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mathcalculus Group Title

⇒ s′(t)=4(2t−3)(8−2t)^3−6(2t−3)^2(8−2t)^2 how can i factor this? I keep factoring incorrectly :(

  • 11 months ago
  • 11 months ago

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  1. Paynesdad Group Title
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    Show me what you have done.

    • 11 months ago
  2. Paynesdad Group Title
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    How many (2t-3) can you pull out?

    • 11 months ago
  3. mathcalculus Group Title
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    I used 2 to factor since it goes in both 4 and -6

    • 11 months ago
  4. mathcalculus Group Title
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    1

    • 11 months ago
  5. Paynesdad Group Title
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    ok so that is part of it

    • 11 months ago
  6. Psymon Group Title
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    \[4(2t-3)(8-2t)^{3}-6(2t-3)^{2}(8-2t)^{2}\] \[(2t-3)(8-2t)^{2}[4(8-2t)-6(2t-3)]\] I'll stop here as to not go all the way, but maybe you can see what I was doing?

    • 11 months ago
  7. mathcalculus Group Title
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    The problem is im not sure how to factor these anymore. i did a similar problem like this.. only they factored out an outer number and here: we factor one of the insides number. im so confused.

    • 11 months ago
  8. Paynesdad Group Title
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    Do you follow what @psymon did?

    • 11 months ago
  9. mathcalculus Group Title
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    yes but why can't we pull out a 2 in the outside since it' can go into 4 and -6?

    • 11 months ago
  10. Paynesdad Group Title
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    ok I have dealt with this before ... it is tough to see. There really are only 2 terms in your original equation...do you see that?

    • 11 months ago
  11. Psymon Group Title
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    Because I rather not do too much at once and worry about mistakes.

    • 11 months ago
  12. mathcalculus Group Title
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    and no i didn't follow... this is where im stuck

    • 11 months ago
  13. Paynesdad Group Title
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    But Psymon could have done that too.

    • 11 months ago
  14. mathcalculus Group Title
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    yes i see

    • 11 months ago
  15. Paynesdad Group Title
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    Look you actually only have two terms with several factors each. Hang on..

    • 11 months ago
  16. Paynesdad Group Title
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    The first term is 4(2t-3)(8-2t)^3

    • 11 months ago
  17. Paynesdad Group Title
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    That is one big term because it is composed of things being multiplied together (called factors)

    • 11 months ago
  18. mathcalculus Group Title
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    ok

    • 11 months ago
  19. Paynesdad Group Title
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    the factors of the first term are 4, (2t-3), and (8-2t)

    • 11 months ago
  20. mathcalculus Group Title
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    right

    • 11 months ago
  21. Paynesdad Group Title
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    So what would be the 2nd term? @mathcalculus

    • 11 months ago
  22. mathcalculus Group Title
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    -6, (2t-3) and (8-2t)

    • 11 months ago
  23. mathcalculus Group Title
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    @Paynesdad

    • 11 months ago
  24. Paynesdad Group Title
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    Good so now think of factoring something easy... like 6x^2-15x ... you factor out the common factors and leave whatever is left ... right?

    • 11 months ago
  25. mathcalculus Group Title
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    yes

    • 11 months ago
  26. Paynesdad Group Title
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    In other words you factor out 3 and x and leave (2x-5) right?

    • 11 months ago
  27. mathcalculus Group Title
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    correct

    • 11 months ago
  28. Paynesdad Group Title
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    So now look at your equation ... what things and how many can you factor out of both terms

    • 11 months ago
  29. mathcalculus Group Title
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    thats what i was saying... 2 goes in for 4 and -6..

    • 11 months ago
  30. Paynesdad Group Title
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    Ok what other factors can you factor out?

    • 11 months ago
  31. mathcalculus Group Title
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    i can factor (2t-3) and (8-2t)

    • 11 months ago
  32. Paynesdad Group Title
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    Right how many 2t-3 can you factor out remember you can only factor out as many as the least number in either of the two terms.

    • 11 months ago
  33. mathcalculus Group Title
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    1

    • 11 months ago
  34. Paynesdad Group Title
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    right and how many 8-2t?

    • 11 months ago
  35. mathcalculus Group Title
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    @Paynesdad right i know i can factor that too but what about the outer numbers. should i factor that firsT?

