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anonymous

  • one year ago

If we divide the polynomial x4 + 4x3 + 2x2 + x + 4 by x2 + 3x, what will be the remainder

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  1. UsukiDoll
    • one year ago
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    \[x^4+4x^3+2x^2+x+4\] by \[x^2+3x\] have you made any attempts to this problem?

  2. anonymous
    • one year ago
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    x4+4x3+2x2+x+4byx2+3x =x4+4x3+2x2+x+4bx2y+3x Combine Like Terms: =x4+4x3+2x2+x+4bx2y+3x =(4bx2y)+(x4)+(4x3)+(2x2)+(x+3x) =4bx2y+x4+4x3+2x2+4x Answer: =4bx2y+x4+4x3+2x2+4x right?

  3. UsukiDoll
    • one year ago
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    whoa man.. direct answers are against openstudy's code of conduct

  4. UsukiDoll
    • one year ago
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    and this is long division... I'm not doing that until the official poster responds

  5. anonymous
    • one year ago
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    i have made attempts and none of the answer are there

  6. UsukiDoll
    • one year ago
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    so you had tried the long division method to this problem?

  7. anonymous
    • one year ago
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    thats not a direct answer i said it step by step and just to let you know i did this on the other post and they said that they under stood it completely now

  8. anonymous
    • one year ago
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    yes sir i have. would you like to see the answer choices

  9. UsukiDoll
    • one year ago
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    yeah I like to see the choices

  10. anonymous
    • one year ago
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    A) x2 + 3x B) 4x + 4 C) x2 + x – 2 D) 4 E) 0 That is all of the answer choices

  11. UsukiDoll
    • one year ago
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    ok... so this problem is asking to divide... so we need long division. |dw:1435149580706:dw|

  12. UsukiDoll
    • one year ago
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    wow.. this is long.. so first of all we need to get rid of the x^4

  13. anonymous
    • one year ago
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    yea it sucks i'm in Algebra 2.

  14. UsukiDoll
    • one year ago
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    I've learned this in Algebra 1 O_O anyway , are you aware of the exponential rules?

  15. anonymous
    • one year ago
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    Don't you combine them? If its not that then I don't know about it

  16. UsukiDoll
    • one year ago
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    one of the exponential rules is \[n^{a+b}=n^{a}n^{b}\] let n = x and a = 2 \[x^{2+b}\] what does b need to equal in order to get \[x^4 \]?

  17. anonymous
    • one year ago
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    2 if i'm not mistaking

  18. UsukiDoll
    • one year ago
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    yes because \[x^{2+2} =x^2x^2\]

  19. UsukiDoll
    • one year ago
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    so we need an x^2

  20. UsukiDoll
    • one year ago
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    |dw:1435149984739:dw|

  21. anonymous
    • one year ago
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    Okay. Inside of the sqrt there is a 2x^2

  22. UsukiDoll
    • one year ago
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    we have to go slowly.. getting rid of the x^4 and x^3 is a great idea at this point

  23. anonymous
    • one year ago
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    Alright so would we divide x^2+3x to x^4

  24. UsukiDoll
    • one year ago
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    we need to multiply (x^2+3x) by x^2

  25. UsukiDoll
    • one year ago
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    |dw:1435150150946:dw|

  26. anonymous
    • one year ago
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    Okay so its going to be x^4+3x^3

  27. UsukiDoll
    • one year ago
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    yeah...one problem we need to switch the signs.

  28. UsukiDoll
    • one year ago
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    then we can use addition.

  29. UsukiDoll
    • one year ago
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    let's just distribute the - -(x^4+3x^3) = -x^4-3x^3

  30. UsukiDoll
    • one year ago
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    |dw:1435150318360:dw|

  31. UsukiDoll
    • one year ago
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    oh the plus sign is invisible for long division. I'll just put it in () |dw:1435150362975:dw|

  32. UsukiDoll
    • one year ago
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    so what's x^4-x^4 and what's 4x^3-3x^3 ?

