The number of marbles in 13 bags # 1 The number of marbles in 13 bags The number of marbles you had before purchasing more bags. The number of marbles in each of the 13 bags.

# 2 The number of children Jamie has. The number of months. The monthly tuition for the science club The one-time materials fee. The total cost for one child to join the science club.

3𝑥 2𝑥 −2 +5 2𝑦 −3 5𝑦 +5 6 𝑥 2 −4𝑥 10 𝑦 2 +10𝑦 +15𝑥 −10 −15𝑦 −15 # 3 Use the area model. 3𝑥 2𝑥 −2 +5 2𝑦 −3 5𝑦 +5 6 𝑥 2 −4𝑥 10 𝑦 2 +10𝑦 +15𝑥 −10 −15𝑦 −15 6 𝑥 2 +11𝑥−10 10 𝑦 2 −5𝑦−15

# 4 Multiply Exponents Multiply Exponents 𝑦 6 ∗ 𝑦 4 = 𝑦 10 𝑧 6 ∗ 𝑧 3 = 𝑧 9 Add Exponents Add Exponents To raise to a power; multiply exponents. To multiply; add exponents.

Simplify the numerator and denominator. # 5 3 𝑎 ∙2 9 27∙ 4 =1 Simplify the radicals. 1 3 𝑎 ∙2∙3 27∙2 =1 Simplify the numerator and denominator. 9 3 𝑎 9 =1 Re-write the equation with all terms in the same base. 3 𝑎 3 2 = 3 0 Use the properties of exponents. 𝑎−2=0 𝑎=2

Use the properties of exponents. # 6 3 𝑑 ∙ 5 3 2 ∙ 45 =3 Simplify the radicals. 3 𝑑 3 2 ∙ 9 =3 3 𝑑 3 2 ∙3 =3 Use the properties of exponents. 3 𝑑 3 3 =3 Use the properties of exponents again. 𝑑−3=1 𝑑=4

# 7 = 𝑔 3 5 𝑎. 7 𝑥 𝑏. 5 𝑔 3 𝑐. 𝑏 𝑥 𝑎

# 8 𝑎. (7 𝑦 3 ) 1 2 𝑏. 𝑘 5 3 𝑐. 𝑝 𝑎 𝑐

Simplify each of the following. # 9 Simplify each of the following. a. 𝑥 1 4 𝑥 3 4 b. 16 𝑚 2 5 𝑛 1 2 8 𝑚 − 3 5 𝑛 3 2 c. 3𝑐 5 6 𝑑 3 27 𝑐 1 6 𝑑 5 When dividing the same base, subtract the exponents. = 𝑥 1 4 − 3 4 = 16 8 𝑚 2 5 − − 3 5 𝑛 1 2 − 3 2 = 3 27 𝑐 5 6 − 1 6 𝑑 3−5 = 𝑥 − 2 4 = 1 9 𝑐 4 6 𝑑 −2 =2 𝑚 5 5 𝑛 − 2 2 = 𝑥 − 1 2 =2 𝑚 1 𝑛 −1 = 𝑐 2 3 9 𝑑 2 = 1 𝑥 1 2 = 2𝑚 𝑛 Do not leave a negative exponent in the answer.

𝑎. real; the 3 is irrational # 10 𝑎. real; the 3 is irrational 𝑏. real; 4− 3 5 =3 2 5 rational 𝑐. real; irrational 𝑑. real; 7 9 + 2 9 = 9 9 =1 rational, integer, whole

2𝑡+8=7+5𝑡−20 20𝑤−10=8+2𝑤−12+12 2𝑡+8=5𝑡−13 20𝑤−10=8+2𝑤 −2𝑤 −2𝑤 −2𝑡 −2𝑡 # 11 Distribute Distribute 2𝑡+8=7+5𝑡−20 20𝑤−10=8+2𝑤−12+12 Combine Like Terms Combine Like Terms 2𝑡+8=5𝑡−13 20𝑤−10=8+2𝑤 −2𝑤 −2𝑤 −2𝑡 −2𝑡 Subtract 2t from each side of the equation. 8=3𝑡−13 18𝑤−10=8 Subtract 2w from each side of the equation. +13 +13 +10 +10 3 3 Divide each side of the equation by 3. 21=3𝑡 18𝑤=18 18 18 Add 13 to each side of the equation. Divide each side of the equation by 18. Add 10 to each side of the equation. 7=𝑡 𝑤=1

