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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 :(

  • one year ago
  • one year ago

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

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

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

    • one year ago
  4. mathcalculus Group Title
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    1

    • one year ago
  5. Paynesdad Group Title
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    ok so that is part of it

    • one year 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?

    • one year 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.

    • one year ago
  8. Paynesdad Group Title
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    Do you follow what @psymon did?

    • one year 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?

    • one year 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?

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

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

    • one year ago
  13. Paynesdad Group Title
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    But Psymon could have done that too.

    • one year ago
  14. mathcalculus Group Title
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    yes i see

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

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

    • one year 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)

    • one year ago
  18. mathcalculus Group Title
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    ok

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

    • one year ago
  20. mathcalculus Group Title
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    right

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

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

    • one year ago
  23. mathcalculus Group Title
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    @Paynesdad

    • one year 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?

    • one year ago
  25. mathcalculus Group Title
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    yes

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

    • one year ago
  27. mathcalculus Group Title
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    correct

    • one year 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

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

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

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

    • one year 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.

    • one year ago
  33. mathcalculus Group Title
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    1

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

    • one year 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?

    • one year ago
  36. mathcalculus Group Title
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    2

    • one year ago
  37. Paynesdad Group Title
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    you can it shouldn't matter.

    • one year 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

    • one year ago
  39. mathcalculus Group Title
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    yes it does.... for ex::

    • one year ago
  40. mathcalculus Group Title
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    see they factored out using the 4

    • one year 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.

    • one year 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 ^_^

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

    • one year 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.,,

    • one year 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

    • one year 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.

    • one year 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

    • one year ago
  48. mathcalculus Group Title
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    @Psymon got it^_^

    • one year ago
  49. mathcalculus Group Title
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    that helped out alot. then what?

    • one year ago
  50. mathcalculus Group Title
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    don't i need to finish?

    • one year ago
  51. Paynesdad Group Title
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    Show me what you have. so far.

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

    • one year ago
  53. Paynesdad Group Title
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    That looks good

    • one year ago
  54. mathcalculus Group Title
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    @Paynesdad

    • one year ago
  55. Paynesdad Group Title
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    now simplify wh

    • one year ago
  56. Paynesdad Group Title
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    is in the backets

    • one year ago
  57. Paynesdad Group Title
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    now simplify what is in the b

    • one year ago
  58. Paynesdad Group Title
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    *brackets

    • one year ago
  59. mathcalculus Group Title
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    what about the left side?

    • one year ago
  60. mathcalculus Group Title
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    add them all up?

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

    • one year ago
  62. mathcalculus Group Title
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    ook i did

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

    • one year ago
  64. Paynesdad Group Title
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    Keep going

    • one year ago
  65. mathcalculus Group Title
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    oh 16-10t+9

    • one year ago
  66. Paynesdad Group Title
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    keep going

    • one year ago
  67. mathcalculus Group Title
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    fine

    • one year ago
  68. Paynesdad Group Title
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    :-)

    • one year ago
  69. mathcalculus Group Title
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    -10t+7

    • one year ago
  70. Paynesdad Group Title
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    errrrrr no re-think that last step

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

    • one year ago
  72. mathcalculus Group Title
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    oops 25

    • one year ago
  73. mathcalculus Group Title
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    -10t+25

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

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

    • one year 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

    • one year 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?

    • one year ago
  78. Paynesdad Group Title
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    yes it is in deed helpful.

    • one year ago
  79. Paynesdad Group Title
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    indeed*

    • one year ago
  80. mathcalculus Group Title
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    the 2?

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

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

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

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

    • one year ago
  85. mathcalculus Group Title
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    yes 2 *-5 =-10

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

    • one year ago
  87. mathcalculus Group Title
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    ok

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

    • one year ago
  89. mathcalculus Group Title
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    -5

    • one year 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.

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

    • one year ago
  92. Paynesdad Group Title
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    Whoa whoa whoa

    • one year ago
  93. Paynesdad Group Title
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    You

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

    • one year ago
  95. mathcalculus Group Title
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    you mean in the brackets t+25

    • one year 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

    • one year ago
  97. mathcalculus Group Title
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    2t-5

    • one year 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?

    • one year ago
  99. mathcalculus Group Title
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    right

    • one year 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

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

    • one year ago
  102. Paynesdad Group Title
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    So

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

    • one year ago
  104. Paynesdad Group Title
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    @mathcalculus

    • one year ago
  105. mathcalculus Group Title
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    3

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

    • one year ago
  107. mathcalculus Group Title
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    1

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

    • one year ago
  109. mathcalculus Group Title
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    right

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

    • one year ago
  111. mathcalculus Group Title
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    ?

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

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

    • one year ago
  114. mathcalculus Group Title
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    this is confusing me more.

    • one year ago
  115. mathcalculus Group Title
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    3?

    • one year ago
  116. Paynesdad Group Title
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    yep three things multiplied together... good

    • one year ago
  117. mathcalculus Group Title
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    ok

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

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

    • one year ago
  120. mathcalculus Group Title
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    okay 4.

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

    • one year ago
  122. mathcalculus Group Title
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    next

    • one year ago
  123. mathcalculus Group Title
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    yeah

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

    • one year ago
  125. mathcalculus Group Title
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    idk

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

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

    • one year ago
  128. mathcalculus Group Title
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    ask*

    • one year 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?

    • one year ago
  130. mathcalculus Group Title
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    0

    • one year ago
  131. mathcalculus Group Title
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    1and 0

    • one year 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

    • one year ago
  133. mathcalculus Group Title
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    omg obvious......

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

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

    • one year ago
  136. mathcalculus Group Title
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    whats next after that.

    • one year 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.!

    • one year 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.

    • one year ago
  139. mathcalculus Group Title
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    im looking for t right.

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

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

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

    • one year 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.

    • one year ago
  144. Paynesdad Group Title
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    Hang on you are thinking of distirbution

    • one year ago
  145. Paynesdad Group Title
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    you don't distribute over multiplication !

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

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

    • one year 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

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

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

    • one year ago
  151. Paynesdad Group Title
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    *than

    • one year ago
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