Quadratics Alg1 Sem2 College Alg Sem1

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.1

x can be any variable say y, or z, whatever you choose its your option
1 -> x⁰ 0th degree horizontic x¹² -> 12th degree duodecic
x -> x¹ 1st degree, liniaric x¹³ -> 13th degree tredecic
x² -> 2nd degree, squaric x¹⁴ -> 14th degree quadecic
x³ -> 3rd degree, cubic x¹⁵ -> 15th degree quindecic
x⁴ -> 4th degree, quartic x¹⁶ -> 16th degree sesdecic
x⁵ -> 5th degree, quintic x¹⁷ -> 17th degree septidecic
x⁶ -> 6th degree, sestic x¹⁸ -> 18th degree octodecic
x⁷ -> 7th degree, septic x¹⁹ -> 19th degree novemdecic
x⁸ -> 8th degree, octic x²⁰ -> 20th degree vigintic
x⁹ -> 9th degree, nonic x²¹ -> 21st degree, unvigintic
x¹⁰ -> 10th degree, decic x²² -> 22nd degree, duovigintic
x¹¹ -> 11th degree undecic x²³ -> 23rd degree, trevigintic

Multiple Choice

1. 1. b(n)=n¹⁷ what degree is it?

• 17th degree, quindecic

• 17th degree, septidecic

• 17th degree, sesdecic

• 16th degree, sesdecic

Multiple Choice

2. 2. k(b)=b¹⁸ what degree is it?

• 18th degree, quindecic

• 18th degree, duodecic

• 18th degree,duovigintic

• 18th degree, octodecic

Multiple Choice

3. 3. g(h)=h⁷ what degree is it?

• 7th degree, septic

• 7th degree, octic

• 7th degree, nonic

• 7th degree, decillic

Multiple Choice

4. 4. q(w)=w¹¹ what degree is it?

• 21th degree, unvigintic

• 21th degree, undecic

• 11th degree, unvigintic

• 11th degree, undecic

Multiple Choice

5. 5. y(x)=x⁸ what degree is it?

• 8th degree decic

• 8th degree octic

• 10th degree octic

• 10th degree decic

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.2

​lets study more of these examples i know this is a different type of math
say we have xⁿ and we want to express something right heres something to express we know we let n be a power base on x so we know when
n=0 its horizontic, n=9 its nonic,
n=1 its linearic, n=10 its decic, and so on but what if we deal with higher order sa
n=2 its squaric, n=20 its vigintic, n=90 its nonagintic,
n=3 its cubic, n=30 its trigintic, n=100 its centic,
n=4 its quartic, n=40 its quadragintic, what if n=97 we can rewrite it as x⁹⁷ since n is
n=5 its quintic, n=50 its quinquagintic, x power we will write since it at 90s place
n=6 its sestic, n=60 its sesagintic, we will add nonagintic when we subtract that
n=7 its septic, n=70 its septuagintic, number by 90 so 97-90=7 since we get 7
n=8 its octic, n=80 its octuagintic, we get septic we remove tic on the side move it on the left we get sepnonagintic i want you to try the rest of these excerises on your own and see how far you get good luck

Multiple Choice

6. 6. w(r)=r⁶⁷ what degree is it?

• 27th degree, sesquadrigintic

• 27th degree, septiquadrigintic

• 27th degree, septitrigintic

• 27th degree, septivigintic

Multiple Choice

7. 7. d(t)=t⁴ what degree is it?

• 44th degree, untrigintic

• 44th degree, quaquadrigintic

• 44th degree, sestrigintic

• 44th degree, septiquadrigintic

Multiple Choice

8. CHALLANGE 8. h(j)=j³⁵ what degree is it?

• 35th degree, quinvigintic

• 35th degree,quintrigintic

• 35th degree, septitrigintic

• 35th degree, septiquadrigintic

Multiple Choice

9. 9. f(a)=a¹⁰⁰ what degree is it?

• 39th degree, trigintic

• 39th degree, novemtrigintic

• 39th degree, quinquadrigintic

• 39th degree, septiquadrigintic

Multiple Choice

10. 10. s(y)=y⁴¹ what degree is it?

• 41th degree, unquintrigintic

• 41th degree, unquadrigintic

• 41th degree, quaquadrigintic

• 41th degree, novemvigintic

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.3

​x -> 1 term -> monomial, 14 term's -> quadecinomial
​x+y -> 2 term's -> binomial, 15 term's -> quindecinomial
​x+y+z -> 3 term's -> trinomial, 16 term's -> sesdecinomial
​x+y+z+a -> 4 term's -> quadrinomial, 27 term's -> septenvigintinomial
5 term's -> quintinomial, 18 term's -> octodecinomial | what ever is bold its an
6 term's -> sestinomial, 19 term's -> novemdecinomial | polynomial
7 term's -> septinomial, 20 term's -> vigintinomial
8 term's -> octinomial, 21 term's -> unvigintinomial
9 term's -> nonomial, 22 term's -> duovigintinomial
10 term's -> decinomial, 23 term's -> trevigintinomial
11 term's -> undecinomial, 24 term's -> quavigintinomial
12 term's -> duodecinomial, 25 term's -> quinvigintinomial
13 term's -> tredecinomial, 26 term's -> sesvigintinomial

Multiple Choice

11. 11. classify x¹⁰+x⁵+1

• decic trinomial

• decic trinomial

• cubic decinomial

• cubic trinomial

Multiple Choice

12. 12. classify x¹⁰+x⁸+x⁶+x⁴+x²+1

• decic decinomial

• decic sestinomial

• decic trinomial

• decic undecinomial

Multiple Choice

13. 13. classify if there are 17 terms and its highest 90th degree

• nonagintic novemdecinomial

• septuagintic novemdecinomial

• nonagintic septendecinomial

• septuagintic septendecinomial

Multiple Choice

14. 14. classify if there are 21 terms and its highest 45th degree

• unvigintic quadecinomial

• quinquadragintic quadecinomial

• quinquadragintic unvigintinomial

• unvigintic unvigintinomial

Multiple Choice

14. 15. classify if there are 16 terms and its highest 37th degree

• septemtrigintic sesdecinomial

• Tretrigintic sesdecinomial

• Tretrigintic tredecinomial

• septemtrigintic tredecinomial

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.4

​a -> 1 term -> monomial, 50 term's -> quinquagintinomial
​a+b -> 2 term's -> binomial, 60 term's -> sesagintinomial
​a+b+c -> 3 term's -> trinomial, 70 term's -> septuagintinomial
4 term's -> quadranomial, 80 term's -> octuagintinomial
5 term's -> quintinomial, 90 term's -> nonagintinomial
6 term's -> sestinomial, 100 term's -> centinomial
7 term's -> septinomial, note when its bold its polynomial, when it has 78 terms we
8 term's -> octinomial, can do the math so since its tens place is 7 its at
9 term's -> noninomial, septuagintinomial then its ones place is 8 its octinomial
10 term's -> decinomial, we put the lower term num variable on the left we get
20 term's -> vigintinomial, octoseptuagintinomial
30 term's -> trigintinomial
40 term's -> quadrigintinomial
then we

Multiple Choice

16. 16. classify if there are 63 terms and its highest 78th degree

• octoseptugintinic octoseptuagintinomial

• tresesagintinic octoseptuagintinomial

• tresesagintinic tresesagintinomial

• octoseptugintinic tresesagintinomial

Multiple Choice

17. 17. classify if there are 89 terms and its highest 100th degree

• novemoctuagintic novemoctuaginomial

• novemoctuagintic centinimial

• centic centinimial

• centic novemoctuaginomial

Multiple Choice

18. 18. classify if there are 63 terms and its highest 84th degree

• quaoctuagintinomial quaoctuagintic

• tresesagintinomial quaoctuagintic

• tresesagintinomial tresesagintic

• quaoctuagintinomial tresesagintic

Multiple Choice

19. 19. classify if there are 100 terms and its highest 100th degree

• centic centinomial

• centic decinomial

• decic decinomial

• decic centinomial

Multiple Choice

20. 20. classify if there are 88 terms and its highest 99th degree

• novemnonagintinic novemnonagintinomial

• octooctuagintinic novemnonagintinomial

• octooctuagintinic octooctuagintinomial

• novemnonagintinic octooctuagintinomial

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.5

​lets classify some polynomials f(x)=5x⁵+9x⁴+20x³+37x²+74x+74 we can find the leading coefficient from this sestinomial for o=the leading coefficient we must look for the degree which in that case is x⁵ we see 5 proof f(x)=5x⁵+9x⁴+20x³+37x²+74x+74 is been multiplied so we can conclude our leading coefficient is 5 to find the constant we must look for something that does not have the variable respected to if f(x)=x no constant but if its different variable then it is an constant we can proove it f(x)=5x⁵+9x⁴+20x³+37x²+74x+74 we can see 74 is the constant so our constant is 74, now that is it with today lecture now i want you to try the rest of these excersizes on your own and see how far you get.