    • 11 months ago
  36. mathcalculus Group Title
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    2

    • 11 months ago
  37. Paynesdad Group Title
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    you can it shouldn't matter.

    • 11 months ago
  38. Paynesdad Group Title
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    so factor out 2(2t-3)(8-2t)^2 and show me what you are left with

    • 11 months ago
  39. mathcalculus Group Title
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    yes it does.... for ex::

    • 11 months ago
  40. mathcalculus Group Title
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    see they factored out using the 4

    • 11 months ago
  41. Paynesdad Group Title
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    That doesn't dispute what I was saying it doesn't matter if you factor out the 2 right now or not.

    • 11 months ago
  42. Psymon Group Title
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    |dw:1375832092218:dw| This is just how I might go about doing these on paper. Looks like you're getting closer to understanding though, @mathcalculus ^_^

    • 11 months ago
  43. mathcalculus Group Title
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    i want to know why can't i in this problem. which confuses so much.

    • 11 months ago
  44. Paynesdad Group Title
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    You can factor out the 2...You can do it now or you can do it in the end ... either way you will end up doing it. I, personally, would do it in the beginning.,,

    • 11 months ago
  45. Paynesdad Group Title
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    @Psymon is choosing to do it at the end...it will get done ... it's just a matter of when

    • 11 months ago
  46. Paynesdad Group Title
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    so factor out 2(2t-3)(8-2t)^2 and show me what you have and I will tell you if you are right.

    • 11 months ago
  47. Psymon Group Title
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    I'm personally too focused on the parenthesis terms and not making mistakes that I kinda just leave the coefficients til the end. Its dividing out those groups that worry me the most :P gtg now, though, good luck @mathcalculus

    • 11 months ago
  48. mathcalculus Group Title
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    @Psymon got it^_^

    • 11 months ago
  49. mathcalculus Group Title
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    that helped out alot. then what?

    • 11 months ago
  50. mathcalculus Group Title
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    don't i need to finish?

    • 11 months ago
  51. Paynesdad Group Title
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    Show me what you have. so far.

    • 11 months ago
  52. mathcalculus Group Title
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    2(2t-3)(8-2t)^2[2(8-2t)-3(2t-3)

    • 11 months ago
  53. Paynesdad Group Title
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    That looks good

    • 11 months ago
  54. mathcalculus Group Title
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    @Paynesdad

    • 11 months ago
  55. Paynesdad Group Title
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    now simplify wh

    • 11 months ago
  56. Paynesdad Group Title
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    is in the backets

    • 11 months ago
  57. Paynesdad Group Title
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    now simplify what is in the b

    • 11 months ago
  58. Paynesdad Group Title
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    *brackets

    • 11 months ago
  59. mathcalculus Group Title
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    what about the left side?

    • 11 months ago
  60. mathcalculus Group Title
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    add them all up?

    • 11 months ago
  61. Paynesdad Group Title
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    no don't just the stuff in the brackets

    • 11 months ago
  62. mathcalculus Group Title
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    ook i did

    • 11 months ago
  63. mathcalculus Group Title
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    i got [16-4t-6t+9]

    • 11 months ago
  64. Paynesdad Group Title
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    Keep going

    • 11 months ago
  65. mathcalculus Group Title
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    oh 16-10t+9

    • 11 months ago
  66. Paynesdad Group Title
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    keep going

    • 11 months ago
  67. mathcalculus Group Title
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    fine

    • 11 months ago
  68. Paynesdad Group Title
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    :-)

    • 11 months ago
  69. mathcalculus Group Title
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    -10t+7

    • 11 months ago
  70. Paynesdad Group Title
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    errrrrr no re-think that last step

    • 11 months ago
  71. mathcalculus Group Title
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    there's no where to go. now what about the left side.

    • 11 months ago
  72. mathcalculus Group Title
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    oops 25

    • 11 months ago
  73. mathcalculus Group Title
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    -10t+25

    • 11 months ago
  74. Paynesdad Group Title
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    So now write it out again (yes the whole thing)

    • 11 months ago
  75. mathcalculus Group Title
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    2(2t-3)(8-2t)^2[-10t+25]

    • 11 months ago
  76. Paynesdad Group Title
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    Good now is there anything else that can be factored out now that you combined that last factor (notice you didn't multiply out that factor before just INSIDE of the parantheses of that last factor

    • 11 months ago
  77. mathcalculus Group Title
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    let's say im loooking for the critical points.... i would still need to follow to this step right?

