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anonymous

  • 5 years ago

Factor 4sˆ5+16s³-32s-20 4s^5 is suppose to be 4s exponent5 but i dont know how to do it

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  1. anonymous
    • 5 years ago
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    What have you done so far?

  2. anonymous
    • 5 years ago
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    nothing

  3. anonymous
    • 5 years ago
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    Well, do you know how to find the factors of each term?

  4. anonymous
    • 5 years ago
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    cf is 4

  5. anonymous
    • 5 years ago
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    That is a start. Pull a factor of 4 from each term and see what you have left.

  6. anonymous
    • 5 years ago
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    4, 8, 5

  7. anonymous
    • 5 years ago
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    No, I mean write the new expression after you factor out a 4

  8. anonymous
    • 5 years ago
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    4(4s^3-8s-5)

  9. anonymous
    • 5 years ago
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    Not quite. You're missing a s^5, but otherwise that's good.

  10. anonymous
    • 5 years ago
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    Now what other factors do some of your terms have in common?

  11. anonymous
    • 5 years ago
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    s

  12. anonymous
    • 5 years ago
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    Ok, so factor those out from the terms that have it

  13. anonymous
    • 5 years ago
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    not sure how

  14. anonymous
    • 5 years ago
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    show me step by step plz and then explain

  15. anonymous
    • 5 years ago
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    How did you factor out the 4?

  16. anonymous
    • 5 years ago
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    4(4s^3-8s-5)

  17. anonymous
    • 5 years ago
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    Right, but I mean what did you do?

  18. anonymous
    • 5 years ago
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    Also that's not right

  19. anonymous
    • 5 years ago
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    divide each term by 4

  20. anonymous
    • 5 years ago
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    You keep forgetting the 4s^5

  21. anonymous
    • 5 years ago
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    then what would it be

  22. anonymous
    • 5 years ago
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    4(s^5 +4s^3 - 8s - 5)

  23. anonymous
    • 5 years ago
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    But as I said, you can factor an s from the first 3 terms inside the parens.

  24. anonymous
    • 5 years ago
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    Or rather, like you said.

  25. anonymous
    • 5 years ago
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    And you would do it the same way. Put an s out front, and divide each term by the s you took away.

  26. anonymous
    • 5 years ago
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    Just don't take one away from the term that doesn't have an s to start with.

  27. anonymous
    • 5 years ago
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    s(^5+4^3-8-5)

  28. anonymous
    • 5 years ago
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    No. What is \[\frac{s^5}{s}\]

  29. anonymous
    • 5 years ago
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    idk im really bad with fractions

  30. anonymous
    • 5 years ago
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    Then you should practice! =) Everything depends on everything else. If you practice your fractions until you are not bad with them, factoring will be a very fast process.

  31. anonymous
    • 5 years ago
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    Lets start now. What is \[\frac{32}{2}\]

  32. anonymous
    • 5 years ago
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    16

  33. anonymous
    • 5 years ago
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    Ok, and if I said \[\frac{2^5}{2}\] ?

  34. anonymous
    • 5 years ago
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    rawr!

  35. anonymous
    • 5 years ago
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    a^5

  36. anonymous
    • 5 years ago
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    What is \(2^5\)

  37. anonymous
    • 5 years ago
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    32

  38. anonymous
    • 5 years ago
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    Interesting. And what was \[\frac{32}{2} \] again?

  39. anonymous
    • 5 years ago
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    16/1

  40. anonymous
    • 5 years ago
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    Right.. Just curious, but what is 2^4?

  41. anonymous
    • 5 years ago
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    rawr! rawr!

  42. anonymous
    • 5 years ago
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    16

  43. anonymous
    • 5 years ago
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    That's interesting. \[\frac{2^5}{2} = 2^4\]

  44. anonymous
    • 5 years ago
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    Does this work with other numbers? \[3^3 = 27\] \[\frac{27}{3} =\ ?\]

  45. anonymous
    • 5 years ago
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    9

  46. anonymous
    • 5 years ago
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    And 9 is \(3^2\)

  47. anonymous
    • 5 years ago
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    Lets see if we can figure out why this might be happening.

