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mathcalculus

  • 2 years 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
    • 2 years ago
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    Show me what you have done.

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

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

  4. mathcalculus
    • 2 years ago
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    1

  5. Paynesdad
    • 2 years ago
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    ok so that is part of it

  6. Psymon
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Do you follow what @psymon did?

  9. mathcalculus
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Because I rather not do too much at once and worry about mistakes.

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

  13. Paynesdad
    • 2 years ago
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    But Psymon could have done that too.

  14. mathcalculus
    • 2 years ago
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    yes i see

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

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

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

  18. mathcalculus
    • 2 years ago
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    ok

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

  20. mathcalculus
    • 2 years ago
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    right

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

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

  23. mathcalculus
    • 2 years ago
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    @Paynesdad

  24. Paynesdad
    • 2 years 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
    • 2 years ago
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    yes

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

  27. mathcalculus
    • 2 years ago
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    correct

  28. Paynesdad
    • 2 years 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
    • 2 years ago
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    thats what i was saying... 2 goes in for 4 and -6..

  30. Paynesdad
    • 2 years ago
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    Ok what other factors can you factor out?

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

  32. Paynesdad
    • 2 years 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
    • 2 years ago
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    1

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

  35. mathcalculus
    • 2 years 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
    • 2 years ago
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    2

  37. Paynesdad
    • 2 years ago
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    you can it shouldn't matter.

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

  39. mathcalculus
    • 2 years ago
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    yes it does.... for ex::

  40. mathcalculus
    • 2 years ago
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    see they factored out using the 4

  41. Paynesdad
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    i want to know why can't i in this problem. which confuses so much.

  44. Paynesdad
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    @Psymon got it^_^

  49. mathcalculus
    • 2 years ago
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    that helped out alot. then what?

  50. mathcalculus
    • 2 years ago
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    don't i need to finish?

  51. Paynesdad
    • 2 years ago
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    Show me what you have. so far.

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

  53. Paynesdad
    • 2 years ago
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    That looks good

  54. mathcalculus
    • 2 years ago
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    @Paynesdad

  55. Paynesdad
    • 2 years ago
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    now simplify wh

  56. Paynesdad
    • 2 years ago
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    is in the backets

  57. Paynesdad
    • 2 years ago
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    now simplify what is in the b

  58. Paynesdad
    • 2 years ago
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    *brackets

  59. mathcalculus
    • 2 years ago
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    what about the left side?

  60. mathcalculus
    • 2 years ago
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    add them all up?

  61. Paynesdad
    • 2 years ago
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    no don't just the stuff in the brackets

  62. mathcalculus
    • 2 years ago
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    ook i did

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

  64. Paynesdad
    • 2 years ago
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    Keep going

  65. mathcalculus
    • 2 years ago
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    oh 16-10t+9

  66. Paynesdad
    • 2 years ago
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    keep going

  67. mathcalculus
    • 2 years ago
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    fine

  68. Paynesdad
    • 2 years ago
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    :-)

  69. mathcalculus
    • 2 years ago
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    -10t+7

  70. Paynesdad
    • 2 years ago
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    errrrrr no re-think that last step

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

  72. mathcalculus
    • 2 years ago
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    oops 25

  73. mathcalculus
    • 2 years ago
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    -10t+25

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

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

  76. Paynesdad
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    yes it is in deed helpful.

  79. Paynesdad
    • 2 years ago
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    indeed*

  80. mathcalculus
    • 2 years ago
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    the 2?

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

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

  83. Paynesdad
    • 2 years ago
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    no Look ---> (-10t+25)

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

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

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

  87. mathcalculus
    • 2 years ago
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    ok

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

  89. mathcalculus
    • 2 years ago
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    -5

  90. Paynesdad
    • 2 years 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
    • 2 years ago
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    2*-5(2t-3)(8-2t)^2[25]

  92. Paynesdad
    • 2 years ago
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    Whoa whoa whoa

  93. Paynesdad
    • 2 years ago
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    You

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

  95. mathcalculus
    • 2 years ago
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    you mean in the brackets t+25

  96. Paynesdad
    • 2 years 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
    • 2 years ago
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    2t-5

  98. Paynesdad
    • 2 years 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
    • 2 years ago
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    right

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

  102. Paynesdad
    • 2 years ago
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    So

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

  104. Paynesdad
    • 2 years ago
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    @mathcalculus

  105. mathcalculus
    • 2 years ago
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    3

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

  107. mathcalculus
    • 2 years ago
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    1

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

  109. mathcalculus
    • 2 years ago
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    right

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

  111. mathcalculus
    • 2 years ago
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    ?

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

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

  114. mathcalculus
    • 2 years ago
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    this is confusing me more.

  115. mathcalculus
    • 2 years ago
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    3?

  116. Paynesdad
    • 2 years ago
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    yep three things multiplied together... good

  117. mathcalculus
    • 2 years ago
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    ok

  118. mathcalculus
    • 2 years ago
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    for a small problem.. it's so long..:?

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

  120. mathcalculus
    • 2 years ago
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    okay 4.

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

  122. mathcalculus
    • 2 years ago
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    next

  123. mathcalculus
    • 2 years ago
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    yeah

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

  125. mathcalculus
    • 2 years ago
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    idk

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

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

  128. mathcalculus
    • 2 years ago
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    ask*

  129. Paynesdad
    • 2 years 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
    • 2 years ago
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    0

  131. mathcalculus
    • 2 years ago
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    1and 0

  132. Paynesdad
    • 2 years ago
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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
    • 2 years ago
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    it's the factoring i need help with. that's all

  136. mathcalculus
    • 2 years ago
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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
    • 2 years 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
    • 2 years ago
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    im looking for t right.

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

  141. mathcalculus
    • 2 years ago
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    but why? the last one?

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

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

  144. Paynesdad
    • 2 years ago
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    Hang on you are thinking of distirbution

  145. Paynesdad
    • 2 years ago
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    you don't distribute over multiplication !

  146. mathcalculus
    • 2 years ago
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    this is taking forever.. for a small problem ___-

  147. mathcalculus
    • 2 years ago
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    thank you for you help @Paynesdad really appreciate it.

  148. Paynesdad
    • 2 years ago
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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
    • 2 years ago
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    No problem ... Good luck in tough subject but honestly

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

  151. Paynesdad
    • 2 years ago
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    *than

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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