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mathcalculus

  • one year ago

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

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  1. Paynesdad
    • one year ago
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    Show me what you have done.

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

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

  4. mathcalculus
    • one year ago
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    1

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

  6. Psymon
    • one year ago
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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?

  7. mathcalculus
    • one year ago
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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.

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

  9. mathcalculus
    • one year ago
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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?

  10. Paynesdad
    • one year ago
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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. Psymon
    • one year ago
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    Because I rather not do too much at once and worry about mistakes.

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

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

  14. mathcalculus
    • one year ago
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    yes i see

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

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

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

  18. mathcalculus
    • one year ago
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    ok

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

  20. mathcalculus
    • one year ago
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    right

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

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

  23. mathcalculus
    • one year ago
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    @Paynesdad

  24. Paynesdad
    • one year ago
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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?

  25. mathcalculus
    • one year ago
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    yes

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

  27. mathcalculus
    • one year ago
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    correct

  28. Paynesdad
    • one year ago
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    So now look at your equation ... what things and how many can you factor out of both terms

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

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

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

  32. Paynesdad
    • one year ago
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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.

  33. mathcalculus
    • one year ago
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    1

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

  35. mathcalculus
    • one year ago
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    @Paynesdad right i know i can factor that too but what about the outer numbers. should i factor that firsT?

  36. mathcalculus
    • one year ago
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    2

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

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

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

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

  41. Paynesdad
    • one year ago
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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.

  42. Psymon
    • one year ago
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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 ^_^

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

  44. Paynesdad
    • one year ago
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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.,,

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

  46. Paynesdad
    • one year ago
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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.

  47. Psymon
    • one year ago
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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

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

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

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

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

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

  53. Paynesdad
    • one year ago
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    That looks good

  54. mathcalculus
    • one year ago
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    @Paynesdad

  55. Paynesdad
    • one year ago
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    now simplify wh

  56. Paynesdad
    • one year ago
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    is in the backets

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

  58. Paynesdad
    • one year ago
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    *brackets

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

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

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

  62. mathcalculus
    • one year ago
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    ook i did

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

  64. Paynesdad
    • one year ago
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    Keep going

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

  66. Paynesdad
    • one year ago
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    keep going

  67. mathcalculus
    • one year ago
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    fine

  68. Paynesdad
    • one year ago
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    :-)

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

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

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

  72. mathcalculus
    • one year ago
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    oops 25

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

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

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

  76. Paynesdad
    • one year ago
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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

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

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

  79. Paynesdad
    • one year ago
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    indeed*

  80. mathcalculus
    • one year ago
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    the 2?

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

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

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

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

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

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

  87. mathcalculus
    • one year ago
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    ok

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

  89. mathcalculus
    • one year ago
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    -5

  90. Paynesdad
    • one year ago
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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.

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

  92. Paynesdad
    • one year ago
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    Whoa whoa whoa

  93. Paynesdad
    • one year ago
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    You

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

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

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

  97. mathcalculus
    • one year ago
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    2t-5

  98. Paynesdad
    • one year ago
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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?

  99. mathcalculus
    • one year ago
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    right

  100. Paynesdad
    • one year ago
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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

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

  102. Paynesdad
    • one year ago
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    So

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

  104. Paynesdad
    • one year ago
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    @mathcalculus

  105. mathcalculus
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    3

  106. Paynesdad
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    nope remember terms are things that are separated by + or - and not in parantheses.

  107. mathcalculus
    • one year ago
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    1

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

  109. mathcalculus
    • one year ago
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    right

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

  111. mathcalculus
    • one year ago
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    ?

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

  113. Paynesdad
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    7(x+1) (x-2) has 3 factors too

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

  115. mathcalculus
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    3?

  116. Paynesdad
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    yep three things multiplied together... good

  117. mathcalculus
    • one year ago
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    ok

  118. mathcalculus
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    for a small problem.. it's so long..:?

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

  120. mathcalculus
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    okay 4.

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

  122. mathcalculus
    • one year ago
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    next

  123. mathcalculus
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    yeah

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

  125. mathcalculus
    • one year ago
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    idk

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

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

  128. mathcalculus
    • one year ago
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    ask*

  129. Paynesdad
    • one year ago
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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?

  130. mathcalculus
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    0

  131. mathcalculus
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    1and 0

  132. Paynesdad
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    right! so look at the original equation...if you set that to 0 you cant solve for t

  133. mathcalculus
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    omg obvious......

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

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

  136. mathcalculus
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    whats next after that.

  137. Paynesdad
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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.!

  138. mathcalculus
    • one year ago
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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.

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

  140. mathcalculus
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    they've multiplied -10 to 2t-5

  141. mathcalculus
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    but why? the last one?

  142. mathcalculus
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    not the 1st, nor the 2nd?

  143. Paynesdad
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    Right that is exactly what you had but YOU factored the -10 out...they didn't.

  144. Paynesdad
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    Hang on you are thinking of distirbution

  145. Paynesdad
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    you don't distribute over multiplication !

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

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

  148. Paynesdad
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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

  149. Paynesdad
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    No problem ... Good luck in tough subject but honestly

  150. Paynesdad
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    you have a better comprehension then 50% of the students I have had. @mathcalculus

  151. Paynesdad
    • one year ago
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    *than

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spraguer (Moderator)
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