  48. anonymous
    • 5 years ago
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    What is \(a^2\)

  49. anonymous
    • 5 years ago
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    Or better still, \(a^3\)

  50. anonymous
    • 5 years ago
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    a*a*a

  51. anonymous
    • 5 years ago
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    Ok, now if we have \[\frac{a\cdot a\cdot a}{a}\] We can cancel one of the a's on top with the a on the bottom. So what would that equal?

  52. anonymous
    • 5 years ago
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    a^2

  53. anonymous
    • 5 years ago
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    Right. So now, coming back to the original question.. What is \[\frac{s^5}{s}\]

  54. anonymous
    • 5 years ago
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    5

  55. anonymous
    • 5 years ago
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    no.

  56. anonymous
    • 5 years ago
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    then just s

  57. anonymous
    • 5 years ago
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    What was \[\frac{a^3}{a}\] again?

  58. anonymous
    • 5 years ago
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    a*a*a

  59. anonymous
    • 5 years ago
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    No, you just said it 6 mins ago. just scroll up

  60. anonymous
    • 5 years ago
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    a^2

  61. anonymous
    • 5 years ago
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    And do you remember why?

  62. anonymous
    • 5 years ago
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    because we cancel out 1

  63. anonymous
    • 5 years ago
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    So now. What is \[\frac{s^5}{s}\] Use the same argument as before

  64. anonymous
    • 5 years ago
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    s^4

  65. anonymous
    • 5 years ago
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    yes, precisely.

  66. anonymous
    • 5 years ago
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    So if we divide s^5 by s, we get s^4

  67. anonymous
    • 5 years ago
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    And if we divide s^3 by s?

  68. anonymous
    • 5 years ago
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    s^2

  69. anonymous
    • 5 years ago
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    And if we divide s by s?

  70. anonymous
    • 5 years ago
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    s

  71. anonymous
    • 5 years ago
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    s/s = s?

  72. anonymous
    • 5 years ago
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    What is 2/2

  73. anonymous
    • 5 years ago
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    1

  74. anonymous
    • 5 years ago
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    What is 5/5

  75. anonymous
    • 5 years ago
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    1

  76. anonymous
    • 5 years ago
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    And w/w?

  77. anonymous
    • 5 years ago
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    w

  78. anonymous
    • 5 years ago
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    Why was 5/5 = 1?

  79. anonymous
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    s/s=1

  80. anonymous
    • 5 years ago
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    why?

  81. anonymous
    • 5 years ago
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    s goes into s 1 time

  82. anonymous
    • 5 years ago
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    that's a good reason.

  83. anonymous
    • 5 years ago
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    So now.

  84. anonymous
    • 5 years ago
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    thats why i said s i thought instead of putting 1 you put the variable

  85. anonymous
    • 5 years ago
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    If we need to factor an s from s^5 +4s^3 - 8s what will our expression be?

  86. anonymous
    • 5 years ago
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    s^4+s^2-8

  87. anonymous
    • 5 years ago
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    not quite.

  88. anonymous
    • 5 years ago
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    But very close!

  89. anonymous
    • 5 years ago
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    i hate this math

  90. anonymous
    • 5 years ago
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    \[s^5 + 4s^3 - 8s\]

  91. anonymous
    • 5 years ago
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    factor out an s, does that effect the numbers out in front?

  92. anonymous
    • 5 years ago
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    no

  93. anonymous
    • 5 years ago
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    so where'd the 4 go when you factored?

  94. anonymous
    • 5 years ago
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    s(s^4+4s^2-8-5)?

  95. anonymous
    • 5 years ago
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    Where'd that 5 come from?

  96. anonymous
    • 5 years ago
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    the original equation

  97. anonymous
    • 5 years ago
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    jumped the gun a bit

  98. anonymous
    • 5 years ago
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    That 5 didn't have an s

  99. anonymous
    • 5 years ago
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    So we cannot factor a s from it

  100. anonymous
    • 5 years ago
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    no

  101. anonymous
    • 5 years ago
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    Our original expression is \[4( s^5 + 4s^3 -8s -5)\] And you said that \[s^5 + 4s^3 -8s = s(s^4 + 4s^2 -8)\] Right?