Multiple Choice

21. 21. describe 8x³+7x⁵+6x³+5x²+4+3x

• leading coefficient 7

  constant 3

• leading coefficient 7

  constant 4

• leading coefficient 8

  constant 3

• leading coefficient 8

  constant 6

Multiple Choice

22. 22. describe 6j⁸+5j⁵+6j³+7j²+2j⁷+3j⁶+6j+2j⁴

• leading coefficient 8

  constant 4

• leading coefficient 8

  constant 0

• leading coefficient 6

  constant 0

• leading coefficient 6

  constant 2

Multiple Choice

23. 23. describe 7y⁵+4y⁴+6y⁷+8y⁸+24+2y²

• leading coefficient 8

  constant 24

• leading coefficient 7

  constant 2

• leading coefficient 8

  constant 0

• leading coefficient 8

  constant 4

Multiple Choice

24. 24. describe 10+9p+8p²+7p⁷+6p⁴+5p⁵+4p⁶+3p³+2p⁸+p⁹+p¹⁰/2

• leading coefficient 1

  constant 10

• leading coefficient 1/2

  constant 10

• leading coefficient 10

  constant 0

• leading coefficient 2

  constant 10

Multiple Choice

25. 25. describe o⁵+o⁴+o³+o²

• leading coefficient 1

  constant 0

• leading coefficient 0

  constant 0

• leading coefficient 1

  constant 1

• leading coefficient 0

  constant 1

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.6

​you might be familliar with combining like terms but heres another one we like tos how a ploblem 2x²+x+x²+6+1+7x+8x² we want to group the variables by power we get 2x²+x+x²+6+1+7x+8x²=8x²+x²+2x²+7x+x+6+1 add them up combine like terms 8x²+x²+2x²+7x+x+6+1=11x²+8x+7 lets try another one this time dealing with negative constants 3x²+4x²+7-8-42x²+50x+50x²+100-25x again rearrange variables by power we get 3x²+4x²+7-8-42x²+50x+50x²+100-25x=3x²+4x²-42x²+50x²+50x-25x+7-8+100 then combine like terms 3x²+4x²-42x²+50x²+50x-25x+7-8+100=15x²+25x+99 lets try another one 7x+32x²+55x-28x²+33x²-100x²-100x+45 again as usual rearrange variables by power we get
7x+32x²+55x-28x²+33x²-100x²-100x+45=32x²-28x²+33x²-100x²+7x+55x-100x+45
=-63x²-38x+45 now i want you to try the rest of the excersize on you own and see how far you get

Multiple Choice

21. CHALLANGE 26. 6f³+43f+6f²-24f-78+13f²-4f³+65

• 21. 63f-13

• 21. 63f

• 49f³-18f²-65f+61

• 21. 2f³+19f²+19f-13

Multiple Choice

22. 27. 5w+6w²-(14+7w)

• 6x²-2x+14

• 6x²+2x+14

• 6x²-2x-14

• 6x²+2x-14

Multiple Choice

23. 28. 7+7t²-(t²+5t+6)

• 6t²-5t+1

• 8t²+5t+13

• 8t²-5t+1

• 6t²+5t+13

Multiple Choice

24. 29. 8+7c²+6c-(c²-7c+5)+8c²+5c+7

• 19x²+8x+15

• 24. 14x²+18x+10

• 7x²+9x+5

• 15x²+19x+8

Multiple Choice

25. CHALLANGE 30. h+√h⁵-√h³+√(36h⁵)+h√4+√(49h³)-√(36h²)

• x+√(13x³)-√(40x⁵)

• 7√x⁵+6√x³-3x

• √(37x⁵)+√(48x³)+x√5-√(36x²)

• x+√(13x³)+√(40x⁵)

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.7

​lets try one example multiplying monomials say i get 6x⁵(5x⁴) first we multiply the constant then do product rule 6x⁵(5x⁴)=30x⁵(x⁴)=30x⁵⁺⁴=30x⁹ lets try the last one this time dealing with roots 5³√x⁵(6*⁵√x²+4*¹¹√x¹⁰-7*⁵√x⁶) well first thing first is to seperate them so we get 5³√x⁵(6*⁵√x²+4*¹¹√x¹⁰-7*⁵√x⁶)=5³√x⁵(6*⁵√x²)
+5³√x⁵(4*¹¹√x¹⁰)+5³√x⁵(-7*⁵√x⁶)=30³√x⁵*⁵√x²+20³√x⁵*¹¹√x¹⁰-35³√x⁵*⁵√x⁶
convert roots to exponitional fractions 30³√x⁵*⁵√x²+20³√x⁵*¹¹√x¹⁰-35³√x⁵*⁵√x⁶
=30x^5/3x^2/5+20x^5/3x^10/11-35x^5/3x^6/5 then use product rule 30x^5/3x^2/5+20x^5/3x^10/11
-35x^5/3x^6/5=30x^5/3+2/5+20x^5/3+10/11-35x^5/3+6/5 then do add both fractions 3 of them 5/3+2/5=25/15+6/15=31/15 5/3+10/11=55/33+30/33=85/33 5/3+6/5
=25/15+18/15=43/15 so 30x^5/3+2/5+20x^5/3+10/11-35x^5/3+6/5=30x^31/15+20x^85/33-35x^43/15 now convert them back to root 30x^31/15+20x^85/33-35x^43/15
=30(¹⁵√x³¹)+20(³³√x⁸⁵)-35(¹⁵√x⁴³) now i want you to try the rest of these excerises on your own good luck. 

Multiple Choice

26. 31. d²(d³+5)

• d⁵+5d²

• 26. d³+5

• d⁵+5d³

• 26. d⁴+5d²

Multiple Choice

27. CHALLANGE 32. 7*⁷√s³(5*²⁰√s¹⁷+14*¹⁷√s¹⁵+68*⁷√(s√s¹³))

• 27. 5*¹⁴⁰√s¹¹⁹+14*¹¹⁹√s¹⁵⁶+68√s³

• 27. 35*¹⁴⁰√s¹¹⁹+98*¹¹⁹√s¹⁵⁶+476√s³

• 27. 35*²⁰√s¹¹⁹+98*¹⁷√s¹⁵⁶+476⁷√s³

• 27. 5*²⁰√s¹¹⁹+14*¹⁷√s¹⁵⁶+68⁷√s³

Multiple Choice

28. 33. w(w+2)

• w+2

• w²

• w

• w²+2w

Multiple Choice

29. 34. 6y⁵(3y²+5y⁷+4y⁵+2y³)

• 29. 3y²+5y⁷+4y⁵+2y³

• 3y¹²+5y¹⁰+4y⁸+2y⁷

• 30y¹²+24y¹⁰+12y⁸+18y⁷

• 30y⁷+24y⁵+12y³+18y²

Multiple Choice

30. CHALLANGE 35. 45a⁴⁵²(23a³⁵⁷+5a²⁴⁴+30a¹⁷⁰+50a¹⁰⁴+100a⁵⁸)

• 1,035a⁸⁰⁹+225a⁶⁹⁶+1,350a⁶²²+2,250a⁵⁵⁶

  +4,500a⁵¹⁰

• 30. 23a⁸⁰⁹+5a⁶⁹⁶+30a⁶²²+50a⁵⁵⁶+100a⁵¹⁰

• 23a^1,009+5a⁸⁹⁶+30a⁸²²+50a⁷⁵⁶

  +100a⁶¹⁰

• 1,035a^1,009+225a⁸⁹⁶+1,350a⁸²²+2,250a⁷⁵⁶

  +4,500a⁶¹⁰

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.8

​say i got an 2 binomials (x+5)(x+8) and we want to expand this expression so lets get to it so first thing first is seperate on the left side we get (x+5)(x+8)=x(x+8)+5(x+8) then separate again x(x+8)+5(x+8)=x(x)+x(8)+5(x)+5(8)
=x²+8x+5x+40=x²+13x+40 lets try another one this time dealing with binomial and trinomial (x²+5)(x²+9x+8) lets seperate on the left side we get (x²+5)(x²+9x+8)
=x²(x²+9x+8)+5(x²+9x+8) then seperate on the right side x²(x²+9x+8)
+5(x²+9x+8)=x²(x²)+x²(9x)+x²(8)+5(x²)+5(9x)+5(8)=x⁴+9x³+8x²+5x²+45x+40
=x⁴+9x³+13x²+45x+40 lets try another one (x+5)(x²+2x+7) first we must seperate on the left (x+5)(x²+2x+7)=x(x²+2x+7)+5(x²+2x+7) now seperate on the right side x(x²+2x+7)+5(x²+2x+7)=x(x²)+x(2x)+x(7)+5(x²)+5(2x)+5(7)=x³+2x²+7x+5x²+10x+35
=x³+7x²+17x+35 now i want you to try the rest of these problems and see how far you get good luck.

Multiple Choice

36. 36. (a+b)(a-b)

• a²+2ab+b²

• a²+b²

• a²-2ab+b²

• a²-b²

Multiple Choice

36. 37. (u+v)²

• u²+2uv+v²

• 36. u²+v²

• u²-2uv+v²

• u²-v²

Multiple Choice

36. CHALLANGE 38. (c+1)³

• c³+3c²+6c+1

• c³+3c²+3c+1

• c²+2c+1

• c³+5c+1

Multiple Choice

36. CHALLANGE 39. (5*³√x²+4*⁴√x³+9*⁹√x⁷)(6*⁵√x³+7*⁶√x⁵+7*⁶√x⁷)

• 60*²⁵√x³⁴+70√x⁴+70*⁷√x¹⁴+48*³⁰√x³⁷

  +56*⁶√x¹⁹+56*⁶√x²³+108*⁴⁵√x¹²⁴+126*¹⁸√x¹⁰⁹

  +126*⁹√x³⁵

• 30*¹⁹√x¹⁵+35³√x²+35*¹¹√x⁶+24*²⁷√x²⁰

  +28*¹⁹√x¹²+28*²³√x¹²+54*⁶²√x⁴⁵+63*²⁹√x¹⁸

  +63*³⁵√x¹⁸

• 30*⁵√x¹⁹+35x³+35*²√x¹¹+24*⁶√x²⁷

  +28*³√x¹⁹+28*³√x²³+54*⁹√x⁶²+63*³√x²⁹

  +63*⁸√x³⁵

• 30*¹⁵√x¹⁹+35√x³+35*⁶√x¹¹+24*²⁰√x²⁷

  +28*¹²√x¹⁹+28*¹²√x²³+54*⁴⁵√x⁶²+63*¹⁸√x²⁹

  +63*¹⁸√x³⁵

Multiple Choice

36. 40. (x-y)²

• 36. x²+2xy+y²

• 36. x²+y²

• x²-2xy+y²

• x²-y²

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.9

say i got an equation (x²+2x)/(3x) and we want to divide it so first i is to seperate them we get (x²+2x)/(3x)=x²/(3x)+2x/(3x) now simplfy the coefficients x²/(3x)+2x/(3x)=x²/(3x)+2x/(3x) next we divide the x we get x²/(3x)+2x/(3x)
=x²x^-1/3+2x¹x^-1/3=x^2-1/3+2x^1-1/3=x/3+2x⁰/3=x/3+2/3=(x+2)/3 since the denominator is the same we put them together lets try another one this time dealing with trinomial (3x³+9x²+9x+7)/(3x) first we have to seperate the coefficients we get (3x³+9x²+9x+7)/(3x)=3x³/(3x)+9x²/(3x)+9x/(3x)+7/(3x) then simplify the coefficients we get 3x³/(3x)+9x²/(3x)+9x/(3x)+7/(3x)=x³/x+3x²/x+3x/x+7/(3x) then we divide the x we get x³/x+3x²/x+3x/x+7/(3x)=x³/x¹+3x²/x¹+3x¹/x¹+7/(3x¹)=x³x^-1+3x²x^-1+3x¹x^-1+7x^-1/3
=x^3-1+3x^2-1+3x^1-1+7x^-1/3=x²+3x+3x⁰+7/(3x)=x²+3x+3+7/(3x) lets try the last one now i want you to try the rest of the excersizes on your own and see how far you get