  33. anonymous
    • one year ago
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    Okay so if we are trying to change the sign wouldn't it be x^8 and x^6

  34. UsukiDoll
    • one year ago
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    whoa no...

  35. UsukiDoll
    • one year ago
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    you only use the exponent rule to figure out what's missing. Like what I did earlier. with the x^4 = x^{2+b}

  36. anonymous
    • one year ago
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    ok. sorry im not much of a math person. I love history

  37. anonymous
    • one year ago
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    Oh Okay.

  38. UsukiDoll
    • one year ago
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    |dw:1435150555870:dw| let's try this again what's x^4-x^4 and 4x^3-3x^3 ?

  39. anonymous
    • one year ago
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    x and x. right.

  40. anonymous
    • one year ago
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    If we divide the polynomial x4 + 4x3 + 2x2 + x + 4 by x2 + 3x, what will be the remainder? Here is the full question

  41. UsukiDoll
    • one year ago
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    ...... let's make it easier I'm going to factor out the x^3 x^3(4-3)

  42. UsukiDoll
    • one year ago
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    well we need to do long division...this is the right method

  43. UsukiDoll
    • one year ago
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    or synthetic division which is faster but I forgot

  44. UsukiDoll
    • one year ago
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    x^4(1-1) x^3(4-3) try solving those two ^

  45. anonymous
    • one year ago
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    You forgot how to do synthetic division

  46. UsukiDoll
    • one year ago
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    I've practiced more with long division...it stuck in my head.

  47. anonymous
    • one year ago
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    Oh okay, and i got x^4-x^4 and 4x^3-3x^3

  48. UsukiDoll
    • one year ago
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    wow @_@!

  49. UsukiDoll
    • one year ago
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    try again... x^4-x^4 is the same as x^4(1-1) what's 1-1 ?

  50. anonymous
    • one year ago
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    sorry i suck at math.

  51. anonymous
    • one year ago
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    1-1=0

  52. UsukiDoll
    • one year ago
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    yes and what's x^4(0) also known as what's x^4 multiplied by 0

  53. anonymous
    • one year ago
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    0

  54. anonymous
    • one year ago
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    cause anything that is multiplyed by 0 is 0

  55. UsukiDoll
    • one year ago
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    yes! so the x^4's are gone! SImilarly what's x^3(4-3) ? what;s 4-3?!

  56. UsukiDoll
    • one year ago
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    what's 4-3?

  57. anonymous
    • one year ago
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    1

  58. UsukiDoll
    • one year ago
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    yes and x^3 times 1 is?

  59. anonymous
    • one year ago
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    x^3

  60. UsukiDoll
    • one year ago
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    yay!

  61. anonymous
    • one year ago
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    lol im getting it slowly but surely

  62. UsukiDoll
    • one year ago
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    |dw:1435151077510:dw| so after we solve that we bring the rest of the terms down

  63. UsukiDoll
    • one year ago
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    now our new goal is to get rid of the x^3

  64. UsukiDoll
    • one year ago
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    remember the exponent rule?

  65. anonymous
    • one year ago
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    na+b=nanb let n = x and a = 2 x2+b what does b need to equal in order to get x4 that one

  66. UsukiDoll
    • one year ago
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    \[x^{2+b} =x^2x^b\]

  67. UsukiDoll
    • one year ago
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    yup now what value of b is needed to produce x^3?

  68. anonymous
    • one year ago
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    1

  69. UsukiDoll
    • one year ago
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    yes.

  70. anonymous
    • one year ago
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    ok so it would be x^3-x^3

  71. anonymous
    • one year ago
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    which would get rid of each other

  72. UsukiDoll
    • one year ago
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    |dw:1435151215690:dw|

  73. UsukiDoll
    • one year ago
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    now this time x(x^2+3x) and switch signs after distributing

  74. anonymous
    • one year ago
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    ok so it is 4x^3

  75. UsukiDoll
    • one year ago
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    |dw:1435151316916:dw|

  76. UsukiDoll
    • one year ago
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    ?!?!?!! |dw:1435151357797:dw|

  77. anonymous
    • one year ago
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    my bad. x^3-3x^2 Right.