  102. anonymous
    • 5 years ago
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    yes what happens to the 5

  103. anonymous
    • 5 years ago
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    We leave it alone. We are going to replace \[s^5 + 4s^3 -8s\] in the original expression with \[s(s^4 + 4s^2 -8)\] What does it look like?

  104. anonymous
    • 5 years ago
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    ?

  105. anonymous
    • 5 years ago
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    What is the original equation (after we factored out the 4)

  106. anonymous
    • 5 years ago
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    4(s^5+4s^3-8s-5)

  107. anonymous
    • 5 years ago
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    Right. So now take all those terms with an s, and replace them with the factored form you found.

  108. anonymous
    • 5 years ago
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    s(s^4+4s^2-8)-5

  109. anonymous
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    Where'd the 4 out front go?

  110. anonymous
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    4(s^4+4s^2-8)-5

  111. anonymous
    • 5 years ago
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    No. Take a second. Look at the original equation \[4(s^5+4s^3-8s-5)\] I want you to retype this equation with a [ and a ] around the terms with an s.

  112. anonymous
    • 5 years ago
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    [4(s^5+4s^3-8s]-5)

  113. anonymous
    • 5 years ago
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    hrm. I think there's a vocab problem

  114. anonymous
    • 5 years ago
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    what is a term?

  115. anonymous
    • 5 years ago
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    each number 4s^5

  116. anonymous
    • 5 years ago
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    close, but not quite

  117. anonymous
    • 5 years ago
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    I realize this has been a very lengthy process, but bear with me for a min

  118. anonymous
    • 5 years ago
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    im really bad at math

  119. anonymous
    • 5 years ago
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    I want you to know. You can be very good at math. You have the skills. You are a very hard worker clearly.

  120. anonymous
    • 5 years ago
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    What is your goal?

  121. anonymous
    • 5 years ago
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    it just doesnt make sense to me

  122. anonymous
    • 5 years ago
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    I know that, but it's not for the reasons you think.

  123. anonymous
    • 5 years ago
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    Somewhere a long time ago, you were learning math. And there were other classes, and due dates, or other life events, and somewhere along the way you missed something here and there. Not a lot, just a little bit. But it was a little bit here, and a little bit there, and slowly over time these little bits add up. They are not hard things to learn. But now you are expected to know them and they are holes in your knowlege. And these holes are making it 100 times harder to learn this new material.

  124. anonymous
    • 5 years ago
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    So lets fill a few holes quickly

  125. anonymous
    • 5 years ago
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    do you mind?

  126. anonymous
    • 5 years ago
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    I promise it will make math less hard

  127. anonymous
    • 5 years ago
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    And I assume that you have a reason for taking the math classes you are taking

  128. anonymous
    • 5 years ago
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    and don't want to give up on those goals

  129. anonymous
    • 5 years ago
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    Believe me. You can learn this.

  130. anonymous
    • 5 years ago
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    im majoring in acct

  131. anonymous
    • 5 years ago
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    That's awsome!

  132. anonymous
    • 5 years ago
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    i drop out in 9th grade got my GED and somehow passed the math test

  133. anonymous
    • 5 years ago
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    12yrs ago i dropped out and got GED 1yr ago

  134. anonymous
    • 5 years ago
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    Sounds like my story ;)

  135. anonymous
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    Except I graduated, but not with terribly good grades. Then screwed around for 15 years before finally going back to school

  136. anonymous
    • 5 years ago
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    i had my son dropped out and raised him and had my daughter and dont want to wait anymore

  137. anonymous
    • 5 years ago
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    That's good!

  138. anonymous
    • 5 years ago
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    yeah so how do i make this easier

  139. anonymous
    • 5 years ago
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    So lets learn some math. =)

  140. anonymous
    • 5 years ago
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    Starting with vocabulary

  141. anonymous
    • 5 years ago
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    Vocabulary is important, because that way when I say something you can know exactly what I mean.