# 12

# 13 23+3𝑥=5𝑥+7 −3𝑥 −3𝑥 23=2𝑥+7 −7 −7 16=2𝑥 2 2 8=𝑥 8 hours

# 14 0.08𝑝=7.50+0.05𝑝 −0.05𝑝 −0.05𝑝 0.03𝑝=7.50 0.03 0.03 250=𝑝

36 total card−14=22 baseball cards # 15 𝑥+ 2𝑥−6 =36 +6 +6 Add 6 to each side of the equation. Combine like terms. 𝑥+2𝑥=42 3 3𝑥=42 3 Divide each side of the equation by 3. x represents the number of football cards! 𝑥=14 36 total card−14=22 baseball cards

Divide each side of the equation by 4. # 16 21+7 𝑥+7 28 𝑥 21 84=𝑥+ 2𝑥−7 +(𝑥+7) Combine Like Terms 2𝑥−7 35 4 4 84=4𝑥 Divide each side of the equation by 4. 2(21)−7 21=𝑥 Draw and label the triangle.

Change the inequality direction when dividing by a negative. # 17 −1 4 1 6 26 4 9 Change the inequality direction when dividing by a negative. 3 56 𝑎. 𝑞+ 1 3 > 1 2 e. d. f. 𝑥+12−2 𝑥−22 >0 5𝑎−3<−8 7<3𝑥−5≤22 2𝑎+9>17 𝑐. −3𝑥≤−9 𝑏. 17≥𝑥−9 −𝟑 −𝟑 3 6 − 2 6 = 1 6 +5 +5 +5 +3 +3 −9 −9 +𝟗 +𝟗 𝑥+12−2𝑥+44>0 12<3𝑥≤27 5𝑎<−5 5 5 2𝑎>8 2 2 − 𝟏 𝟑 − 𝟏 𝟑 3 3 3 𝑥≥3 26≥𝑥 −𝑥+56>0 −56 −56 𝑎<−1 𝑎>4 −𝑥>−56 −1 −1 4<𝑥≤9 𝑞> 1 6 𝑥<56

The plant height must be 79 inches or taller. # 18 (4) Clear the denominator by multiplying each side of the equation by 4. 70+71+72+𝑥 4 ≥73 (4) 70+71+72+𝑥≥292 Combine Like Terms. 213+𝑥≥292 Subtract 213 from each side of the equation. −213 −213 𝑥≥79 The plant height must be 79 inches or taller.

Mrs. Hawk can have no more than 28 questions. # 19 Clear the denominator by multiplying each side of the equation by 6. (6) 11+10+13+14+14+𝑥 6 ≤15 (6) 11+10+13+14+14+𝑥≤90 Combine Like Terms. 62+𝑥≤90 Subtract 62 from each side of the equation. −62 −62 𝑥≤28 Mrs. Hawk can have no more than 28 questions.

(𝑚) 𝑏. 𝑘𝑥−𝑏𝑓= 𝑓𝑦 𝑚 (𝑚) 𝑎. 𝑎−𝑞=𝑎+𝑠𝑥 −𝑎 −𝑎 𝑚(𝑘𝑥−𝑏𝑓)=𝑓𝑦 𝑓 𝑓 −𝑞=𝑠𝑥 𝑠 𝑠 # 20 (𝑚) 𝑏. 𝑘𝑥−𝑏𝑓= 𝑓𝑦 𝑚 (𝑚) 𝑎. 𝑎−𝑞=𝑎+𝑠𝑥 −𝑎 −𝑎 𝑚(𝑘𝑥−𝑏𝑓)=𝑓𝑦 𝑓 𝑓 −𝑞=𝑠𝑥 𝑠 𝑠 − 𝑞 𝑠 =𝑥 𝑚(𝑘𝑥−𝑏𝑓) 𝑓 =𝑦 (𝑚) 𝑑. 𝑘𝑥−𝑏𝑓= 𝑓𝑦 𝑚 (𝑚) 𝑐. 𝑎−𝑞=𝑎+𝑠𝑥 −𝑎 −𝑎 −𝑞=𝑠𝑥 −1 −1 𝑚(𝑘𝑥−𝑏𝑓)=𝑓𝑦 𝑘𝑥−𝑏𝑓 𝑘𝑥−𝑏𝑓 𝑞=−𝑠𝑥 𝑚= 𝑓𝑦 𝑘𝑥−𝑏𝑓

