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anonymous

  • one year ago

given the function f(x)=4x^8+3x^6+2x^4+2 what is the value of f(1)

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  1. UsukiDoll
    • one year ago
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    f(x) =f(1) means that our x =1 so we plug in x =1 throughout the function \[\LARGE f(x) =4x^8+3x^6+2x^4+2\]

  2. UsukiDoll
    • one year ago
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    so at f(1) our function should look like this \[\LARGE f(1) =4(1)^8+3(1)^6+2(1)^4+2 \]

  3. anonymous
    • one year ago
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    @UsukiDoll so the answers 11?

  4. UsukiDoll
    • one year ago
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    YEAH 1 to any exponent number will just be a number so \[\LARGE f(1) =4+3+2+2 =11\]

  5. anonymous
    • one year ago
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    could you help me with a few more please ?

  6. UsukiDoll
    • one year ago
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    ack one to any exponent number will just be 1 so then we have 4(1) which is 4 3(1) which is 3 2(1) which is 2 si 4+3+2+2 is 11. I could, but I am on a netbook, and it's a bit buggy, so you'll have to be a bit patient with me.

  7. anonymous
    • one year ago
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    thats fine , im a very patient person ! Given the function f(x)=x3−8x2+21x−18 what are the factors of of f(x) ?

  8. UsukiDoll
    • one year ago
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    \[\LARGE f(x) =x^3-8x^2+21x-18\] like this?

  9. anonymous
    • one year ago
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    yes

  10. UsukiDoll
    • one year ago
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    hmmm...I wish I could use factor by grouping...

  11. anonymous
    • one year ago
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    i wish i was good at algebra :(

  12. UsukiDoll
    • one year ago
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    ok factor by grouping is out of the question. I got to use something that I haven't done in years. Using the rational root theorem.

  13. anonymous
    • one year ago
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    okay?

  14. UsukiDoll
    • one year ago
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    it's like setting that problem to 0 and breaking this function into at least 3 parts

  15. UsukiDoll
    • one year ago
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    \[\LARGE 0 =x^3-8x^2+21x-18\]

  16. UsukiDoll
    • one year ago
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    sigh... ok... we have to find all factors of 18 which is 1 2 3 6 9 18 and each of them has + or - signs so we got a bunch of choices to choose from 1 2 3 6 9 18 -1 -2 -3 -6 -9 -18

  17. UsukiDoll
    • one year ago
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    so we need to plug those values in until we get our 0... so we need to let x =2, x = 3

  18. UsukiDoll
    • one year ago
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    \[\LARGE f(2) =(2)^3-8(2)^2+21(2)-18\] can you evaluate this function?

  19. UsukiDoll
    • one year ago
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    the reason why I ask this is that we need an x that can produce a 0 and then afterwards we have to use the Factor theorem which states that if p/q is that root in the polynomial then it can be divided by qx-p

  20. anonymous
    • one year ago
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    i think so , is in -2?

  21. UsukiDoll
    • one year ago
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    shouldn't be -2.

  22. anonymous
    • one year ago
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    0?

  23. UsukiDoll
    • one year ago
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    yes so now that we got our x or in that theorem p/q which is 2/1 (yeah I forgot to write the roots in fraction form), Then it's possible to divide that original function with x-2. That would give us the second polynomial we need

  24. UsukiDoll
    • one year ago
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    |dw:1437005051966:dw| sorry it's hard to draw on a touchpad mouse

  25. UsukiDoll
    • one year ago
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    so now we need to use long division . so let's start with |dw:1437005170681:dw| that portion only.

  26. UsukiDoll
    • one year ago
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    what we need to do is to get rid of that x^3 so I have to multiply a variable with x-2 so which variable do we need?

  27. UsukiDoll
    • one year ago
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    what produces x^3 \[\large x(x^{?}) \rightarrow x^{1+?}\]

  28. UsukiDoll
    • one year ago
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    similar to solveing 1 + x =3

  29. anonymous
    • one year ago
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    2

  30. anonymous
    • one year ago
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    a. (x - 3)(x + 2)(x + 3) b.. (x + 3)(x + 2)(x + 3) c. (x + 3)(x - 2)(x - 3) d. (x - 3)(x - 2)(x - 3) these are the answer choices !!