  142. anonymous
    • 5 years ago
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    can we finish the problem we were working on first so i can close my homework this is the last problem

  143. anonymous
    • 5 years ago
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    I was trying to.

  144. anonymous
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    But you weren't understanding what I was saying

  145. anonymous
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    k

  146. anonymous
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    So vocabulary.

  147. anonymous
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    First off, we have a sum. This is a sum: a + b + c

  148. anonymous
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    Very simple I know

  149. anonymous
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    right

  150. anonymous
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    This sum has how many terms?

  151. anonymous
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    A term is just the things you are summing together.

  152. anonymous
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    How many terms are in the sum: a+b+c ?

  153. anonymous
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    3

  154. anonymous
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    And how many terms are in the sum a*b + c +d*e

  155. anonymous
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    4

  156. anonymous
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    How many different things are we adding together?

  157. anonymous
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    a*b+c c+d*e so 2?

  158. anonymous
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    (a*b) + (c) + (d*e)

  159. anonymous
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    so 3

  160. anonymous
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    Yes. There are 3 things being added together. Those things are called terms.

  161. anonymous
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    Now this is a product: a*b

  162. anonymous
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    right

  163. anonymous
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    What are the factors of this product?

  164. anonymous
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    a b

  165. anonymous
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    So how many factors are there?

  166. anonymous
    • 5 years ago
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    2

  167. anonymous
    • 5 years ago
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    Ok now 5 - 3a + 8c Is that a product, or a sum?

  168. anonymous
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    difference of a sum

  169. anonymous
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    I'd rather say it was a sum with one negative term

  170. anonymous
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    How many terms are in this sum?

  171. anonymous
    • 5 years ago
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    2

  172. anonymous
    • 5 years ago
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    How many different things are being added (and subtracted) together?

  173. anonymous
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    5-3a 3a+8c

  174. anonymous
    • 5 years ago
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    Why do you keep breaking it up with repeated parts?

  175. anonymous
    • 5 years ago
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    Is 5 - 3 + 2 = 5 - 3 and 3+2 ?

  176. anonymous
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    no

  177. anonymous
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    So what do you mean 5-3a 3a + 8c ?

  178. anonymous
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    5 - 3 + 2 Has how many terms?

  179. anonymous
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    2

  180. anonymous
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    (5) - (3) + (2)

  181. anonymous
    • 5 years ago
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    Or if you prefer (5) + (-3) + (2)

  182. anonymous
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    How many things are being added together?

  183. anonymous
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    3

  184. anonymous
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    yes

  185. anonymous
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    5 - 4 + 2 + 6

  186. anonymous
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    How many terms?

  187. anonymous
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    so each number is a term

  188. anonymous
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    no, each thing you are adding together as part of the sum is a term

  189. anonymous
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    a + b + c Has no numbers, but still has 3 terms

  190. anonymous
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    so 4a+6b+8c is 3 or 6 terms

  191. anonymous
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    3 terms.

  192. anonymous
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    k

  193. anonymous
    • 5 years ago
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    But! 4*3c + 7b + 4*12*7a Also has 3 terms

  194. anonymous
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    What is the second term of the above expression?

  195. anonymous
    • 5 years ago
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    7b

  196. anonymous
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    right.

  197. anonymous
    • 5 years ago
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    Ok, now going back to products

  198. anonymous
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    7b is a product

  199. anonymous
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    you know that 7b is the same as 7*b yes?

  200. anonymous
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    I think you're being trolled...

  201. anonymous
    • 5 years ago
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    yes or 7(b)

  202. anonymous
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    Yes, exactly

  203. anonymous
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    Ok, so how many factors are in the product 7b?

  204. anonymous
    • 5 years ago
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    2

  205. anonymous
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    good

  206. anonymous
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    Ok, so what is \(7bc^2\) a sum, or a product?

  207. anonymous
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    product

  208. anonymous
    • 5 years ago
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    Ok, here's a trickier one. 4(a + 3) Sum or a product?