    • 11 months ago
  78. Paynesdad Group Title
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    yes it is in deed helpful.

    • 11 months ago
  79. Paynesdad Group Title
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    indeed*

    • 11 months ago
  80. mathcalculus Group Title
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    the 2?

    • 11 months ago
  81. Paynesdad Group Title
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    Nope there is something you can pull out of the last factor...

    • 11 months ago
  82. mathcalculus Group Title
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    the (8-2t)^2?

    • 11 months ago
  83. Paynesdad Group Title
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    no Look ---> (-10t+25)

    • 11 months ago
  84. mathcalculus Group Title
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    oh i thought you meant only from looking at the left side.. :?

    • 11 months ago
  85. mathcalculus Group Title
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    yes 2 *-5 =-10

    • 11 months ago
  86. Paynesdad Group Title
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    There is no side it is just one big thing...

    • 11 months ago
  87. mathcalculus Group Title
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    ok

    • 11 months ago
  88. Paynesdad Group Title
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    well what are you factoring out of the last factor?

    • 11 months ago
  89. mathcalculus Group Title
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    -5

    • 11 months ago
  90. Paynesdad Group Title
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    ok so do that and combine it with the 2 in the front and re-write it one last time...promise.

    • 11 months ago
  91. mathcalculus Group Title
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    2*-5(2t-3)(8-2t)^2[25]

    • 11 months ago
  92. Paynesdad Group Title
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    Whoa whoa whoa

    • 11 months ago
  93. Paynesdad Group Title
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    You

    • 11 months ago
  94. Paynesdad Group Title
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    You factored out a -5 and had what left in the parantheses

    • 11 months ago
  95. mathcalculus Group Title
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    you mean in the brackets t+25

    • 11 months ago
  96. Paynesdad Group Title
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    No it was (-10t+25) if you just had this and factored out -5 what would the result be

    • 11 months ago
  97. mathcalculus Group Title
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    2t-5

    • 11 months ago
  98. Paynesdad Group Title
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    right welll you can't just throw that out...So now you should have all together...\[2(2t-3)(8-2t)^2(-5)(2t-5)\] right?

    • 11 months ago
  99. mathcalculus Group Title
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    right

    • 11 months ago
  100. Paynesdad Group Title
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    Now normally we combine our any numerical factors so we would bring the -5 to the front and multiply it by 2 and get -10

    • 11 months ago
  101. Paynesdad Group Title
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    So the final answer is \[-10(2t-3)(8-2t)^2(2t-5)\]

    • 11 months ago
  102. Paynesdad Group Title
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    So

    • 11 months ago
  103. Paynesdad Group Title
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    Looking at the final answer ... how many terms are there?

    • 11 months ago
  104. Paynesdad Group Title
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    @mathcalculus

    • 11 months ago
  105. mathcalculus Group Title
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    3

    • 11 months ago
  106. Paynesdad Group Title
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    nope remember terms are things that are separated by + or - and not in parantheses.

    • 11 months ago
  107. mathcalculus Group Title
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    1

    • 11 months ago
  108. Paynesdad Group Title
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    For example 5x+2y +16x^2 is three terms

    • 11 months ago
  109. mathcalculus Group Title
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    right

    • 11 months ago
  110. Paynesdad Group Title
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    Right only 1... Now how many factors in that 1 terms?

    • 11 months ago
  111. mathcalculus Group Title
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    ?

    • 11 months ago
  112. Paynesdad Group Title
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    Factors are things that are multiplied together so 6xy has 3 factors.

    • 11 months ago
  113. Paynesdad Group Title
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    7(x+1) (x-2) has 3 factors too

    • 11 months ago
  114. mathcalculus Group Title
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    this is confusing me more.

    • 11 months ago
  115. mathcalculus Group Title
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    3?

