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anonymous

  • one year ago

Part A: Write the expression x2 + 5x + 6 as a product of two linear expressions. Show your work and justify each step. (5 points) Part B: Rewrite x2 - 4x + 4 as a square of a linear expression. (3 points) Part C: Do the expressions in parts A and B have a common factor? Justify your answer. (2 points)

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  1. wolf1728
    • one year ago
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    Part A Basically isn't this just asking to factor the equation? (x +2) * (x +3)

  2. anonymous
    • one year ago
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    yes but i have been having big problems trying to solve this things, i shouldn't be up crying over this....

  3. wolf1728
    • one year ago
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    Part B (x -2) * (x -2)

  4. anonymous
    • one year ago
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    but how are you getting these...

  5. wolf1728
    • one year ago
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    I guess you don't know how to factor.

  6. anonymous
    • one year ago
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    sadly no...

  7. wolf1728
    • one year ago
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    Okay let's take x2 - 4x + 4 in this case a=1 b=-4 and c=4 (where a, b and c are the factors of each term of the equation) Do you understand that?

  8. anonymous
    • one year ago
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    so in this case, a= 2 b=5 and c=6 ???

  9. wolf1728
    • one year ago
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    If you are referring to part A then a would equal 1 and you have the rest correct.

  10. anonymous
    • one year ago
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    sorry, i missed that.. i understand that part. thank you! but can you explain part B?

  11. wolf1728
    • one year ago
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    Okay for part B the equation is x2 - 4x + 4 and a =1 b=-4 and c=4

  12. anonymous
    • one year ago
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    so its just like part a?

  13. wolf1728
    • one year ago
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    Yes they both require you to factor the equation

  14. anonymous
    • one year ago
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    seems like i was stressing over nothing! somehow you made it clear to me, thank you so much

  15. wolf1728
    • one year ago
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    Okay then if you think you know it, then I am glad I helped you out. Do you have any problems with part C?

  16. anonymous
    • one year ago
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    im having trouble finding the gcf

  17. anonymous
    • one year ago
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    im new to this, its hard with no teacher or anyone to explain..

  18. wolf1728
    • one year ago
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    As for Part C I don't think that the equations have a common factor. Unless they mean both equations have an x^2 in there and the factor of x^2 is 1.

  19. anonymous
    • one year ago
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    thats the part i was mostly confused on.

  20. wolf1728
    • one year ago
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    Well that's my answer. I don't know if it the correct answer but it's the best I could come up with.

  21. anonymous
    • one year ago
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    thank you, do you mind if you can walk me through some other ones?

  22. UsukiDoll
    • one year ago
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    I don't see a common factor for part c either

  23. UsukiDoll
    • one year ago
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    for A (x +2) * (x +3) for B (x -2) * (x -2) unless the signs were swapped in B, there isn't a common factor.

  24. UsukiDoll
    • one year ago
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    and congrats @wolf1728 for reaching 95SS :)

  25. wolf1728
    • one year ago
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    Thanks UsukiDoll :-)

  26. anonymous
    • one year ago
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    Thank you both.

  27. UsukiDoll
    • one year ago
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    you're welcome :)

  28. wolf1728
    • one year ago
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    u r welcome

  29. anonymous
    • one year ago
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    can you help me on 3 more questions?

  30. wolf1728
    • one year ago
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    Three more??? Well okay. :-)

  31. anonymous
    • one year ago
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    yes.. but this is the last time having to deal with this topic...

  32. anonymous
    • one year ago
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    A function is shown below: f(x) = x3 + 2x2 - x - 2 Part A: What are the factors of f(x)? Show your work. (3 points) Part B: What are the zeros of f(x)? Show your work. (2 points) Part C: What are the steps you would follow to graph f(x)? Describe the end behavior of the graph of f(x). (5 points)

  33. anonymous
    • one year ago
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    for part a.... a=1 b=2 c=-2 not 100% sure...

  34. wolf1728
    • one year ago
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    Part A a=1 b=2 c=-1 d=-2

  35. wolf1728
    • one year ago
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    You are now factoring a cubic equation

  36. anonymous
    • one year ago
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    totaly just misssed the "x" there.. sorry about that

  37. anonymous
    • one year ago
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    totally**

  38. UsukiDoll
    • one year ago
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    factor by grouping for the first part rational root test for the second part

  39. anonymous
    • one year ago
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    im new that this.. @UsukiDoll

  40. wolf1728
    • one year ago
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    and it's been a while since I factored a cubic equation

  41. UsukiDoll
    • one year ago
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    sorry x.x

  42. anonymous
    • one year ago
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    B should be \[(x-2)^2 \] because that is in square of a linear expression formio

  43. UsukiDoll
    • one year ago
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    (x-2)^2 and (x-2)(x-2) mean the same thing.

  44. UsukiDoll
    • one year ago
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    For (x-2)^2 that exponent number is 2 meaning write (x-2) 2 times (x-2)(x-2) .

  45. anonymous
    • one year ago
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    how did you get that

  46. UsukiDoll
    • one year ago
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    I was talking to Shalante earlier...