  209. anonymous
    • 5 years ago
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    sum

  210. anonymous
    • 5 years ago
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    If I tell you that b = a+3 Wouldn't that expression be 4b ?

  211. anonymous
    • 5 years ago
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    is 4b a sum or a product?

  212. anonymous
    • 5 years ago
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    product

  213. anonymous
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    If 4b is a product, then 4(a+3) must be a product, cause they're basically the same expression

  214. anonymous
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    ok

  215. anonymous
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    So if 4b is a product, what are its factors?

  216. anonymous
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    3,1

  217. anonymous
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    No. The factors of 4b are 4, and b

  218. anonymous
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    oh yeah sry

  219. anonymous
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    So if b = a+3 What are the factors of the product 4(a+3)

  220. anonymous
    • 5 years ago
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    a 3

  221. anonymous
    • 5 years ago
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    What are the factors of 4b again?

  222. anonymous
    • 5 years ago
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    4 b

  223. anonymous
    • 5 years ago
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    Ok, in the expression (the product) 4(a+3) What are the factors (things being multiplied together to form the product)?

  224. anonymous
    • 5 years ago
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    4 a 4 3

  225. anonymous
    • 5 years ago
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    No.

  226. anonymous
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    4(a+3) is 4 times (a+3) What two things are being multiplied!!?

  227. anonymous
    • 5 years ago
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    4 (a+3)

  228. anonymous
    • 5 years ago
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    yes

  229. anonymous
    • 5 years ago
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    The product 4(a+3) has 2 factors. They are 4 and (a+3)

  230. anonymous
    • 5 years ago
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    One of those factors is a sum. Which one is a sum?

  231. anonymous
    • 5 years ago
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    (a+3)

  232. anonymous
    • 5 years ago
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    What are the terms of the sum (a+3)

  233. anonymous
    • 5 years ago
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    a 3

  234. anonymous
    • 5 years ago
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    Correct

  235. anonymous
    • 5 years ago
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    Lets try this one

  236. anonymous
    • 5 years ago
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    (x+1)(x-2) A product? or a sum?

  237. anonymous
    • 5 years ago
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    (x+1) sum (x-2) difference

  238. anonymous
    • 5 years ago
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    no.

  239. anonymous
    • 5 years ago
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    It is a product. (x+1)(x-2) is (x+1) TIMES (x-2) It is two things multiplied together.

  240. anonymous
    • 5 years ago
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    oh ok i get it

  241. anonymous
    • 5 years ago
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    Each of those two things by themselves is a sum, but when you multiply them together it's a product.

  242. anonymous
    • 5 years ago
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    What is 4(s^5 + 4s^3 - 8s -5) A product? or a sum?

  243. anonymous
    • 5 years ago
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    product

  244. anonymous
    • 5 years ago
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    yay!

  245. anonymous
    • 5 years ago
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    What are the factors of this product?

  246. anonymous
    • 5 years ago
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    4 s^5+4s^3-8s-5

  247. anonymous
    • 5 years ago
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    yes!

  248. anonymous
    • 5 years ago
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    Now, the second factor has many terms with s in them

  249. anonymous
    • 5 years ago
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    I want you to find each term with a factor of s

  250. anonymous
    • 5 years ago
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    s^5 4s^3 8s

  251. anonymous
    • 5 years ago
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    Yes, now. Factor an s out of each of those terms, this will make a product where one factor is an s, and the other factor is the sum of the terms that you took the s from.

  252. anonymous
    • 5 years ago
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    s+4^3-8

  253. anonymous
    • 5 years ago
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    I'm not sure where to begin there. You were doing well up until that point, so something about my instructions must not have made sense.

  254. anonymous
    • 5 years ago
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    i just dont understand what you mean

  255. anonymous
    • 5 years ago
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    4 + 16 + 12 Factor a 4 from each of those terms.