    • 11 months ago
  116. Paynesdad Group Title
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    yep three things multiplied together... good

    • 11 months ago
  117. mathcalculus Group Title
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    ok

    • 11 months ago
  118. mathcalculus Group Title
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    for a small problem.. it's so long..:?

    • 11 months ago
  119. Paynesdad Group Title
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    Well actually 4...-10, 2t-3, 8-2t, and 2t-5

    • 11 months ago
  120. mathcalculus Group Title
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    okay 4.

    • 11 months ago
  121. Paynesdad Group Title
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    ...I know ... do you want to know why this form of the equation is better?

    • 11 months ago
  122. mathcalculus Group Title
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    next

    • 11 months ago
  123. mathcalculus Group Title
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    yeah

    • 11 months ago
  124. Paynesdad Group Title
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    ok it has to do with this rule...if xy=0 what do you know?

    • 11 months ago
  125. mathcalculus Group Title
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    idk

    • 11 months ago
  126. mathcalculus Group Title
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    i have to keep going... it's getting too late for one problem:/

    • 11 months ago
  127. mathcalculus Group Title
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    is it possible to show me so that i can how...

    • 11 months ago
  128. mathcalculus Group Title
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    ask*

    • 11 months ago
  129. Paynesdad Group Title
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    well If I am thinking of two numbers and I tell you that they multiply to 0 can you tell me what one of the numbers I am thinking of?

    • 11 months ago
  130. mathcalculus Group Title
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    0

    • 11 months ago
  131. mathcalculus Group Title
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    1and 0

    • 11 months ago
  132. Paynesdad Group Title
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    right! so look at the original equation...if you set that to 0 you cant solve for t

    • 11 months ago
  133. mathcalculus Group Title
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    omg obvious......

    • 11 months ago
  134. Paynesdad Group Title
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    but if you get it into this form \[-10(2t-3)(8-2t)^2(2t-5)=0\]

    • 11 months ago
  135. mathcalculus Group Title
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    it's the factoring i need help with. that's all

    • 11 months ago
  136. mathcalculus Group Title
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    whats next after that.

    • 11 months ago
  137. Paynesdad Group Title
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    ok I just wanted you to see how it helps you in actual calc problems but I understand...next? no next ...\[-10(2t-3)(8-2t)^2(2t-5)\] is completely factored you did it.!

    • 11 months ago
  138. mathcalculus Group Title
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    are you sure? look how simple this person did this problem ds(t) / dt = 4 (2t - 3) (8 - 2t)^3 - 6 (2t - 3)^2 (8 - 2t)^2 = 0 ==> factoring: (2t - 3) (8 - 2t)^2 (4 (8 - 2t) - 6 (2t -3)) = 0 ==> (2t -3) (8 - 2t)^2 (50 - 20t) = 0 so the three critical points are: t = 3/2, 4, 5/2.

    • 11 months ago
  139. mathcalculus Group Title
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    im looking for t right.

    • 11 months ago
  140. mathcalculus Group Title
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    they've multiplied -10 to 2t-5

    • 11 months ago
  141. mathcalculus Group Title
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    but why? the last one?

    • 11 months ago
  142. mathcalculus Group Title
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    not the 1st, nor the 2nd?

    • 11 months ago
  143. Paynesdad Group Title
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    Right that is exactly what you had but YOU factored the -10 out...they didn't.

    • 11 months ago
  144. Paynesdad Group Title
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    Hang on you are thinking of distirbution

    • 11 months ago
  145. Paynesdad Group Title
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    you don't distribute over multiplication !

    • 11 months ago
  146. mathcalculus Group Title
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    this is taking forever.. for a small problem ___-

    • 11 months ago
  147. mathcalculus Group Title
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    thank you for you help @Paynesdad really appreciate it.

    • 11 months ago
  148. Paynesdad Group Title
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    5(79)(3)(2)...you can muliply the 5 times any one of those numbers but not all of them ...I would ppick the last one but we could multiply any of them just not all of them

    • 11 months ago
  149. Paynesdad Group Title
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    No problem ... Good luck in tough subject but honestly

    • 11 months ago
  150. Paynesdad Group Title
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    you have a better comprehension then 50% of the students I have had. @mathcalculus

    • 11 months ago
  151. Paynesdad Group Title
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    *than

    • 11 months ago
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