Substitute the values for x and y into the equation. # 21 Substitute the values for x and y into the equation. 𝐴𝑥+3𝑦=48 𝐴(−3)+3(14)=48 Now solve for A. −3𝐴+42=48 −42 −42 −3𝐴=6 −3 −3 𝐴=−2

Substitute the values for x and y into the equation. # 22 Substitute the values for x and y into the equation. 2𝑥+𝐷𝑦=13 2(−18)+𝐷(7)=13 Now solve for D. −36+7𝐷=13 +36 +36 7𝐷=49 7 7 𝐷=7

b. Solve the equation for d. c. Solve the equation for b. # 23 Marcy has $150 to buy packages of hot dogs and hamburgers for her booth at the carnival. At her local grocery store she found packages of hot dogs cost $6 and packages of hamburgers cost $20. a. Write an equation that can be used to find the possible combination of hot dog and hamburger packages Marcy can buy using her budget of exactly $150. (Hint: Use d to represent the number of hot dog packages and b to represent the number of hamburger packages.) b. Solve the equation for d. c. Solve the equation for b. c. 𝑏= 150−6𝑑 20 a. 6𝑑 + 20𝑏 = 150 (or equivalent) 𝑏. 𝑑= 150−20𝑏 6 (or equivalent)

# 24 Andrew needs exactly 120 pencils and markers for his class to play the review game he created. His supply closet has boxes of pencils with 12 in each and boxes of markers with 6 in each. Write an equation that can be used to find the number of boxes of pencils and boxes of markers Andrew will need to take to reach his total of 120. Re-write the equation so that is could be used to find the number of pencil boxes. Re-write the equation so that is could be used to find the number of marker boxes. a. If p = the number of pencil boxes and m = the number of marker boxes, 12𝑝+6𝑚=120 b. 𝑝=− 1 2 𝑚+10 (or equivalent) c. 𝑚=−2𝑝+20 (or equivalent)

# 25 This is not a function because –4 (domain) is assigned to more than one value in the range. This is a function because every element of the domain is assigned to exactly one element in the range. This is a function because every element of the domain is assigned to exactly one element in the range. This is not a function because –3 is assigned to multiple values in the range.

# 26 Subtract 1 from each side of the equation. 13=0.3𝑚+1 −1 −1 0.3 0.3 12=0.3𝑚 Divide each side of the equation by 0.3. 40=𝑚 40 miles

# 27 185=25𝑥 +10 Subtract 10 from each side of the equation. −10 −10 175=25𝑥 25 25 Divide each side of the equation by 25. 7=𝑥 7 months

𝑎. 3 𝑥 3 +2 𝑥 2 −3𝑥−2 +( 𝑥 2 +5𝑥−6) 3 𝑥 3 +3 𝑥 2 +2𝑥−8 # 28 Combine Like Terms… 𝑎. 3 𝑥 3 +2 𝑥 2 −3𝑥−2 +( 𝑥 2 +5𝑥−6) 3 𝑥 3 +3 𝑥 2 +2𝑥−8 Distribute 𝑏. 12 𝑦 2 +5𝑦−6 − 9 𝑥 2 +3𝑥−4 +11 Combine Like Terms… 12 𝑦 2 +5𝑦−6 + −9 𝑥 2 −3𝑥+4 +11 12 𝑦 2 −9 𝑥 2 +5𝑦−3𝑥+9

𝑑. 6𝑚−2 2 𝑚 2 −3𝑚+2 𝑐. 2𝑘−5 3𝑘+6 2 𝑚 2 6𝑚 −3𝑚 −2 +2 2𝑘 3𝑘 −5 +6 12 𝑚 3 # 28 𝑑. 6𝑚−2 2 𝑚 2 −3𝑚+2 𝑐. 2𝑘−5 3𝑘+6 2 𝑚 2 6𝑚 −3𝑚 −2 +2 2𝑘 3𝑘 −5 +6 12 𝑚 3 −18 𝑚 2 +12𝑚 6 𝑘 2 −15𝑘 −4 𝑚 2 +6𝑚 −4 +12𝑘 −30 12 𝑚 3 −22 𝑚 2 +18𝑚−4 6 𝑘 2 −3𝑘−30