  256. anonymous
    • 5 years ago
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    4(1+4+3)

  257. anonymous
    • 5 years ago
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    yes! perfect

  258. anonymous
    • 5 years ago
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    4a + 3a + 2a Factor an a from each of those terms

  259. anonymous
    • 5 years ago
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    a(4+3+2)^3

  260. anonymous
    • 5 years ago
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    What is the ^3 for?

  261. anonymous
    • 5 years ago
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    Why didn't you do the same thing you did for the 4?

  262. anonymous
    • 5 years ago
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    i thought for the a's

  263. anonymous
    • 5 years ago
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    It wasn't 4(1+4+3)^3

  264. anonymous
    • 5 years ago
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    Just put an 'a' out front, and then divide each term by a.

  265. anonymous
    • 5 years ago
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    ok

  266. anonymous
    • 5 years ago
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    So what is it?

  267. anonymous
    • 5 years ago
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    s(^5 +4^3-8)

  268. anonymous
    • 5 years ago
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    No. We went over this before. s^5 divided by s is?

  269. anonymous
    • 5 years ago
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    s^4

  270. anonymous
    • 5 years ago
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    s(s^4+4s^2-8)

  271. anonymous
    • 5 years ago
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    Yes

  272. anonymous
    • 5 years ago
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    Now take the terms in the original equation that you factored that s from, and put square brackets [] around them

  273. anonymous
    • 5 years ago
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    [4(s^4+4s^3-8s-5]

  274. anonymous
    • 5 years ago
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    a) that is not the original equation b) the 4 out front is not one of the terms you took an s from

  275. anonymous
    • 5 years ago
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    If you do not understand the instructions why don't you ask to clarify?

  276. anonymous
    • 5 years ago
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    4s^5+16s^3-32s-20 this is original equation where do i put the brackets i got s from all except the -20

  277. anonymous
    • 5 years ago
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    That is a very well thought out question. Thank you. I would like you to go to the form of the expression where we factored out the 4 already. Then place the brackets such that they surround ONLY the terms (from the sum) that you took the s from, and no others.

  278. anonymous
    • 5 years ago
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    Realize that you haven't actually factored the s out yet. I just want you do denote which terms you will be taking the s from by placing them in the brackets

  279. anonymous
    • 5 years ago
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    s(s^4+4s^2-8) this is where i factored the s

  280. anonymous
    • 5 years ago
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    just guessing is [4+4][2-8] the answer you are looking for?

  281. anonymous
    • 5 years ago
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    Ok This is the expression I want you to work with: \[4(s^5 + 4s^3 - 8s -5)\] Put square brackets around the terms that have s in them

  282. anonymous
    • 5 years ago
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    4([s^5+4s^3-8s]-5)

  283. anonymous
    • 5 years ago
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    yes!

  284. anonymous
    • 5 years ago
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    Now, do you agree that s^5+4s^3-8s = s(s^4 + 4s^3 - 8) And that s^5+4s^3-8s is what is inside the brackets?

  285. anonymous
    • 5 years ago
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    yes

  286. anonymous
    • 5 years ago
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    Good. Then delete what is inside the brackets and replace it with the version where the s is factored out. You should not change anything not inside the brackets.

  287. anonymous
    • 5 years ago
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    Bah!

  288. anonymous
    • 5 years ago
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    You didn't check my work ;p I made a mistake

  289. anonymous
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    It should be s(s^4 + 4s^2 - 8)

  290. anonymous
    • 5 years ago
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    I had a 3 on the power of the middle term.

  291. anonymous
    • 5 years ago
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    4([s^4+4s^2-8]-5)

  292. anonymous
    • 5 years ago
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    where'd the s out in front go?

  293. anonymous
    • 5 years ago
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    4s([s^4+4s^2-8]-5)

  294. anonymous
    • 5 years ago
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    you changed something outside the brackets

  295. anonymous
    • 5 years ago
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    so then explain where im suppose to put the s that was in front

  296. anonymous
    • 5 years ago
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    Exactly where it was in the factored version. s(s^4 + 4s^2 -8)

  297. anonymous
    • 5 years ago
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    4(s[s^4+4s^2-8]-5)

  298. anonymous
    • 5 years ago
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    Delete what's in the brackets. Take that whole expression, and put it in the brackets instead.