Multiple Choice

41. 41. (5w⁴+20w³+60w²+120w+120)/(5w²)

• w³+4w²+12w

• w²+4w+12

• w²+4w+12+24/w+24/w²

• w⁴+4w³+12w²+24w+24

Multiple Choice

42. CHALLANGE 42. (5*³√k²-7*⁷√k³)/(8*⁷√k⁵)

• (²¹√k-*⁷√k²)/8

• (5*²¹√k-7*⁷√k²)/8

• 5/(8*²¹√k)-7/(8*⁷√k²)

• (²¹√k+*⁷√k²)/8

Multiple Choice

43. 43. (y²+1)/y

• y+1

• y²+1/y

• y+1/y

• y²+1

Multiple Choice

44. CHALLANGE 44. ²⁵√(425²⁵b⁶⁰⁰+625²⁵b⁶²⁵)/(25b²²)

• 44. 25b⁶⁰³+17b⁵⁷⁸

• 44. 425²⁵b⁵⁷⁸+625²⁵b⁶⁰³/25

• 44. 25b⁵⁷⁸+17b⁶⁰³

• 44. 25b³+17b²

Multiple Choice

45. 45. √(n²⁴+n¹²+n⁶)/⁴√n⁸

• n¹⁶+n⁴+1/n²

• n⁸+n²+1/x

• n⁴+1/n²+1/n⁵

• n¹⁰+n⁴+n

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.10

​lets say i got a trinomial divided by binomial example (x²+438x+437)/(x+1) now lets work this one together we seperate the numberator which will look something like this first off we divide x² by the highest degree of denominator on x x²/x=x now we can use that x and multiply the numerator we get x(x+1)=x²+x now we can subtract the numerator by x²+x we get (x²+438x+437-x²-x)/(x+1)=(437x+437)/(x+1) save the x for later for now lets divide 437x by x so 437x/x=437 then mutliply it by x+1 437(x+1)=437x+437 we can subtract again (437x+437-437x-437)/(x+1)=0/x+1
now we add those parts together x and 437 we get x+437 so in conclusion (x²+438x+437)/(x+1)=x+437 now i want you to try the rest of the problems on your own and see how for you get.

Multiple Choice

46. CHALLANGE 46. (10s³+7s²+13s-20)/(2s²+3s+5)

• 5s-4

• 5s+4

• 5s-4+(9-7s)/(2s²+3s+5)

• 5s-4+(9+7s)/(2s²+3s+5)

Multiple Choice

47. 47. (s²-4)/(s+2)

• s-2

• s+2

• s-4

• s+4

Multiple Choice

48. CHALLANGE 48. (d⁵+5d⁴+10d³+10d²+5d+1)/(d³+3d²+3d+1)

• d²+3d+3

• d²+3d+1

• d²+1

• d²+2d+1

Multiple Choice

49. 49. (f²+6f+9)/(f+3)

• f+3

• f+2

• f+1

• f

Multiple Choice

50. CHALLANGE 50. (x⁴+3x⁴y²+6x³y²+2x³y+6x²y²+2xy³+3x²y⁴+6xy⁴+y⁴)/(x+y)²

• 50. x²+3x²y²+3xy²+y²

• 50. x²+3x²y²+6xy²+y²

• 50. x²+6x²y²+3xy²+y²

• 50. x²+6x²y²+6xy²+y²

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.11

​lets say i want to factor 50x⁴+20x³+10x² so what we can do is to find the lowest term in this equation which is x² so we separate it we get 50x⁴+20x³+10x²
=x²(50x²+20x+10) now find the lowest coefficient and see if others are factorable if not reduce it until you get the exact one x²(50x²+20x+10)=x²(10*5x²+10*2x+10*1)
=10x²(5x²+2x+1) lets try another one 15x²⁰+6x⁸+30x¹⁸+21x²+54x¹⁶+18x⁴+9x¹² so what we can do is to again find the lowest term which is x² once again so 15x²⁰+6x⁸+30x¹⁸
+21x²+54x¹⁶+18x⁴+9x¹²=x²(15x¹⁸+6x⁶+30x¹⁶+21+54x¹⁴+18x²+9x¹⁰) now find the lowest coefficient which is 6 lets see if all is the factor of 6 x²(15x¹⁸+6x⁶+30x¹⁶+21+54x¹⁴+18x²
+9x¹⁰) sems like 15 21 and 9 is not the factor of 6 so we 6/2 3 much better now we can move 3 on left side we get x²(15x¹⁸+6x⁶+30x¹⁶+21+54x¹⁴+18x²+9x¹⁰)=3x²(5x¹⁸+2x⁶
+10x¹⁶+7+18x¹⁴+6x²+3x¹⁰) lets arrange it from from highest degree to lowest degree we get 3x²(5x¹⁸+2x⁶+10x¹⁶+7+18x¹⁴+6x²+3x¹⁰)=3x²(5x¹⁸+10x¹⁶+18x¹⁴+3x¹⁰+2x⁶+6x²+7)
now i want you to try the rest of the exercises on your own and see how far you get

Multiple Choice

51. 51. 8h²+4h

• 4h²(2h²+1)

• 4h(2h²+1)

• 4h(2h+1)

• 4h²(2h+1)

Multiple Choice

52. 52. 49w⁴+98w³+35w²

• 49(w²+2w+5)

• 7(7w²+14w+5)

• 49w²(w²+2w+5)

• 7w²(7w²+14w+5)

Multiple Choice

53. CHALLANGE 53. 4b^1,000+36b⁸⁹⁵+96b⁷³²+100b⁵⁹⁵+732b⁴²⁰+896b²⁶⁷+1,000b¹⁰¹

• 4b¹⁰¹(b⁸⁹⁹+9b⁷⁹⁴+24b⁶³¹+25b⁴⁹⁴+183b³¹⁹

  +224b¹⁶⁶+250)

• 4b^1,000(b⁸⁹⁹+9b⁷⁹⁴+24b⁶³¹+25b⁴⁹⁴+183b³¹⁹

  +224b¹⁶⁶+250)

• 4b¹⁰⁰(b⁸⁹⁹+9b⁷⁹⁴+24b⁶³¹+25b⁴⁹⁴+183b³¹⁹

  +224b¹⁶⁶+250)

• 4b¹³⁹(b⁸⁹⁹+9b⁷⁹⁴+24b⁶³¹+25b⁴⁹⁴+183b³¹⁹

  +224b¹⁶⁶+250)

Multiple Choice

54. CHALLANGE 54. 54*³√o⁵+81*√o³+27*⁷√o²

• 27*⁷√o²(2*²¹√o²⁰+3*¹⁴√o¹¹)

• 27*⁷√o²(2*²¹√o²⁷+3*¹⁴√o¹⁶)

• 27*⁷√o²(2*²¹√o²⁹+3*¹⁴√o¹⁷)

• 27*⁷√o²(2*²¹√o²⁹+3*¹⁴√o¹⁷+1)

Multiple Choice

55. 55. 70a³⁰+80x²⁵+90x²¹+40x¹⁷+60x¹³+100x¹⁰

• 10a¹⁰(7a²+8√a³+9*¹⁰√a¹¹+4*¹⁰√a⁷+6*¹⁰√a³

  +10)

• 10a¹⁰(7a³+8√a⁵+9*¹⁰√a²¹+4*¹⁰√a¹⁷

  +6*¹⁰√a¹³

  +10a)

• 10a¹⁰(7a²⁰+8a¹⁵+9a¹¹+4a⁷+6a³+10)

• 10a¹⁰(7a²⁰+8a¹⁵+9a¹¹+4a⁷+6a³+1)

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.12

​lets i got d²+5d+6=0 and we want to find the quadratic formula well for ax²+bx+c=0 we use x=-(b±√(b²-4ac))/2a so lets plug those values in d²+5d+6=0 => 1*d²+5d+6=0 a=1, b=5, c=6 now lets plug that into formula x=-(b±√(b²-4ac))/2a=-(5±√(5²-4*1*6))/(2*1)=-(5±√(25-24))/2=-(5±√1)/2=-(5±1)/2 so now lets separate x=-(5+1)/2=-6/2=-3 and x=-(5-1)/2=-4/2=-2 for u=-(b+√(b²-4ac))/2a, and
v=-(b-√(b²-4ac))/2a lets say I want to factor it well we use the formula for ax²+bx+c
a(x-u)(x-v) if there is something that is not whole all we can do for example u is not whole all we get is a(x-u)(x-v)=(ax-au)(x-v) let's try that let u=-2, v=-3 we can plug those in (d-(-2))(d-(-3))=(d+2)(d+3) so, in conclusion, d²+5d+6=(d+2)(d+3) now I want you to try the rest of the problems on your own and see how far you get.