𝑒. a, b and c are polynomials # 29 𝑎. (3𝑥+5)+(−8𝑥−10) 𝑑. (3𝑥+5) (−8𝑥−10) −𝟓𝒙−𝟓 𝑏. 3𝑥+5 −(−8𝑥−10) 𝟑𝒙+𝟓 −𝟐(𝟒𝒙−𝟓) 3𝑥+5 +(+8𝑥+10) 𝟏𝟏𝒙+𝟏𝟓 𝑐. 3𝑥+5 (−8𝑥−10) 𝑒. a, b and c are polynomials −24𝑥 2 −40𝑥 −30𝑥 −50 −𝟐𝟒 𝒙 𝟐 −𝟕𝟎𝒙−𝟓𝟎

# 30 𝒈 𝒇 𝒙 =𝟏.𝟎𝟖(𝟎.𝟕𝒙+𝟐𝟑

𝑔(𝑓 3 ) 𝑓(𝑔 𝑥 ) =𝑔(3 3 +2) =𝑓(2 𝑥 2 ) =𝑔(11) =3 2 𝑥 2 +2 =2 11 2 # 31 𝑔(𝑓 3 ) 𝑓(𝑔 𝑥 ) =𝑔(3 3 +2) =𝑓(2 𝑥 2 ) =𝑔(11) =3 2 𝑥 2 +2 =2 11 2 =6 𝑥 2 +2 =2(121) =242

# 32 a. There are no games left in the warehouse after 27 shipments. b. He has 9,000 games in the warehouse before making any shipments. c. The slope represents the decrease in the number of games in the warehouse per shipment. d. Domain 0≤𝑥≤27 and Range 0≤𝑦≤9,000

a. Domain 𝟎≤𝒙≤𝟖 and Range 𝟎≤𝒚≤𝟏𝟗𝟐 # 33 192 a. Domain 𝟎≤𝒙≤𝟖 and Range 𝟎≤𝒚≤𝟏𝟗𝟐 b. Both intercepts are 0. This means no cookies can be made with no cups of flour.

# 34 Elissa uses the function 𝑃 ℎ =15ℎ to represent the money she makes working at the party store for h hours each week. The function 𝐶 ℎ =25ℎ+30 represents the money she makes cleaning houses in the neighborhood. Write a function T to describe the total amount of money Elissa makes each week working h hours. 𝑇 ℎ =40ℎ+30

# 35 Henry predicts the amount of wheat, in bushels, produced on his farm 𝑡 years from now can be modeled by the function 𝑊 𝑡 =70𝑡+900. The price, in dollars, per bushel of wheat 𝑡 years from now can be modeled by the function 𝑃 𝑡 =0.25𝑡+6. Write a function 𝐼 to describe the total income Henry predicts for producing wheat 𝑡 years from now. 𝐼 𝑡 =𝑊 𝑡 ∙𝑃 𝑡 =17.5 𝑡 2 +645𝑡+5400

Slope is the average rate of change for a linear function. # 36 The graph below shows the average Valentine’s Day spending between 2003 and 2012. Slope is the average rate of change for a linear function. (2009, 102) (2004, 100) (2005, 98) (2007, 120) (2010, 103) (2010, 103) 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 𝑜𝑟 𝑓 𝑏 −𝑓(𝑎) 𝑏−𝑎 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 𝑜𝑟 𝑓 𝑏 −𝑓(𝑎) 𝑏−𝑎 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 𝑜𝑟 𝑓 𝑏 −𝑓(𝑎) 𝑏−𝑎 120−98 2007−2005 103−102 2010−2009 103−100 2010−2004 = 22 2 = 1 1 = 3 6 = 1 2 =1 =11 What is the average rate of change in spending between 2005 and 2007? What is the average rate of change in spending between 2004 and 2010? What is the average rate of change in spending between 2009 and 2010? $11 per year $0.50 per year $1 per year

Graph the line with a y-intercept of 2 and a slope of 𝟓 𝟐 . # 37 Graph the line with a y-intercept of 2 and a slope of 𝟓 𝟐 . Translate the line 4 units left.