  31. UsukiDoll
    • one year ago
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    yes so we need to distribute (multiply) x^2 x^2(x-2) which is x^3-2x^2 however we need to switch the signs when dividing with polynomials so -x^3+2x^2

  32. UsukiDoll
    • one year ago
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    |dw:1437005603335:dw|

  33. UsukiDoll
    • one year ago
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    |dw:1437005651594:dw|

  34. UsukiDoll
    • one year ago
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    ok... so now I need to get rid of the -6x^2 and I have (x-2) so which choice do we need. If I can get a copycat version of -6x^2 that would be great.

  35. UsukiDoll
    • one year ago
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    |dw:1437005826577:dw|

  36. UsukiDoll
    • one year ago
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    so right now I have x -2 what do I need to distribute to have -6x^2? I need to have at least one x

  37. anonymous
    • one year ago
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    i dont know ?

  38. UsukiDoll
    • one year ago
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    [?] (x-2) well we need to produce a -6x^2 so we need at least one x but what number do we need (and it has to be negative because we already have a negative number inside the divison and we want that to go) ?x(x-2)

  39. anonymous
    • one year ago
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    -3?

  40. UsukiDoll
    • one year ago
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    no I would have -3x^2 and our goal is to get rid of the -6x^2 similar to -7+7 =0

  41. anonymous
    • one year ago
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    i dont know :(

  42. UsukiDoll
    • one year ago
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    well there is a -6 in the division. ?x(x-2) perhaps using that number -6 would work -6x(x-2) try distribute and tell me if you can get a -6x^2 ?

  43. UsukiDoll
    • one year ago
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    we also need a x^2 so \[x \cdot x =x^{1+1}\] what's 1+1?

  44. anonymous
    • one year ago
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    2

  45. UsukiDoll
    • one year ago
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    yes so we have our x^2 x(x-2) when we distribute... but we also need a number so we can create -6x^2 as well... if we have -6 already in the division, then we need that same number -6x(x-2) so using the distributive property -6x^2+12x switching the signs due to dividing polynomials 6x^2-12x

  46. UsukiDoll
    • one year ago
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    |dw:1437006719993:dw|

  47. UsukiDoll
    • one year ago
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    at this point since all variables are used up, we just need a number

  48. UsukiDoll
    • one year ago
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    ?(x-2) what number do I need to produce a 9x?

  49. anonymous
    • one year ago
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    11? or 3?

  50. UsukiDoll
    • one year ago
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    no.... 9 times ? = 9x

  51. UsukiDoll
    • one year ago
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    OH MY I just typed it XD

  52. UsukiDoll
    • one year ago
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    we need a 9 9(x-2) so that's 9x-18 switching the signs -9x+18

  53. UsukiDoll
    • one year ago
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    |dw:1437007066892:dw|

  54. UsukiDoll
    • one year ago
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    so our remainder is 10 but that's not important. What's important is that we have found the second polynomial which is x^2-6x+9 so from the rational root test we had x-2 (x-2)(x^2-6x+9) the second polynomial is obtained through long division can you factor (x^2-6x+9) ?

  55. anonymous
    • one year ago
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    (x-3)^2

  56. UsukiDoll
    • one year ago
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    nice! (x-2)(x-3)^2 or (x-2)(x-3)(x-3) now we can solve for x (x-2)(x-3)(x-3) =0 since one factor is repeating we just have two cases x-3=0 and x-2=0 what is our x values for those 2 equations?

  57. anonymous
    • one year ago
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    3,2?

  58. UsukiDoll
    • one year ago
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    yes 3 and 2 so that is our roots.

  59. anonymous
    • one year ago
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    so the answers ( x - 3)(x - 2)(x - 3)

  60. anonymous
    • one year ago
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    can you help me with 4 or 5 more that is much easier than that one ? 1.A polynomial is to be constructed that has 5 turning points. What is the minimum degree of the polynomial?