  78. UsukiDoll
    • one year ago
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    something is off.. are you sure the x^2 part of your original question is correct?

  79. anonymous
    • one year ago
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    yes it is ill post the whole question in here

  80. anonymous
    • one year ago
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    If we divide the polynomial x4 + 4x3 + 2x2 + x + 4 by x2 + 3x, what will be the remainder?

  81. UsukiDoll
    • one year ago
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    let's try continuing and see what happens..

  82. anonymous
    • one year ago
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    Alright

  83. UsukiDoll
    • one year ago
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    |dw:1435151602784:dw|

  84. anonymous
    • one year ago
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    so the x^3 would cancel and it would be -x^2

  85. UsukiDoll
    • one year ago
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    I see the question is asking for the remainder XD. we're not done xD we're doing it right heheeh so what's x^3-x^3 and 2x^2-3x^2 ... yes x^3 is gone and we will have -x^2

  86. anonymous
    • one year ago
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    so i did that part right

  87. UsukiDoll
    • one year ago
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    |dw:1435151714443:dw|

  88. UsukiDoll
    • one year ago
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    the last part is a digit, but it has to be negative

  89. UsukiDoll
    • one year ago
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    because when switching the sign it will be positive.. and I have a negative -x^2 right at me

  90. anonymous
    • one year ago
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    what do yo mean by that

  91. UsukiDoll
    • one year ago
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    we've reached the end of the problem.. after all the exponents are used up, it's down to digits. like 1 2 3 4 ...

  92. UsukiDoll
    • one year ago
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    there can only be 1 that will produce x^2

  93. anonymous
    • one year ago
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    ok so the only one that can be right is 4 then. cause i have 0 and 4 as a answer

  94. UsukiDoll
    • one year ago
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    0 is no remainder.. I doubt that will happen

  95. anonymous
    • one year ago
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    A) x2 + 3x B) 4x + 4 C) x2 + x – 2 D) 4 E) 0 Those are the answer

  96. UsukiDoll
    • one year ago
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    |dw:1435151939146:dw|

  97. UsukiDoll
    • one year ago
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    it's not going to be 0 . trust me.. I did the final step in my head.

  98. UsukiDoll
    • one year ago
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    quick question what's -1(x^2) ?

  99. anonymous
    • one year ago
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    -x^2 right

  100. UsukiDoll
    • one year ago
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    yeah.. noticed that I needed the -1 to produce the -x^2-3x but I need to switch signs because we already have a -x^2 present!

  101. anonymous
    • one year ago
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    ok so it becomes positive

  102. UsukiDoll
    • one year ago
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    |dw:1435152046028:dw|

  103. UsukiDoll
    • one year ago
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    yup you just need to compute x+3x and 4+0 that's your remainder

  104. anonymous
    • one year ago
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    SO it would be 4

  105. UsukiDoll
    • one year ago
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    oh I saw what happened I confused regular division with polynomial division. In polynomial division we can have bigger or smaller than the original numbers... can't have that in regular division xD (that was for the x^2 remark) you're missing x+3x!!!!!!!!

  106. UsukiDoll
    • one year ago
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    x(1+3) what's 1+3 ?

  107. anonymous
    • one year ago
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    4

  108. UsukiDoll
    • one year ago
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    yeah and 4(x) is ?

  109. anonymous
    • one year ago
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    4x

  110. UsukiDoll
    • one year ago
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    yeah with 4x and 4 combined together we have 4+4x

  111. UsukiDoll
    • one year ago
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    |dw:1435152315128:dw|

  112. anonymous
    • one year ago
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    Ok so thats the remainder

  113. UsukiDoll
    • one year ago
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    anything left over on the bottom is the remainder.. what aloud did was way off.

  114. UsukiDoll
    • one year ago
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    so that's done... yay x)

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