Graph the line with a y-intercept of −3 and a slope of 𝟐 𝟏 . # 38 Graph the line with a y-intercept of −3 and a slope of 𝟐 𝟏 . Translate the line 3 units up.

𝑎. 𝑎 𝑛 =7+(𝑛−1)(−3) 𝑏. 𝑎 𝑛 =−11+(𝑛−1)(8) Initial value # 39 Initial value 𝑎. 𝑎 𝑛 =7+(𝑛−1)(−3) Common difference 𝑏. 𝑎 𝑛 =−11+(𝑛−1)(8)

𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 Find the slope… Find the y-intercept… 𝑦=𝑚𝑥+𝑏 # 40 Find the slope… Find the y-intercept… 𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 𝑦=𝑚𝑥+𝑏 151=6 24 +𝑏 151−79 24−12 = 72 12 = 6 1 =6 151=144+𝑏 −144 −144 7=𝑏 The special rate would be $6. The shipping would be $7.

𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 Find the slope… Find the y-intercept… 𝑦=𝑚𝑥+𝑏 # 41 Find the slope… Find the y-intercept… 𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 𝑦=𝑚𝑥+𝑏 40−30 6−2 = 10 4 = 5 2 30= 5 2 2 +𝑏 30=5+𝑏 −5 −5 25=𝑏 He would make 2.5 cards per hour. He started with 25 cards in the box.

The slope of the line of best fit is 𝟏 𝟑 and it’s y-intercept is 0. # 42 The slope of the line of best fit is 𝟏 𝟑 and it’s y-intercept is 0. 𝒇 𝒙 = 𝟏 𝟑 𝒙 Every 3 days, 1 additional golf cart is repaired

The slope of the line of best fit is −𝟓 and it’s y-intercept is 90. # 43 The slope of the line of best fit is −𝟓 and it’s y-intercept is 90. 𝒇 𝒙 =−𝟓𝒙+𝟗𝟎 Every day there are 5 less golf carts sold

Multiply each term in the first equation by −𝟑. # 44 Multiply each term in the first equation by −𝟑. (−𝟑) (−𝟑) (−𝟑) 𝑎. 2𝑥+5𝑦=20 6𝑥+15𝑦=15 −6𝑥−15𝑦=−60 6𝑥+15𝑦=15 0𝑥+0𝑦=−45 No Solution

Infinitely many solutions # 44 Multiply each term in the first equation by 𝟒. 𝑏. 9𝑥+18𝑦=6 12𝑥+24𝑦=8 Multiply each term in the second equation by −𝟑. 36𝑥+72𝑦=24 −36𝑥−72𝑦=−24 0𝑥+0𝑦=0 Infinitely many solutions

𝑐. 5𝑥+4𝑦=6 6𝑥+𝑦=13 (19) (19) 30𝑥+24 − 29 19 =36 (19) 30𝑥+24𝑦=36 # 44 Multiply each term in the first equation by 𝟔. 𝑐. 5𝑥+4𝑦=6 6𝑥+𝑦=13 Multiply each term in the second equation by −𝟓. (19) (19) 30𝑥+24 − 29 19 =36 (19) 30𝑥+24𝑦=36 −30𝑥−5𝑦=−65 570𝑥−696=684 +696 +696 0𝑥+19𝑦=−29 19 19 570𝑥=1380 570 570 𝑦=− 29 19 𝑥= 46 19

28−16=12 blocks 𝑥+𝑦=28 0.38𝑥+1.56𝑦=24.80 𝑦=28−𝑥 0.38𝑥+1.56 28−𝑥 =24.80 # 45 28−16=12 blocks Let x be the number of bricks and y the number of blocks. 𝑥+𝑦=28 0.38𝑥+1.56𝑦=24.80 𝑦=28−𝑥 Substitute 0.38𝑥+1.56 28−𝑥 =24.80 Distribute 0.38𝑥+43.68−1.56𝑥=24.80 Combine Like Terms −1.18𝑥+43.68=24.80 −43.68 −43.68 −1.18 −1.18 Subtract 43.68 from each side of the equation. −1.18𝑥=−18.88 Divide each side of the equation by -1.18. 𝑥=16