  299. anonymous
    • 5 years ago
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    Sure, that is at least correct.

  300. anonymous
    • 5 years ago
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    If you want, you can switch the brackets to parenthesis.

  301. anonymous
    • 5 years ago
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    is that the right answer

  302. anonymous
    • 5 years ago
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    almost. We still have more we can factor.

  303. anonymous
    • 5 years ago
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    \[4(s (s^4 + 4s^2 - 8) - 5)\] There are two terms that have an \(s^2\)

  304. anonymous
    • 5 years ago
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    4(s[s^4 + 4s^2 -8)]-5) is this what it should look like with the brackets

  305. anonymous
    • 5 years ago
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    yes, except for the ) after the 8

  306. anonymous
    • 5 years ago
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    sry forgot to take that out

  307. anonymous
    • 5 years ago
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    Which two terms in this expression have an s still?

  308. anonymous
    • 5 years ago
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    s^4 s^2

  309. anonymous
    • 5 years ago
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    s^4 and 4s^2

  310. anonymous
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    The 4 is part of the term

  311. anonymous
    • 5 years ago
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    yes forgot the 4

  312. anonymous
    • 5 years ago
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    Ok, so you can factor an s^2 from each of those terms s^4 + 4s^2 = s^2( ? + ? )

  313. anonymous
    • 5 years ago
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    show me

  314. anonymous
    • 5 years ago
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    How did you factor out the 4, or the s, or anything else you've factored so far?

  315. anonymous
    • 5 years ago
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    2(s^2+4

  316. anonymous
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    I can only assume you made some typos there, but you meant to have s^2(s^2 + 4)

  317. anonymous
    • 5 years ago
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    yes ty

  318. anonymous
    • 5 years ago
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    Ok, so replace those two terms (ONLY) with the newly factored version of the expression

  319. anonymous
    • 5 years ago
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    The two terms in the big expression we are working on.

  320. anonymous
    • 5 years ago
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    4(s[s^2+4-8]-5)

  321. anonymous
    • 5 years ago
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    where did the s^2 out in front go?

  322. anonymous
    • 5 years ago
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    And the parenthases?

  323. anonymous
    • 5 years ago
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    s^4 + 4s^3 = s^2(s^2 + 4) So take this [s^4 + 4s^3] and put in this [s^2(s^2 + 4)]

  324. anonymous
    • 5 years ago
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    Err I miss typed that exponent again, but you know what I mean.

  325. anonymous
    • 5 years ago
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    s^4 + 4s^2 = s^2(s^2 + 4) Take out [ s^4 + 4s^2 ] Put in [ s^2(s^2 + 4) ]

  326. anonymous
    • 5 years ago
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    4(s[s^2(s^2 + 4) -8]-5)

  327. anonymous
    • 5 years ago
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    yes. exactly. the end.

  328. anonymous
    • 5 years ago
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    There are no terms that have any common factors, so we cannot factor anything else out anywhere.

  329. anonymous
    • 5 years ago
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    4(s[s^2(s^2 + 4) -8]-5) this is the final answer

  330. anonymous
    • 5 years ago
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    Yes. Now I'm going to go do my math homework and go to bed before I have to get up in 5 hours.

  331. anonymous
    • 5 years ago
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    i typed in the answer we came up with and it is wrong the answer is 4(s^5+4s^3-8s-5) i had that 2 and 1/2 hrs ago

  332. anonymous
    • 5 years ago
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    That's the problem with having a computer grade. The form we came up with is correct. You can factor it a lot of different ways. But "Factor" doesn't make it clear which bits to factor, or how far to factor.

  333. anonymous
    • 5 years ago
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    At a minimum the instructions should have been to factor using the greatest common factor.

  334. anonymous
    • 5 years ago
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    then at least you would know that the factorization you wanted would include all those terms.

  335. anonymous
    • 5 years ago
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    and also you learned a lot more about factoring from the extra practice.

  336. anonymous
    • 5 years ago
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    yes i did ty for everything

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