๐‘ฅ + 1 37) (๐‘ฅ + 3๐‘ฆ)2 โˆ’ 16 Do Do ๐Ÿ๐Ÿ” โˆ’ (๐’™ + ๐Ÿ‘๐’š)๐Ÿ

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38) 8๐‘ฅ๐‘ค + 9๐‘ฅ2 + 35๐‘ฅ๐‘ฆ2 + 28๐‘ฆ2๐‘ค + ๐‘ฅ2 39) 10๐‘ฅ2 + 5๐‘ฅ๐‘ฆ โˆ’ 20 40) 18 โˆ’ 2๐‘ฅ2 Start Do Ans: (๐Ÿ“๐’™ + ๐Ÿ’๐’˜)(๐Ÿ๐’™ + ๐Ÿ•๐’š๐Ÿ) 41) 25๐‘ฅ2 = 36 42) (๐‘ฅ โˆ’ 3)4 + 4(๐‘ฅ โˆ’ 3)2 43) (3๐‘ฅ โˆ’ 2)3 โˆ’ 3๐‘ฅ + 2 Ex 44 Solve and check. a) Do b) Do

(2๐‘ฅ โˆ’ 3)(๐‘ฅ โˆ’ 1) = 3 (๐‘ฅ โˆ’ 5)(๐‘ฅ + 4) = 2(๐‘ฅ โˆ’ 5) c) Do d) Do e) PP

๐‘ฅ2 + 5๐‘ฅ

6= 4 (3๐‘ฅ โˆ’ 4)(5๐‘ฅ + 1)(2๐‘ฅ โˆ’ 7) = 0 (1119๐‘ฅ โˆ’ 1)(777๐‘ฅ + 19) = 0

Ans: ๐’™ =๐Ÿ

๐Ÿ๐Ÿ๐Ÿ๐Ÿ—, โˆ’

๐Ÿ๐Ÿ—

๐Ÿ•๐Ÿ•๐Ÿ•

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f) g) h) ๐‘ฅ(12 โˆ’ ๐‘ฅ) = 32 152 = (๐‘ฅ + 3)2 + ๐‘ฅ2 4๐‘ฅ3 + 12๐‘ฅ2 โˆ’ 9๐‘ฅ โˆ’ 27 = 0 Ex 45 PP Grade the solution. 81๐‘ฅ2 โˆ’ 16 = (9๐‘ฅ + 4)(9๐‘ฅ โˆ’ 4) = (3๐‘ฅ + 2)(3๐‘ฅ โˆ’ 2) Ex 46 Fill in the boxes to create a perfect square trinomial. 64๐‘ฅ2 ยฑ 81๐‘ฆ2 Ex 47 Consider (29๐‘ฅ + 7)(29๐‘ฅ โˆ’ 14) = 0, (29๐‘ฅ + 7)(29๐‘ฅ โˆ’ 14) = 1, and (29๐‘ฅ + 7)(29๐‘ฅ โˆ’ 14) = ๐‘ฅ.

Sec 5.8: Applications of Quadratic Equations Ex 48 (#6) The product of two consecutive odd integers is 1 less than 4 times their sum. Find the two integers. Define variable and set up. PP-solve. Ans: 7 and 9 OR -1 and 1

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Ex 49 (#12) One number is 2 more than twice another. Their product is 2 more than twice their sum. Find the numbers. Ex 50 (#14) The length of a rectangle is 3 more than twice the width. The area is 44 square inches. Find the dimensions. Ex 51 (#18) The hypotenuse of a right triangle is 15 inches. One of the legs is 3 inches more than the other. Find the lengths of the two legs.

Pythagorean Theorem In any right triangle, if ๐‘ is the length of the hypotenuse and ๐‘Ž and ๐‘ are the lengths of the two legs, then ๐‘Ž2 + ๐‘2 = ๐‘2.

๐‘Ž

๐‘ ๐‘

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Ex 52 (#34) A rocket is fired vertically into the air with a speed of 240 feet per second. Its height at time ๐‘ก seconds is given by โ„Ž(๐‘ก) = โˆ’16๐‘ก2 + 240๐‘ก. At what time(s) will the rocket be the following number of feet above the ground? a) PP 704 feet b) 896 feet c) Why do parts a and b have two answers? d) How long will the rocket be in the air? e) When the equation for part d is solved, one of the answers is ๐‘ก = 0. What does this represent? Ex 53 (#26) A company manufactures flash drives for home computers. It knows from experience that the number of drives it can sell each day, ๐‘ฅ, is related to the price ๐‘ per drive by the equation ๐‘ฅ = 800 โˆ’ 100๐‘. At what price should the company sell the flash drives if it wants the daily revenue to be $1200? The equation for revenue is ๐‘… = ๐‘ฅ๐‘.

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Revisit example 1a

Multiplicity Ex 54 PP You are standing on the edge of a cliff near Acapulco, overlooking the ocean. The place where you stand is 180 meters from the ocean. You drop a pebble into the water. (Dropping the pebble implies that there is no initial velocity, so ๐‘ฃ = 0.) How many seconds will it take to hit the water? How

far has the pebble dropped after 3 seconds? Use the formula ๐‘บ = โˆ’๐Ÿ“๐’•๐Ÿ + ๐’—๐’• + ๐’‰, where ๐‘† = the height of the object ๐‘ฃ = the upward velocity in meters/second ๐‘ก = the time of flight in seconds โ„Ž = the height above level ground from which the object is thrown

Discriminant โ€˜Problems from Factoring Assignment (Due:________________)