Multiple Choice

56. 56. p²-49

• p(p+14)

• p(p+7)

• (p+14)(p-14)

• (p+7)(p-7)

Multiple Choice

57. 57. 4c²-36c+72

• (c-9)(c-2)

• 4(c-6)(c-3)

• (c-6)(c-3)

• 5\4(c-9)(c-2)

Multiple Choice

58. 58. 3h²+h-14

• (3h-7)(h+2)

• 3(h-7)(h+2)

• 3(h+7)(h-2)

• (3h+7)(h-2)

Multiple Choice

59. CHALLANGE 59. 10c⁵⁰⁰-29c²⁵⁰-21

• 59. (c¹²⁵+√(7/2))²(c¹²⁵+√(3/5))²

• 59. (c¹²⁵+√(7/2))(c¹²⁵-√(7/2))(c¹²⁵+√(3/5))(c¹²⁵-√(3/5))

• 59. (c¹²⁵+√(7/2))²(c²⁵⁰+3/5)

• 59. (c¹²⁵+√(7/2))(c¹²⁵-√(7/2))(c²⁵⁰+3/5)

Multiple Choice

60. CHALLANGE 60. f^1,000+2f⁵⁰⁰+1

• 60. (f⁵⁰⁰-1)²

• 60. (f²⁵⁰+1)⁴

• 60. (f²⁵⁰+1)²(f²⁵⁰-1)²

• 60. (f⁵⁰⁰+1)²

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.13

​say i got a polynomial 48xy+24y+24x+12 and we want to factor such a polynomial so first is the factor the left side we get 48xy+24y+24x+12=24y(2x+1)+24x+12 now facotr the right side we get 24y(2x+1)+24x+12=24y(2x+1)+12(2x+1) seems promising now we can construct factors we get 24y(2x+1)+12(2x+1)=(24y+12)(2x+1) continiue factoring we get (24y+12)(2x+1)=12(2y+1)(2x+1) now lets try the another one 30x³+35x²+12x+14 again to factor this polynomial we factor the left hand side first we get 30x³+35x²+12x+14=5x²(6x+7)+12x+14 now factor the right side 5x²(6x+7)+12x+14
=5x²(6x+7)+2(6x+7) now construct factors we get 5x²(6x+7)+2(6x+7)=(5x²+2)(6x+7) lets try last one x⁶+2x⁴+x² first factor out the lowest term which is x² we get x⁶+2x⁴+x²=x²(x⁴+2x²+1) seems promising we can substitute with u so let u=x² in we get x²(x⁴+2x²+1)=u(u²+2u+1)=u(u+1)² bring back the u=x² we get u(u+1)²=x²(x²+1)² so i want you to try the rest of the excerises on your own and see how far you get note you do not have to use u sub on outer shell if you dont have to.

Multiple Choice

61. 61. x¹⁵+5x¹⁰+10x⁵+50

• 61. (x¹⁰+10)(x⁵+5)

• 61. (x²+5)(x+10)

• 61. (x¹⁰+5)(x⁵+10)

• 61. (x²+10)(x+5)

Multiple Choice

62. CHALLANGE 62. 12*⁵⁶√h¹⁰³+6*⁴⁰√h⁶¹+18*⁸√h⁹+24*⁸⁴√h⁹⁵+12*⁶⁰√h⁴⁹+36*⁷√h⁵+36*¹²√h⁵+18*⁵√h²+54

• 62. (6*⁷√h⁵+3*⁵√h²+27)(⁸√h⁹+2*¹²√h⁵+2)

• 62. (6*⁷√h⁵+3*⁵√h²+54)(⁸√h⁹+2*¹²√h⁵+1)

• 62. (12*⁷√h⁵+6*⁵√h²+18)(⁸√h⁹+2*¹²√h⁵+3)

• 62. (6*⁷√h⁵+3*⁵√h²+9)(⁸√h⁹+2*¹²√h⁵+6)

Multiple Choice

63. 63. s⁷+2s⁵+s³

• (s⁷+s⁵)(s³+1)

• (s⁶+s⁴)(s+1)

• (s⁶+2s⁴)(s+1)

• (s⁵+s³)(s²+1)

Multiple Choice

64. 64. t¹⁵+2t¹⁰+t⁵

• 64. (t¹⁰+1)(t⁵+1)

• 64. (t¹⁰+t⁵)(t⁵+1)

• 64. (t¹⁰+t⁷)(t⁵+1)

• 64. (t¹⁰+t⁵)(t³+1)

Multiple Choice

65. 65. w²⁵+w²²+w¹⁵+w¹²

• 65. (w²⁵+w¹⁵)(w⁵+w²)

• 65. (w²⁰+w¹⁰)(w⁵+w²)

• 65. (w²⁰+w¹⁰)(w⁵+w³)

• 65. (w²⁵+w¹⁵)(w⁵+w³)

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.14

​say i got a trinomial f(x)=4x³+6x²-2x and we want to find the all values btwn x=-3 and x=3 lets do this starting with -3 so f(-3)=4(-3)³+6(-3)²-2(-3)=4(-27)+6(9)+6
=-108+54+6=-48 so f(-3)=-48 try f(-2)=4(-2)³+6(-2)²-2(-2)=4(-8)+6(4)+4=-32+24+4=-4 so f(-2)=-4, try f(-1)=4(-1)³+6(-1)²-2(-1)=4(-1)+6(1)+2=-4+6+2=4 so f(-1)=4 try f(0)=4(0)³+6(0)²-2(0) hers the lession about 0 when n>0 then 0ⁿ=0 , when n=0 then 0ⁿ=1, when n<0 0ⁿ=∞, so 4(0)³+6(0)²-2(0)=4(0)+6(0)-0=0+0-0=0 try f(1)=4(1)³+6(1)²-2(1)=4(1)+6(1)-2(1)=4+6-2=8 try f(2)=4(2)³+6(2)²-2(2)=4(8)+6(4)-4=32+24-4=52 try f(3)=4(3)³+6(3)²-2(3)=4(27)+6(9)-6=108+54-6=156 so we can complete the table in x,f(x) form -3,-48 -2,-4 -1,4 0,0 1,8 2,52 3,156 now i want you to try the rest of the excersizes on your own and see how far you get

Multiple Choice

66. 66. t(s)=s²+2s+1 find the value of t(s) when s is btwn -5,5 in s,t form

• s,t

  -5,16

  -4,9

  -3,4

  -2,1

  -1,0

  0,1

  1,4

  2. 2,9

  3. 3,16

  4. 4,25

  5. 5,36

• s,t

  -5,-4

  -4,-3

  -3,-2

  -2,-1

  -1,0

  0,1

  1,2

  2. 2,3

  3. 3,4

  4. 4,5

  5. 5,6

• s,t

  -5,25

  -4,16

  -3,9

  -2,4

  -1,1

  0,0

  1,1

  2. 2,4

  3. 3,9

  4. 4,16

  5. 5,25

• s,t

  s,t

  -5,-5

  -4,-4

  -3,-3

  -2,-2

  -1,-1

  0,0

  1,1

  2. 2,2

  3. 3,3

  4. 4,4

  5. 5,5

Multiple Choice

67. 67. f(w)=5w²+25w+30 when w is -btwn 6,-2 in w,f form

• w,f

  -6,12

  -5,6

  -4,2

  -3,0

  -2,0

• w,f

  -6,42

  -5,30

  -4,20

  -3,12

  -2,6

• w,f

  -6,60

  -5,30

  -4,10

  -3,0

  -2,0

• w,f

  -6,210

  -5,150

  -4,100

  -3,60

  -2,30

Multiple Choice

68. 68. d(t)=t⁴+4t³+12t²+24t+24 when t is btwn -3,3 in t,d form

• t,d

  -3,33

  -2,-16

  -1,9

  0,0

  1,24

  2,168

  3,393

• t,d

  -3,33

  -2,16

  -1,9

  0,0

  1,24

  2,168

  3,393

• t,d

  -3,33

  -2,16

  -1,9

  0,24

  1,65

  2,168

  3,393

• t,d

  -3,33

  -2,-16

  -1,9

  0,24

  1,65

  2,168

  3,393

Multiple Choice

69. 69. t(y)=y⁵+5y+9 when y is btwn -2,2 in y,t form

• y,t

  -2,-33

  -1,3

  0,9

  1,15

  2,51

• y,t

  -2,33

  -1,3

  0,9

  1,15

  2,51

• y,t

  -2,32

  -1,3

  0,9

  1,15

  2,51

• y,t

  -2,-32

  -1,3

  0,9

  1,15

  2,51

Multiple Choice

67. 70. r(t)=t²-4 when t is btwn -2,2 in t,r form

• -2,-4

  -1,-3

  0,0

  1,3

  2,4

• -2,-4

  -1,-3

  0,0

  1,-3

  2,-4

• -2,0

  -1,3

  0,4

  1,3

  2,0

• -2,0

  -1,-3

  0,-4

  1,-3

  2,0

​say i got an equation 5x²+2x+1 and we want to find the axis of symmetry where
x=-b/(2a) from ax²+bx+c so lets use that formula x=-2/(2*5)=-2/10=-1/5 lets try another one 4x²+8x+24 lets use x=-b/(2a) again we get x=-8/(2*4)=-8/8=-1 lets try another one 6x²+24x+21 lets use x=-b/(2a) again we get x=-24/(2*6)=-24/12=-2 lets try another one 3x²+24x+48 lets use x=-b/(2a) again we get x=-24/(2*3)=-24/6=-4 lets try another one 7x²+63x+108 lets use x=-b/(2a)=-63/(2*7)=-63/14=-9/2 lets try another one 