300 adult tickets were sold. # 46 300 adult tickets were sold. 300 can now be substituted into either equation for a. Let a = # adult tickets and c = # of child tickets 20𝑎+10𝑐=15,000 𝑐=3𝑎 Use substitution to replace c with 3a. 𝑐=3𝑎 20𝑎+10 3𝑎 =15,000 𝑐=3(300) 20𝑎+30𝑎=15,000 𝑐=900 50𝑎=15,000 50 50 900 child tickets were sold. 𝑎=300

+ Multiply the first equation by −𝟒𝟑. # 47 Multiply the first equation by −𝟒𝟑. Let S = # of small dogs and L = # of large dogs 𝑆+𝐿=22 43𝑆+75𝐿=1234 −43𝑆−43𝐿=−946 + 43𝑆+75𝐿=1234 Since we are looking for L, let’s eliminate S. Add the second equation. 32𝐿=288 32 32 𝐿=9 9 large dogs were groomed.

# 48 where C is the cost and d is the number of disks 13.75𝑑=3255+2.9𝑑 −2.9𝑑 −2.9𝑑 10.85𝑑=3255 10.85 10.85 𝑑=300

𝑎. −2𝑥−3𝑦=−4 12𝑥+3𝑦=1 𝑏. −5𝑥+35𝑦=−40 5𝑥+4𝑦=20 # 49 Answers will vary. Below are sample answers. 𝑎. −2𝑥−3𝑦=−4 12𝑥+3𝑦=1 We can multiply the first equation by – 1. 𝑏. −5𝑥+35𝑦=−40 5𝑥+4𝑦=20 We can multiply the first equation by – 5.

# 50 𝑎. 7𝑥+6𝑦>30 −7𝑥 −7𝑥 6𝑦>−7𝑥+30 6 6 6 𝑦>− 7 6 𝑥+5

# 50 𝑏. 9𝑥+4𝑦≥28 −9𝑥 −9𝑥 4𝑦≥−9𝑥+28 4 4 4 𝑦≥− 9 4 𝑥+7

# 50 𝑐. 3𝑥+7𝑦<14 −3𝑥 −3𝑥 7𝑦<−3𝑥+14 7 7 7 𝑦<− 3 7 𝑥+2

# 50 𝑑. 7𝑥+14𝑦≤3 −7𝑥 −7𝑥 14𝑦≤−7𝑥+3 14 14 14 𝑦≤− 7 14 𝑥+ 3 14 𝑦≤− 1 2 𝑥+ 3 14

−5𝑥−4𝑦≤6 +5𝑥 +5𝑥 −4𝑦≤5𝑥+6 −4 −4 −4 𝑦≥− 5 4 𝑥−1 1 2 8𝑥−16𝑦>24 −8𝑥 # 51 −5𝑥−4𝑦≤6 +5𝑥 +5𝑥 −4𝑦≤5𝑥+6 −4 −4 −4 𝑦≥− 5 4 𝑥−1 1 2 8𝑥−16𝑦>24 −8𝑥 −8𝑥 −16𝑦>−8𝑥+24 −16 −16 −16 𝑦< 1 2 𝑥−1 1 2

𝑥+𝑦>1 −𝑥 −𝑥 𝑦>−𝑥+1 5𝑥+2𝑦<10 −5𝑥 −5𝑥 2𝑦<−5𝑥+10 2 2 2 # 51 𝑥+𝑦>1 −𝑥 −𝑥 𝑦>−𝑥+1 5𝑥+2𝑦<10 −5𝑥 −5𝑥 2𝑦<−5𝑥+10 2 2 2 𝑦<− 5 2 𝑥+5

# 52 175 feet 25 seconds

# 53 about 4.5 seconds about 26 feet

What are the factors of −𝟔 that also combine to +𝟏? # 54 FACTOR 𝑥 2 +𝑥−6 What are the factors of −𝟔 that also combine to +𝟏? FACTOR 𝑥 −2 𝑥 𝑥 2 −2𝑥 +3 3𝑥 −6 −6 𝑥 2 𝒙+𝟑