  47. anonymous
    • one year ago
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    gotcha! i wish i could understand this, the first problem i got, but now this is totally different

  48. UsukiDoll
    • one year ago
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    I know... I got distracted... one sec while I cool off before going berserk.

  49. anonymous
    • one year ago
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    okay, thank you for your help!

  50. wolf1728
    • one year ago
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    UsukiDoll it seems you know how to factor a cubic equation?

  51. UsukiDoll
    • one year ago
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    \[f(x) = x^3+2x^2-x-2\]

  52. wolf1728
    • one year ago
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    You mentioned grouping

  53. UsukiDoll
    • one year ago
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    I can't do anymore of this... I know it, but I'm being picked on.

  54. anonymous
    • one year ago
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    no one is picking on you...

  55. wolf1728
    • one year ago
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    I found a method for factoring a cubic http://www.wikihow.com/Factor-a-Cubic-Polynomial it is a 12 step process!

  56. UsukiDoll
    • one year ago
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    oh no way... why do that when for the first two terms we can take out a x^2 and for the last two terms, take out the negative sign

  57. UsukiDoll
    • one year ago
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    \[\large f(x) = x^3+2x^2-x-2 \] on the first two terms \[\large x^3+2x^2 \] we noticed that there is a x^2 in common. we can see this easier by expanding. the exponent 3 means write x 3 times. similarly the exponent 2 means write x 2 times \[\large xxx+2xx \] we can yank out 2 x's \[\large xx(x+2) \] \[\large x^2( x+2) \]

  58. UsukiDoll
    • one year ago
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    so for the second half we just factor out the negative \[\large f(x) = x^3+2x^2-x-2 \] \[\large -x-2 \] \[\large -(x+2) \]

  59. anonymous
    • one year ago
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    @UsukiDoll It tells us to write in squares (x-2)^2 is a square. This is new in algebra It like basic math telling us to write the solution of 5/8 +10/8 in mixed number form instead of reciprocal terms.

  60. UsukiDoll
    • one year ago
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    \[\large f(x) = x^3+2x^2-x-2 \] through factor by grouping we have \[\large x^2(x+2)-(x+2)\]

  61. UsukiDoll
    • one year ago
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    @Shalante can you stop interrupting me please?

  62. anonymous
    • one year ago
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    part b is worth 3 points. guess he wont get all the credit. I would have replied earlier, but my laptop went ham

  63. UsukiDoll
    • one year ago
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    anyway, we're still not done with the factor by grouping. We have one common term to yank out \[\large f(x) = x^2(x+2)-(x+2)\] so we notice that there is a (x+2) in common uh oh. I forgot about one thing. there is usually a 1 written in front of the -(x+2) \[\large f(x) = x^2(x+2)-1(x+2)\] so now yanking out the (x+2) \[\large f(x) = (x+2)(x^2-1) \]

  64. UsukiDoll
    • one year ago
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    we can factor the \[\large x^2-1\] using difference of squares \[(a^2-b^2) = (a+b)(a-b)\] letting a = x and b =1 \[(x^2-1^2) = (x+1)(x-1)\]

  65. UsukiDoll
    • one year ago
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    \[\large f(x) = (x+2)(x+1)(x-1) \] that's the final factored form

  66. anonymous
    • one year ago
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    thank you so much for explaining this to me.

  67. UsukiDoll
    • one year ago
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    so now we have to find our 0's... we just have to solve for x in 3 cases x+2 = 0, x+1=0, x-1 =0

  68. anonymous
    • one year ago
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    now what about the steps you would follow to graph f(x)? and Describe the end behavior of the graph of f(x)

  69. UsukiDoll
    • one year ago
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    did you solve for x first in the previous part? there should be 3

  70. anonymous
    • one year ago
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    2,1,1????

  71. UsukiDoll
    • one year ago
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    close.... solve for x x+1= 0 subtract 1 on both sides

  72. anonymous
    • one year ago
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    some how im getting 2,1 and then -1 but im over thinking this and getting more confused

  73. UsukiDoll
    • one year ago
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    it's correct now x =2,1,-1

  74. wolf1728
    • one year ago
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    -2 -1 and1

  75. UsukiDoll
    • one year ago
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    oh wait I missed. ugh yeah wolf is right -2 -1 1

  76. UsukiDoll
    • one year ago
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    I overlooked by accident.

  77. wolf1728
    • one year ago
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    you did the factoring though

  78. anonymous
    • one year ago
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    made the same mistake.

  79. UsukiDoll
    • one year ago
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    It was an accident... geez.

  80. wolf1728
    • one year ago
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    but you factored it correctly Usuki :-)

  81. anonymous
    • one year ago
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    its late, im sure we all make mistakes.. its okay.. your doing a good job!

  82. wolf1728
    • one year ago
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    well I'm gonna get going

  83. UsukiDoll
    • one year ago
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    I'm out too. the nitpicking towards me is too much.

  84. anonymous
    • one year ago
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    nobody is nitpicking, your helping me out alot

  85. UsukiDoll
    • one year ago
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    tell that to @Shalante then. I'm through with this.

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