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.15

Multiple Choice

71. 71. 6p²+5p+8 find axis of symmetry

• -5/6

• -5/12

• -5/24

• -5/48

Multiple Choice

72. 72. 5j²+5j+5 find axis of symmetry

• 1

• 1/2

• -1/2

• -1

Multiple Choice

73. 73. 2k²+7k+6 find axis of symmetry

• 7/2

• -7/4

• 7/8

• -7/16

Multiple Choice

74. 74. m²-5m+4 find axis of symmetry

• 5/2

• -5/2

• 5

• -5

Multiple Choice

75. 75. 6k²+6k+6 find axis of symmetry

• 1/2

• 1

• -1

• -1/2

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.16

​say i got an equation 5x²+2x+1 and i want to convert that into into vertex form from ax²+bx+c to a(x-h)²+k well lets use a b c and see if we figure it out so first tking first is to seperate c into n+k where c=n+k we get ax²+bx+c=ax²+bx+n+k then factor a b and n by a we get ax²+bx+n+k=a(x²+bx/a+n/a)+k lets solve for h using vertex formula a(x-h)²+k first expand the equation we get a(x-h)²+k=a(x-h)(x-h)+k=a(x²-2xh+h²)+k now set that equal to a(x²+bx/a+n/a)+k we get a(x²-2xh+h²)+k=a(x²+bx/a+n/a)+k we can cancel the terms as highlighed here a(x²-2xh+h²)+k=a(x²+bx/a+n/a)+k => -2xh+h²=bx/a+n/a we can group these in 1st term we get -2xh=bx/a and 0th term we get h²=n/a lets try solving for h tht has only a, b, and c, on 0th term h²=n/a h=√(n/a) oh no theres an n how we gonna solve it any how lets try 1st term -2xh=bx/a solve for h we get h=-b/(2a) thats good now we can solve for n where n/a=h² plug h=-b/(2a) in we get n/a=(-b/(2a))²=b²/(4a²) then solve for n we get n=b²/(4a) plug that into c=n+k we get c=b²/(4a)+k now solve for k we get from c=b²/(4a)+k to k=c-b²/(4a) now we got h=-b/(2a) and k=c-b²/(4a) we can plug them into vertex formula a(x-h)²+k=a(x-(-b/(2a)))²+c-b²/(4a)
=a(x+b/(2a))²+c-b²/(4a) now we got a new formula a(x+b/(2a))²+c-b²/(4a) now we can just plug them in we always need to remember and test this formula if it works or not lets plug 5x²+2x+1 into vertex formula we have a=5, b=2, c=1 since it is ax²+bx+c lets use the vertex formula to plug the values in a(x+b/(2a))²+c-b²/(4a)
=5(x+2/(2*5))²+1-2²/(4*5)=5(x+1/5)²+1-4/(4*5)=5(x+1/5)²+1-1/5=5(x+1/5)²+4/5 take any trinomial that has a highest degree of 2 and see if it works and it actually works make sure to double check as i encourage you to do so try the excersizes on your own and see how far you get good luck.

Multiple Choice

76. 76. find the vertex form of 7c²+14c+7

• 7(c+1)²

• (c+1)²

• 7(c+1)²+7

• (c+1)²+1

Multiple Choice

77. 77. find the vertex form of 4t²+56t+24

• 77. 4(t+7)²-168

• 77. 4(t+7)²-172

• 77. 4(t+7)²-196

• 77. 4(t+7)²-200

Multiple Choice

78. 78. find the vertex form of 2h²+3h+4

• 2(h-3/4)²+23/8

• 2(h-3/2)²+23/2

• 2(h+3/2)²+23/2

• 2(h+3/4)²+23/8

Multiple Choice

79. CHALLANGE 79. find the vertex form of 9o⁴/5-4o²/3+9/8

• 9(o²-10/27)²/5+569/648

• 9(o²+10/27)²/5+569/648

• 9(o-10/27)²/5+569/648

• 9(o+10/27)²/5+569/648

Multiple Choice

80. 80. find the vertex form of 6d²/5+4d/3+7/8

• 80. 6(d+5/9)²/5+109/216

• 80. 6(d+9/5)²/5+178/216

• 80. 6(d+5/9)²/5+178/216

• 80. 6(d+9/5)²/5+109/216

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.17

​say i got the vertex equation 5(x+3)²+10 and we want to find the minimum value of a function for to do that is to use the axis of symmetry which is x=-b/(2a) we can use 2 vertex's and set them equal we get a(x-h)²+k=a(x+b/(2a))²+c-b²/(4a) we know h=-b/(2a) we can plug the values in a=5 -h=3 k=10 right now -h is what we need so lets solve for h we get h=-3 lets plug h into x we get 5(x+3)²+10=5(-3+3)²+10=5(0)²+10=10 we get our minimum value or another way to do that is to let x=h a(x-h)²+k=a(h-h)²+k=a(0)²+k=k this is how you manipulate minimum value of y axis or other words range since the parabola is continious in y axis above k we get k<y<∞ when a>0, if a=0 then y=k, if a<0 then -∞<y<k for all conditions x has all numbers which means -∞<x<∞ you may not see it but it continius on and on and on as you keep going it doesnt stop which means any parabolas is always continious if you have a standard quadratic form use this k=c-b²/a do not use c only use the k formula good luck

Multiple Choice

81. 81. find the domain of y=7x²+10x+5

• -∞<x<∞

• 7<x<10

• 5<x<10

• x>∞

Multiple Choice

82. 82. true or false values in any quadratic and linear equations in domain have al solutions within y=f(x)

• true

• false

Multiple Choice

83. 83. what is the range of ax²+bx+c

• if a>0 then

  y>c-b^2/(4a)

  else if a=0 then

  y=c-b^2/(4a)

  else if a<0 then

  y<c-b^2/(4a)

• if a>0 then

  y=c-b^2/(4a)

  else if a=0 then

  y>c-b^2/(4a)

  else if a<0 then

  y<c-b^2/(4a)

• if a>0 then

  y<c-b^2/(4a)

  else if a=0 then

  y>c-b^2/(4a)

  else if a<0 then

  y=c-b^2/(4a)

• if a>0 then

  y0c-b^2/(4a)

  else if a=0 then

  y<c-b^2/(4a)

  else if a<0 then

  y>c-b^2/(4a)

Multiple Choice

84. 84. what is the range of a(x-h)²+k

• if a>0 then

  y<k

  else if a=0 then

  y>k

  else if a<0 then

  y=k

• if a>0 then

  y=k

  else if a=0 then

  y<k

  else if a<0 then

  y>k

• if a>0 then

  y=k

  else if a=0 then

  y>k

  else if a<0 then

  y<k

• if a>0 then

  y>k

  else if a=0 then

  y=k

  else if a<0 then

  y<k

Multiple Choice

85. 85. find the range y=7x²+10x+5

• y<10/7

• y>10/7

• y<25/7

• y>25/7

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.18

​if i have a equation 6(x-7)²+42 and i want to convert it into a standard quadratic form lets create the formula for it so let a(x-h)²+k=ax²+bx+c lets solve for the left hand side a(x-h)²+k=ax²+bx+c => a(x-h)(x-h)+k=ax²+bx+c =>
a(x²-2xh+h²)+k=ax²+bx+c => ax²-2xah+ah²+k=ax²+bx+c lets use
a(x-h)²+k=ax²-2xah+ah²+k to convert the vertex into standard quadratic form let a=6,h=7, k=42 from 6(x-7)²+42 for a(x-h)²+k for our current function so
ax²-2xah+ah²+k=6x²-2x*6*7+6*7²+42=6x²-84x+294+42=6x²-84x+336 lets try another one 5(x-2)²-15 let 5(x-2)²-15=a(x-h)²+k so a=5 h=2, and k=-15 now use a formula
ax²-2xah+ah²+k=5x²-2x*5*2+5*2²-15=5x²-20x+5*4-15=5x²-20x+20-15=5x²-20x+5 lets try the last one 5(x-5/7)²/2+7/8 set 5(x-5/7)²/2+7/8=a(x-h)²+k so a=5/2, h=5/7, and k=7/8 lets use ax²-2xah+ah²+k lets plug 5(x-5/7)²/2+7/8=a(x-h)²+k them in we get a=5/2, h=5/7, and k=7/8 lets plug them in ax²-2xah+ah²+k=5x²/2-2x5/2*5/7+5/2*(5/7)²
=5x²/2-25/7x+125/98 now i want you to try the rest of the excerises on your own.