What are the factors of +𝟑𝟎 that also combine to +𝟏𝟑? # 55 FACTOR 2𝑥 2 +13𝑥+15 What are the factors of +𝟑𝟎 that also combine to +𝟏𝟑? FACTOR 2𝑥 +3 𝑥 2𝑥 2 3𝑥 +5 10𝑥 +15 +30 𝑥 2 𝒙+𝟓

What are the factors of 𝟏𝟎𝟓 that also combine to +𝟐𝟐? # 56 FACTOR 3𝑥 +7 𝑎. 3 𝑥 2 +22𝑥+35=0 𝑥 3 𝑥 2 7𝑥 𝑥+5=0 3𝑥+7=0 15𝑥 +35 −5 −5 −7 −7 +5 3 3 3𝑥=−7 𝑥=−5 105 𝑥 2 What are the factors of 𝟏𝟎𝟓 that also combine to +𝟐𝟐? 𝑥=− 7 3

𝑏. 2 𝑥 3 +4 𝑥 2 +2𝑥=0 2𝑥( 𝑥 2 +2𝑥+1)=0 2𝑥=0 𝑥 2 +2𝑥+1=0 𝑥=0 # 56 𝑏. 2 𝑥 3 +4 𝑥 2 +2𝑥=0 2𝑥( 𝑥 2 +2𝑥+1)=0 2𝑥=0 𝑥 2 +2𝑥+1=0 𝑥=0 (𝑥+1)(𝑥+1)=0 𝑥+1=0 −1 −1 𝑥=−1

𝑑. 45 𝑥 2 −20=0 5 9 𝑥 2 −4 =0 5 3𝑥−2 3𝑥+2 =0 𝑐. 𝑥 2 −6𝑥−27=0 3𝑥−2=0 # 56 𝑑. 45 𝑥 2 −20=0 5 9 𝑥 2 −4 =0 5 3𝑥−2 3𝑥+2 =0 𝑐. 𝑥 2 −6𝑥−27=0 3𝑥−2=0 3𝑥+2=0 +2 +2 −2 −2 𝑥−9 𝑥+3 =0 3 3 3 3 3𝑥=2 3𝑥=−2 𝑥−9=0 𝑥+3=0 𝑥= 2 3 𝑥=− 2 3 +9 +9 −3 −3 𝑥=9 𝑥=−3

𝑒. 5 𝑥 3 −20𝑥=0 𝑥(5 𝑥 2 −20)=0 𝑥𝑥=0 5 𝑥 2 −4 =0 5 𝑥−2 𝑥+2 =0 𝑥−2=0 # 56 𝑒. 5 𝑥 3 −20𝑥=0 𝑥(5 𝑥 2 −20)=0 𝑥𝑥=0 5 𝑥 2 −4 =0 5 𝑥−2 𝑥+2 =0 𝑥−2=0 𝑥+2=0 +2 +2 −2 −2 𝑓. 𝑥 2 +4𝑥−5=0 𝑥=2 𝑥=−2 𝑥−1 𝑥+5 =0 𝑥−1=0 𝑥+5=0 +1 +1 −5 −5 𝑥=1 𝑥=−5

What are the factors of −𝟖 that also combine to −𝟐? # 57 𝑥 +2 FACTOR 𝑎. 3 𝑥 2 −6𝑥−24 𝑥 𝑥 2 +2𝑥 3( 𝑥 2 −2𝑥−8) −4 −4𝑥 −8 𝟑(𝒙+𝟐)(𝒙−𝟒) −8 𝑥 2 What are the factors of −𝟖 that also combine to −𝟐?

This is difference of squares. # 57 Notice that EVERY term is a perfect square and the operation is subtraction. 𝑏. 4 𝑥 2 −9 This is difference of squares. (𝟐𝒙−𝟑)(𝟐𝒙+𝟑)

What are the factors of 𝟏𝟒 that also combine to 𝟗? # 57 7𝑥 +2 FACTOR 𝑐. 7 𝑥 2 +9𝑥+2 𝑥 7𝑥 2 +2𝑥 +1 7𝑥 +2 (𝟕𝒙+𝟐)(𝒙+𝟏) +14 𝑥 2 What are the factors of 𝟏𝟒 that also combine to 𝟗?