  61. UsukiDoll
    • one year ago
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    yes it is (x-3)(x-3)(x-2).... ummm I really don't know how that next questino would work. .__.

  62. anonymous
    • one year ago
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    okay what about 2.Given the polynomial f(x)=6x5−4x3+4x+10 What one of the following is a possible rational root?

  63. UsukiDoll
    • one year ago
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    we have to do the rational root test again so we find all factors of 10

  64. UsukiDoll
    • one year ago
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    so all factors of 10 is 1 2 5 10 but we also have to consider the negative versions as well so -1, -2, -5, -10 b

  65. UsukiDoll
    • one year ago
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    try -1 and see if it works \[\LARGE f(-1)=6(-1)^5-4(-1)^3+4(-1)+10\]

  66. UsukiDoll
    • one year ago
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    if our result isn't 0, we need another root...

  67. anonymous
    • one year ago
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    i got 4

  68. UsukiDoll
    • one year ago
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    yeah time to toss that one out

  69. UsukiDoll
    • one year ago
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    so 1 and -1 are fails. try 2 and -2

  70. anonymous
    • one year ago
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    it wasnt zero there either

  71. UsukiDoll
    • one year ago
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    5 and -5

  72. anonymous
    • one year ago
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    nope

  73. UsukiDoll
    • one year ago
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    then the last two choices are -10 , 10 either one of them works or a math error happened. I do know that 1,-1 don't work at all

  74. anonymous
    • one year ago
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    that didnt work either !

  75. UsukiDoll
    • one year ago
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    right... none of the roots worked which means there aren't any with the rational root test.

  76. anonymous
    • one year ago
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    okay lets just skip that one , i have two more . Given the following polynomial find the maximum possible number of turning points. f(x)=x8+x14+x6+x13

  77. UsukiDoll
    • one year ago
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    well the answer to that previous question should be none because the rational root test failed and we have proven that by finding all roots and testing them out. And, again I don't know what a turning point is

  78. anonymous
    • one year ago
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    okay then , one more?Given the function f(x)=x3+10x2+31x+30 what one the following is a factor of f(x) ?

  79. UsukiDoll
    • one year ago
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    the rational root test is used again and this time it's going to work all factors of 30 1 2 3 5 6 10 15 30 -1 -2 -3 -5 -6 -10 -15 -30

  80. UsukiDoll
    • one year ago
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    try x=-2

  81. UsukiDoll
    • one year ago
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    \[\LARGE f(-2) = (-2)^3+10(-2)^2+31(-2)+30\]

  82. anonymous
    • one year ago
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    i got 0!

  83. UsukiDoll
    • one year ago
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    woo HOO! which means we an divide that polynomial by qx-p from the Factor theorem which is p/q=-2/1 x-(-2) = x+2

  84. UsukiDoll
    • one year ago
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    |dw:1437009518274:dw|

  85. UsukiDoll
    • one year ago
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    so we need to produce an x^3 but we have already done that earlier x^2(x+2) x^3+2x^2 switching signs -x^3-2x^2

  86. UsukiDoll
    • one year ago
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    |dw:1437009682298:dw|

  87. UsukiDoll
    • one year ago
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    so now we need an 8x^2 so we need an 8x 8x(x+2) =8x^2+16x switching signs -8x^2-16x

  88. UsukiDoll
    • one year ago
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    |dw:1437009861228:dw|

  89. UsukiDoll
    • one year ago
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    well this works out nicely .. if we use the number 15 everything cancels out and there is no remainder so we have found our second polynomial (x+2)(x^2+8x+15)

  90. UsukiDoll
    • one year ago
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    can you factor x^2+8x+15?

  91. anonymous
    • one year ago
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    (x+3) (x+5)

  92. UsukiDoll
    • one year ago
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    mhm so now we have (x+2)(x+3)(x+5) we can solve for x to obtain the roots x+2=0 x+3=0 x+5=0

  93. anonymous
    • one year ago
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    the answer was just (x+2) ! THANK YOU , you're a life saver!!

  94. UsukiDoll
    • one year ago
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    you're welcome :)

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