Multiple Choice

86. 86. convert 7(h-7)²+7 into standard quadratic form

• 7h²+98h+7

• 7h²-98h+7

• 7h²+98h+350

• 7h²-98h+350

Multiple Choice

87. 87. convert 6(d-5)²+89 into standard quadratic form

• 6d²-60d+240

• 6d²-60d+239

• 6d²+60d+239

• 6d²+60d+240

Multiple Choice

88. 88. convert 8(b-4)²+57 into standard quadratic form

• 8b²+64b+185

• 8b²-64b+185

• 8b²-32b+185

• 8b²+32b+185

Multiple Choice

89. 89. convert 6(s-2)²-20 into standard quadratic form

• 89. 6s²+24s+4

• 89. 6s²-24s-4

• 89. 6s²-24s+4

• 89. 6s²+24s-4

Multiple Choice

90. CHALLANGE 90. convert -2(d⁵+5)²-78 into standard quadratic form

• 90. -2d¹⁰-20d⁵-128

• 90. -2d¹⁰+20d⁵-128

• 90. -2d¹⁰+20d⁵+128

• 90. -2d¹⁰-20d⁵+128

​say i got a equation f(g)=6(g+7)²-100 and i want to transform this equation to h(g) which is h(g)=f(g-2) so what we can do is to use the same interpretation we get from f(g)=6(g+7)²-100 to h(g)=f(g-2)=6((g-2)+7)²-100=6(g+5)²-100 now lets try another one f(n)=8(n-2)²+32 and we want to transform this graph into h(n) which is h(n)=f(n-3) keep using the same interpretation we get f(n)=8(n-2)²+32 to
h(n)=f(n-3)=8((n-3)-2)²+32=8(n-5)²+32 lets try another one this time its standard quadratic form f(j)=4j²+8j+6 first is to convert this standard quadratic form into vertex quadratic form using this formula a(x+b/(2a))²+c-b²/(4a) lets plug the values in a=4, b=8, c=6 since were using j as our variable a(j+b/(2a))²+c-b²/(4a)= 4(j+8/(2*4))²+6-8²/(4*4)=4(j+8/(2*4))²+6-8²/(4*4)=4(j+8/8)²+6-64/16=
4(j+1)²+6-4=4(j+1)²+2 now i want you to try the rest of the excersizes on your own and see how far you get

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.19

Multiple Choice

91. 91. if f(n)=5n³+35n+7, and h(n)=f(n-5) whats h(n)

• 5n²-410n-793

• 5n²+75n²+410n-793

• 5n²+410n-793

• 5n²-75n²+410n-793

Multiple Choice

92. 92. if m(s)=s²-7s+3, and h(s)=m(s-5)+7s+5 whats h(s)

• ms+5m+7s+5

• ms-5m+7s+5

• s²-17s+63

• s²+17s+63

Multiple Choice

93. CHALLANGE 93. f(s)=s⁶+6s+6 h(s)=f(s+5)+s² whats h(s)

• s⁶+30s⁵+375s⁴+2500s³+9376s²+18750s

  +15625

• s⁶+32s⁵+379s⁴+2503s³+9376s²+18754s

  +15628

• s⁶+30s⁵+375s⁴+2500s³+9375s²+18750s

  +15625

• s⁶+30s⁵+375s⁴+2500s³+9376s²+18750s

  +15625

Multiple Choice

94. 94. h(d)=d²-5d+6 and f(d)=h(d+7) whats f(d)

• 94. d²+9d+20

• 94. d²-5d+13

• 94. d²-5d+6

• 94. d²+9d+6

Multiple Choice

95. 95. s(q)=q²+2q+1, c(q)=s²(q)+5s(q)+7 find c(q)

• 95. q⁴+4q³+11q²+14q+13

• 95. q⁴+4q³+6q²+4q+1

• 95. q⁴+4q³+11q²+14q+6

• 95. q⁴+4q³+6q²+4q+6

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.20

​say i got a vertex (0,-4) and a pt (4,0) and we want to find a function so first thing first is to make a function its ok to use a as a factor to a parabola so we will make one for vertex pt (h,k) we can rewrite a function as a(x-h)²+k since we got (0,-4) we get h=0 k=-4 so lets plug the values in we get a(x-h)²+k=a(x-0)²-4=ax²-4 since when x=4 y=0 let y=ax²-4 like you did to all functions so 0=a4²-4 => 0=16a-4 4=16a a=1/4 lets plug a in we get y=ax²-4=x²/4-4 lets try a last one i got a vertex (2,-5) and a pt (10,-1) and we want to find a function so first is to make a function use a to solve any points on vertexes let h=2, and k=-5 right from the vertex point a(x-h)²+k
=a(x-2)²-5 now we have to solve for a using a pt that is not a vertex like we did so x=10, y=-1 so let y=a(x-2)²-5 -1=a(10-2)²-5 4=a(8)² 4=64a a=1/16 lets plug a in we get a(x-2)²-5=(x-2)²/16-5 now i want you yo try the rest of the excerises on your own and see how far you get.

Multiple Choice

96. 96. we got a vertex (-7,0), and a pt (-5,-4) what is the function that fits with 2 points

• -(x+5)²-4

• -(x+7)²

• (x+7)²

• (x+5)²-4

Multiple Choice

97. 97. we got a vertex (8,-9) and a pt (10,1) what is the function that fits with 2 points

• 5(x-8)²/2-9

• (x-8)²-9

• -(x-8)²-9

• -5(x-8)²/2-9

Multiple Choice

98. 98. we got a vertex (5,-20) and a pt (3,0) what is the function that fits with 2 points

• 5(x-5)²-20

• 4(x+5)²-20

• 20(x-5)²-20

• 10(x+5)²-20

Multiple Choice

99. 99. we got a vertex (3/2,7/8) and a pt (1/2,2/8) what is the function that fits with 2 points

• 7/8-5(x-3/2)²/8

• 5(x-3/2)²/8+7/8

• 5(x+3/2)²/8+7/8

• 7/8-5(x+3/2)²/8

Multiple Choice

100. CHALLANGE 100. we got a vertex (4/7,-5/8) and a pt (0,0) what is the function that fits with 2 points a=245/128

• 49(x+4/7)²/8-5/8

• 49(x-4/7)²/8-5/8

• 245(x+4/7)²/128-5/8

• 245(x-4/7)²/128-5/8

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.21

​say i got an equation 5x²-20x+c and we want to find the value of c that makes it perfect square well we have to use ax²+bx+c as our example let c be a perfect square a(x²+bx/a+c/a) lets use u and v factors to solve a ploblem a(u+v)²
=a(x²+bx/a+c/a) expand u and v a(u²+2uv+v²)=a(x²+bx/a+c/a) solve for u and x we get a(u²+2uv+v²)=a(x²+bx/a+c/a) => u²+2uv+v²=x²+bx/a+c/a lets use row by row method u²=x² 2uv=bx/a v²=c/a lets solve for u on the left doing row by row u²=x² u=x then let's do the second one 2uv=bx/a subsitive u=x in 2xv=bx/a now solve for v 2v=b/a v=b/(2a) let's do the third one v²=c/a subsitute v=b/(2a) in (b/(2a))²=c/a => b²/(4a²)=c/a solve for c b²/(4a)=c now our perfect square is b²/(4a) our new formula 5x²-20x+c lets plug s c in c=b²/(4a)=(-20)²/(4*5)
=400/20=20 plug that in that squaric trinomial we get 5x²-20x+c=5x²-20x+20 now i want you to try the rest of the excerises on your own and see how far you get.

Multiple Choice

101. 101. 7a²/2+35a/2+c whats the value of c to complete the square

• 175/4

• 175/8

• 175/16

• 175/32

Multiple Choice

102. 102. (24n²/7+192n/7+c whats the value of c to complete the square

• 102. 256/7

• 102. 328/7

• 102. 384/7

• 102. 400/7

Multiple Choice

103. 103. 25x²/7-20x/21+c whats the value of c to complete the square

• 103. 20/63

• 103. 10/63

• 103. 32/63

• 103. 26/63

Multiple Choice

104. 104. 5n²/12+10n/27+c whats the value of c to complete the square

• 104. 20/251

• 104. 20/217

• 104. 20/223

• 104. 20/243

Multiple Choice

105. 105. 8o²/9-24o/5+c whats the value of c to complete the square

• 105. 217/57

• 105. 823/45

• 105. 617/68

• 105. 526/75

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.22

​say i got points (6,7),(7,0), and (9,0) with the equation a(x-u)(x-v) well 2 solutions are clear enough when x=u a(x-u)(x-v)=0 and when x=v a(x-u)(x-v)=0 so (u,0) and (v,0) is defined so for (7,0), and (9,0) u=7, and v=9 so we can fill out the infoirmation we got a(x-u)(x-v)=0 => a(x-7)(x-9)=0 since y=7 then x=6 we can plug those values in as y=a(x-u)(x-v) let 7=a(6-7)(6-9) => 7=a(-1)(-3) => 7=3a a=7/3 lets plug a in we get a(x-7)(x-9)=7(x-7)(x-9)/3 lets try last one i got points (1,0)(3,0)(2,-1) again use the same interpratation a(x-u)(x-v)=0, then (u,0), and (v,0) is defined plug in values we get u=1, v=3 plug them into the function we get a(x-u)(x-v)=0, a(x-1)(x-3)=0 since y=-1, when x=2 we can plug in those values in as y=a(x-u)(x-v) let -1=a(2-1)(2-3)
=> -1=a(-1) a=1 plug it in we get a(x-1)(x-3)=(x-1)(x-3) now i want you to try the rest of the excerises on your own and see how far you get good luck

Multiple Choice

106. 106. ​a parabola hits points (5/2,-1/4),(3,0), and (2,0)

• (x-5/2)(x-2)/2

• (x-5/2)(x-2)

• (x-3)(x-2)

• 1(x-3)(x-2)/2

Multiple Choice

107. 107. ​a parabola hits points (6,-2),(8,0), and (4,0)

• (x-4)(x-8)/2

• (x-8)(x-6)(x-4)/2

• (x-4)(x-8)

• (x-8)(x-6)(x-4)

Multiple Choice

108. 108. ​a parabola hits points (8,-1/4),(7,0), and (9,0)

• (x-7)(x-9)

• (x-7)(x-8)(x-9)/4

• (x-7)(x-8)(x-9)

• (x-7)(x-9)/4

Multiple Choice

109. 109. ​a parabola hits points (4/7,-2),(0,0), and (8/7,0)

• 109. 49x(x-8/7)/16

• 109. 49x(x-8/7)/8

• 109. 49x(x-4/7)/8

• 109. 49x(x-4/7)/16

Multiple Choice

110. 110. ​a parabola hits points (4/5,-1/20),(3/5,0), and (1,0)

• -4(x-3/5)(x-1)/5

• (x-3/5)(x-1)

• 5(x-3/5)(x-1)/4

• 4(x-3/5)(x-1)/5

say i got the standard quadratic form 5x²+2x+10 and to find the 0 within what x value we need we'll just create a formula out of ax²+bx+c lets solve for x ax²+bx+c=0 let c=n+k we know k=c-b²/(4a) so c=n+c-b²/(4a) solve for n n=b²/(4a) we get c=n+k=b²/(4a)+c-b²/(4a) i know its unfactored but bear with me as we go along so ax²+bx+c=ax²+bx+b²/(4a)+c-b²/(4a)=a(x²+bx/a+b²/(4a²))+c-b²/(4a) lets use the vertex formula to solve a ploblem a(x-h)²+k=a(x²+bx/a+b²/(4a²))+c-b²/(4a)=> a(x²-2xh+h²)+k=a(x²+bx/a+b²/(4a²))+c-b²/(4a) use k=c-b²/(4a) if you havent remember it i encourage you to rewind to find the solution lets continiue a(x²-2xh+h²)+c-b²/(4a)=a(x²+bx/a+b²/(4a²))+c-b²/(4a) cancel stuff accept x² we get a(x²-2xh+h²)+c-b²/(4a)
=a(x²+bx/a+b²/(4a²))+c-b²/(4a) => x²-2xh+h²=x²+bx/a+b²/(4a²) we know h=-b/(2a) we can plug those values in on the left we get x²-2xh+h²=x²-2x(-b/(2a))+(-b/(2a))²=x²+bx/a+b²/(4a²))
now check x²+bx/a+b²/(4a²)=x²+bx/a+b²/(4a²) seems perfect to factor lets factor this squaric trinomial x²+bx/a+b²/(4a²)=(x+b/(2a))² now we can set variable in we get a(x²+bx/a+b²/(4a²))+c-b²/(4a)=a(x+b/(2a))²+c-b²/(4a) now solve for 0 a(x+b/(2a))²+c-b²/(4a)=0 => a(x+b/(2a))²=b²/(4a)-c => (x+b/(2a))²=b²/(4a²)-c/a => x+b/(2a)=±√(b²/(4a²)-c/a) set a into perfect square 4a² on c we get x+b/(2a)=±√(b²/(4a²)-c/a=x+b/(2a)=±√(b²/(4a²)-4ac/(4a²))=x=-b/(2a)±√((b²-4ac)/(4a²))=
-b/(2a)±√(b²-4ac)/(2a)=(-b±√(b²-4ac))/(2a) now with our quadratic formula we solved for we can finally plug the values in so for 5x²+2x+10=0 set a=5, b=2, c=10 then x=(-b±√(b²-4ac))/(2a)=(-2±√(2²-4*5*10))/(2*5)
=(-2±√(4-200))/10=(-2±√(-196))/10 since the root part is <0 there is no solution but good luck on the next page.

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.23

Multiple Choice

111. 111. find n if 6n²+4n+1=0

• n₁=-(2+√2)/3, n₂=-(2-√2)/3

• n₁=-(2+√2)/12, n₂=-(2-√2)/12

• n₁=-(2+√2)/6, n₂=-(2-√2)/6

• NO SOLUTION

Multiple Choice

112. 112. find h if 2j²-7j+8=0

• j₁=15/2, j₂=-1/2

• j₁=(7+√13)/4, j₂=(7-√13)/4

• j₁=(7+√13)/2, j₁=(7-√13)/2

• NO SOLUTION

Multiple Choice

113. 113. find c if c²-5c+6=0

• c₁=6, c₂=1

• c₁=3, c₂=2

• c₁=6, c₂=4

• NO SOLUTION

Multiple Choice

114. 114. find g if 7g²/3+49g/36-35/18=0

• g₁=14/9, g₂=-35/12

• g₁=2/3, g₂=-5/4

• g₁=-14/9, g₂=35/12

• NO SOLUTION

Multiple Choice

115. CHALLANGE 115. find d if 6d²/13-34d/21+14/123=0

• d₁=200/127+√(22,598,692/7,231,577), d₂=200/127-√(22,598,692/7,231,577)

• d₁=213/119+√(23,404,578/7,349,395), d₂=213/119-√(23,404,578/7,349,395)

• d₁=221/126+√(23,776,261/7,194,608), d₂=221/126-√(23,776,261/7,194,608)

• NO SOLUTION

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.24

​say i got an equation 3(x-5)²-7 and i want to find when our equation going to be 0 within x so we setup our vertex equation a(x-h)²+k and set that equal to 0 a(x-h)²+k=0 now solve for x a(x-h)²+k=0 => a(x-h)²=-k => (x-h)²=-k/a => x-h=±√(-k/a) x=h±√(-k/a) now we solve for x now plug the values in we get a=3 h=5, and k=-7 now plug them into formula x=5±√(7/3) now we can seperate x₁=5+√(7/3), and x₂=5-√(7/3) seems straight forward lets try another another one 7(x-3)²-6 lets use the formula to find the 0 x=h±√(-k/a)=3±√(6/7) lets seperate them x₁=3+√(6/7),
x₂=3-√(6/7) lets try a another one 6(x+4)²+44 lets use the formula x=h±√(-k/a) so a=6 ,h=-4, and k=44 x=-4±√(-44/6)=-4±√(-22/3) oh no this seems like there is no solution to the equation so answer is no solution since we have negative root remember if negative root do not count as an solution so i want you to try the rest of the ploblem on your own and see how far you get.

Multiple Choice

116. 116. find w if 5(w-2)²+3=0

• w₁=2+√3, w₂=2-√3

• w₁=2+√(3/5), w=2-√(3/5)

• w=√5-2, w=-2-√5

• NO SOLUTION

Multiple Choice

117. 117. find m if 7(m-3/2)²/4-8=0

• 117. m₁=3/2+32√(1/7), m₂=3/2-32√(1/7)

• 117. m₁=3/2+4√(2/7), m₂=3/2-4√(2/7)

• m₁=3/2+16√(2/7), m₂=3/2-16√(2/7)

• NO SOLUTION

Multiple Choice

118. 118. find j if 2(j-5/2)²/3-32/67=0

• j₁=√(67/48)-5/2, j₂=-5/2-√(67/48)

• j₁=√(48/67)-5/2, j₂=-5/2-√(48/67)

• j₁=5/2+√(48/67), j₂=5/2-√(48/67)

• NO SOLUTION

Multiple Choice

119. 119. find m if 6/7-4(m-8/5)²/5=0

• m₁=8/5+√(14/15), m₂=8/5-√(14/15)

• m₁=-8/5+√(15/14), m₂=-8/5-√(15/14)

• m₁=8/5+√(15/14), m₂=8/5-√(15/14)

• NO SOLUTION

Multiple Choice

120. 120. find h if 7/9-6(h-5/7)²/5=0

• h₁=5/7+√(35/54), h₂=5/7-√(35/54)

• h₁=5/7+√(54/35), h₂=5/7-√(54/35)

• h₁=-5/7+√(54/35), h₂=-5/7-√(54/35)

• NO SOLUTION

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.25

​lets say i got an equation 7x²+5x+8 and i want to find the discriminant of it but first lets explain how discriminant works well discriminant is the test if the root has a positve or negative value if its positive then it will return a real number if its negative it will return an imaginary root which is ALWAYS REMEMBER i=√-1 which cannot be an solution lets find the discriminant out of the quadratic formula x=(-b±√(b²-4ac))/2a we can see the root part we can set that to d=b²-4ac so if d>0 then it has 2 solution when d=0 then it has 1 solution and when d<0 then it has no solution got it lets use it to solve the discriminant so we do have the equation 7x²+5x+8 we can plug them in we get a=7, b=5, and c=8 lets plug them in d=b²-4ac => d=5²-4*7*8=25-224=-199 so this one has no solution lets try another one 8x²+8x-4 like always use discriminant d=b²-4ac=8²-4*8*-4=64+128=192 seems like it has 2 solutions now i want you to try the rest of the excerises on your own and see how far you get

Multiple Choice

121. 121. how many solutions when 6j²+35j=0

• 0

• 1

• 2

Multiple Choice

122. 122. how many solutions when 6y²+21y+70=0

• 0

• 1

• 2

Multiple Choice

123. 123. how many solutions when 5h²/3+20h/7+60/49

• 0

• 1

• 2

Multiple Choice

124. 124. how many solutions when 3h²+38h+1,000=0

• 0

• 1

• 2

Multiple Choice

125. 125. how many solutions when 5b²/7-32b/7+25/49=0

• 0

• 1

• 2

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.26

​say i got an equation o²-7o+10 and we want to factor such an equation one way to do it is to use quadratic formula x=(-b±√(b²-4ac))/(2a) note also focus on the discriminant in bolded one any how o=(-(-7)±√((-7)²-4*1*10)/(2*1)=(7±√(49-40))/2
=(7±√(49-40))/2=(7±√(9))/2 note discriminant>0 since our discriminant=9 so there are 2 solution if you check the discriminant so we use (x-x₁)(x-x₂) lets continiue factoring (7±√(9))/2=(7±3)/2 then we setup the solution o₁=(7+3)/2=10/2=5, o₂=(7-3)/2=4/2=2 lets plug them in this time using o we get (o-5)(o-2) so in factored conclusion o²-7o+10=(o-5)(o-2) lets try he last one s²-8s+16 to factor this squaric trinomal as always use the quadratic formula also use the discriminant method for this one x=(-b±√(b²-4ac))/(2a) s=(-(-8)±√((-8)²-4*1*16))/(2*1)
=(8±√(64-64))/2=(8±√0)/2 note we have a discriminant=0 so there is one solution we rewrite as (x-x₁)² lets continue solving (8±√0)/2=8/2=4 so s₁=4 so we get (s-4)² so in factored conclusion s²-8s+16=(s-4)² now i want you to try the excersies on your own and see how far you get

Multiple Choice

126. 126. factor g²-15g+56

• 126. (g+7)(g+8)

• 126. (g+7)(g-8)

• 126. (g-7)(g+8)

• 126. (g-7)(g-8)

Multiple Choice

127. 127. factor n²+3n-10

• 127. (n+5)(n-2)

• 127. (n-5)(n+2)

• 127. (n-5)(n-2)

• (n+5)(n+2)

Multiple Choice

128. 128. factor t²-9t/5+18/25

• 128. (t+6/5)(t-3/5)

• 128. (t-6/5)(t+3/5)

• 128. (t-6/5)(t-3/5)

• 128. (t+6/5)(t+3/5)

Multiple Choice

129. CHALLANGE 129. factor t²-27t/20+9/20

• 129. (t-5/4)(t-4/5)

• 129. (t+7/20)(t-13/20)

• 129. (t-3/4)(t-3/5)

• 129. (t+5/4)(t+4/5)

Multiple Choice

130. 130. factor h²-16h+63

• 130. (h+49)(h-63)

• 130. (h-49)(h-63)

• 130. (h+7)(h-9)

• 130. (h-7)(h-9)

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.27

​say i got an equation 5y²+10y+5 and i want to factor it so what we do is the use a quadratic formula which is x=(-b±√(b²-4ac))/(2a) lets use the same method y=(-10±√(10²-4*5*5))/(2*5)=(-10±√(100-100))/(10)=(-10±√0)/(10)=-10/10=-1 since our discrimant=0 we will use (bx-x₁)² let y₁=-1 a=5 since a is leading coefficient we get 5(y+1)² so in factored inconclusion 5y²+10y+5=5(y+1)² lets try another one 40h²-22h-8 to factor this squaric trinomial 40h²-22h-8 use the quadratic formula x=(-b±√(b²-4ac))/(2a) h=(-(-22)±√((-22)²-4*40*-8))/(2*40)
=(22±√(484+1,280))/80=(22±√1,764)/80 since our discriminant>0 we will use (x-x₁)(x-x₂) h=(22±√1,764)/80=(22±42)/80=(11±21)/40, h₁=(11+21)/40=32/40=4/5, and h₂=(11-21)/40=-10/40=-1/4 lets plug those values in we get (h-4/5)(h+1/4) since we got fractions we must seperate them we get (h-4/5)(h+1/4) => since leading coefficient a=40 we get 40(h-4/5)(h+1/4) remove all denominators we get 40(h-4/5)(h+1/4)=2(h-4)(h+1)

Multiple Choice

131. 131. factor 3j²-48

• 131. (j-4)²

• 131. (j-4)(j+4)

• 131. 3(j-4)²

• 131. 3(j-4)(j+4)

Multiple Choice

132. 132. factor 6o²/7+3o/7-20/21

• 132. (6o-5)(3o+4)/21

• 132. (6o-5)(3o+4)/18

• 132. 6(7o-5)(7o+8)/343

• 132. (7o-5)(7o+8)/49

Multiple Choice

133. 133. factor 7d²/20-427d/480+245/480

• 133. 7(3d-5)(8d+7)/480

• 133. 7(3d-5)(8d+7)/576

• 133. 7(3d-5)(8d-7)/576

• 133. 7(3d-5)(8d-7)/480

Multiple Choice

134. CHALLANGE 134. factor f²/4+35f/1440-5/6

• 134. (3f-320)(8f+9)/3,456

• 134. (f-320)(2f+9)/4

• 134. (9f-16)(8f+15)/288

• 134. (3f-4)(4f+3)/48

Multiple Choice

135. CHALLANGE 135. factor 24t¹⁰/7+4t⁵/7-160/7

• 135. 4(2t⁵-5)(3t⁵+8)/7

• 135. 4(14t⁵-5)(21t⁵+8)/49

• 135. 4(2t⁵-35)(3t⁵+56)

• 135. 24(2t⁵-5)(3t⁵+8)/7

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.28

x can be any variable say y, or z, whatever you choose its your option
1 -> x⁰ 0th degree horizontic x^-12 -> -12th degree antiduodecic
x -> x^-1 -1st degree, antiliniaric x^-13 -> -13th degree antitredecic
x^-2 -> -2nd degree, antisquaric x^-14 -> -14th degree antiquadecic
x^-3 -> -3rd degree, anticubic x^-15 -> -15th degree antiquindecic
x^-4 -> -4th degree, antiquartic x^-16 -> -16th degree antisesdecic
x^-5 -> -5th degree, antiquintic x^-17 -> -17th degree antiseptidecic
x^-6 -> -6th degree, antisestic x^-18 -> -18th degree antioctodecic
x^-7 -> -7th degree, antiseptic x^-19 -> -19th degree antinovemdecic
x^-8 -> -8th degree, antioctic x^-20 -> -20th degree antivigintic
x^-9 -> -9th degree, antinonic x^-21 -> -21st degree, antiunvigintic
x^-10 -> -10th degree, antidecic x^-22 -> -22nd degree, antiduovigintic
x^-11 -> -11th degree antiundecic x^-23 -> -23rd degree, antitrevigintic

Multiple Choice

136. 136. classify 5/x⁷+28/x⁹+34/x¹⁵-45/x²⁰

• antiseptic quadranomial

• antivigintic quadranomial

• antiseptic trinomial

• antivigintic trinomial

Multiple Choice

136. 137. classify 6/x¹⁰+11/x¹⁵-16/x²⁰

• antivigintic trinomial

• antiseptic quadranomial

• antidecic trinomial

• antidecic quadranomial

Multiple Choice

136. 138. classify the polynomial if there are 87 terms and highest degree is -7

• antiseptic septinomial

• antiseptic octuagintinomial

• antiseptic septenoctuagintinomial

• antiseptic octonomial

Multiple Choice

136. 139. classify the polynomial if there are 56 terms and highest degree is -14

• antiquadecic quntinomial

• antiquadecic sesquinquagintinomial

• antiquadecic quinquagintinomial

• antiquadecic sesagintinomial

Multiple Choice

136. 140. classify the polynomial if there are 20 terms and highest degree is -20

• antivigintic vigintinomial

• antivigintic decinomial

• antidecic decinomial

• antidecic vigintinomial

​Quadratics Alg1 Sem2, College Alg Sem1 Pt.29

​lets study more of these examples i know this is a different type of math
say we have xⁿ and we want to express something right heres something to express we know we let n be a power base on x so we know when
n=0 its horizontic, n=-9 its andtionic,
n=-1 its antilinearic, n=-10 its antidecic, and so on but what if we deal with higher order
n=-2 its antisquaric, n=-20 its antivigintic, n=-90 its antinonagintic,
n=-3 its anticubic, n=-30 its antitrigintic, n=-100 its anticentic,
n=-4 its antiquartic, n=-40 its antiquadragintic, say i got power n=-47 and we want
n=-5 its antiquintic, n=-50 its antiquinquagintic, to classify the power it is on so our
n=-6 its antisestic, n=-60 its antisesagintic, tens is 4 which is antiquadragintic
n=-7 its antiseptic, n=-70 its antiseptuagintic, and ones is 7 which is antiseptic so
n=-8 its antioctic, n=-80 its antioctuagintic, putting them together we get antiseptenquadragintic anti means opposite if you use 2 then it is not opposite so i want you to try the rest of the excerises on your own and see how far you get

Multiple Choice

136. 141. classify the polynomial if there are 72 terms and highest degree is -58

• octoquinquagintic duoseptuagintinomial

• duoquinquagintic septuagintinomial

• antiduoquinquagintic septuagintinomial

• antioctoquinquagintic duoseptuagintinomial

Multiple Choice

136. 142. classify the polynomial if there are 46 terms and highest degree is -98

• antioctononagintic sesaquadraintinomial

• antinonagintic sesaquadraintinomial

• antinovemnonagintic sesaquadraintinomial

• antioctoctuagintic sesaquadraintinomial

Multiple Choice

136. 143. classify the polynomial if there are 1 terms and highest degree is -74

• antisesaintinomial monomial

• antiquasesaintinomial monomial

• antiquasesaintinomial trinomial

• antisesaintinomial binomial

Multiple Choice

136. 144. classify the polynomial if there are 99 terms and highest degree is -100

• anticentic novemnonagintinomial

• novemnonagintic

  centinomial

• novemnonagintic novemnonagintinomial

• centic novemnonagintinomial

Multiple Choice

136. 145. classify the polynomial if there are 100 terms and highest degree is -100

• decic decinomial

• antidecic decinomial

• anticentic centinomial

• centic centinomial

Multiple Choice

146. 146. when x(u)=u²+2u+7 and y(u)=u²+5u+3 when will they equal to each other find the value of u

• 4/3

• 3/4

• 4

• 3

Multiple Choice

147. 147. when f(t)=2t²+5t+7 and d(t)=t²+1 when will they be equal to each other find the value of t

• t₁=5, t₂=6

• t₁=-2, t₂=-3

• t₁=-5, t₂=-6

• t₁=2, t₂=3

Multiple Choice

148. CHALLANGE 148. when c(r)=r⁸+5r⁴+3 and s(r)=-3 when will they be equal to each other find the value of r

• r₁=-3 r₂=-2

• r₁=√3 r₂=√2 r₃=-√3 r₄=-√2

• r₁=3 r₂=2

• N/A

Multiple Choice

149. CHALLANGE 149. when a(s)=s¹⁰+7s⁵+23 and b(s)=2s¹⁰+s⁵+32 when will they be equal to each other find the value of s s¹⁰-6s⁵+9

• s₁=⁵√3 s₂=⁵√2 s₃=-⁵√3 s₄=-⁵√2

• s₁=⁵√3 s₂=-⁵√3

• s₁=⁵√3 s₂=⁵√2

• s₁=⁵√3 s₂=⁵√6 s₃=-⁵√3 s₄=-⁵√6

Multiple Choice

150. CHALLANGE 150. when b(r)=4/⁵√r+5/¹⁰√r+10 and c(r)=3/⁵√r+10/¹⁰√r+4

• r₁=59,049 r₂=1,024

• r₁=1/59,049 r₂=1/1,024

• r₁=1/243 r₂=1/32

• r₁=243 r₂=32