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ashking1

  • one year ago

Find the zeros of the polynomial function and state the multiplicity of each. f(x) = 4(x + 7)2(x - 7)3

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  1. ashking1
    • one year ago
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    these are the choices 4, multiplicity 1; -7, multiplicity 3; 7, multiplicity 3 -7, multiplicity 3; 7, multiplicity 2 4, multiplicity 1; 7, multiplicity 1; -7, multiplicity 1 -7, multiplicity 2; 7, multiplicity 3 can someone help me please

  2. amoodarya
    • one year ago
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    do you mean ?\[f(x) = 4(x + 7)^2(x - 7)^3\]

  3. phi
    • one year ago
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    you should put in ^ to show exponents. I assume this is f(x) = 4(x + 7)^2(x - 7)^3 \[ f(x) = 4(x + 7)^2(x - 7)^3 \]

  4. ashking1
    • one year ago
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    yes

  5. phi
    • one year ago
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    what happens if you erase the x and put in 7 in its place in (x-7) what number do you get ?

  6. ashking1
    • one year ago
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    0

  7. ashking1
    • one year ago
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    3 times

  8. phi
    • one year ago
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    yes. they use the word "multiplicity" so look for an answer that says 7 multiplicity 3

  9. ashking1
    • one year ago
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    -7, multiplicity 3; 7, multiplicity 2 so would this be the answer

  10. phi
    • one year ago
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    it is easy to get confused. 4(x + 7)^2(x - 7)^3 we did (x-7) . what x makes that zero?

  11. ashking1
    • one year ago
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    what do you mean by that like its the second x the one that says in (x-7) and if x is 7 it will equal 0 right ?

  12. phi
    • one year ago
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    (x-7) is zero when x=7 and multiplicity 3 because we have (x-7)^3

  13. phi
    • one year ago
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    you picked *** 7, multiplicity 2 so would this be the answer***

  14. ashking1
    • one year ago
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    no

  15. ashking1
    • one year ago
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    it would be -7, multiplicity 2; 7, multiplicity 3

  16. phi
    • one year ago
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    yes

  17. ashking1
    • one year ago
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    thanks btw can you help me with one more

  18. phi
    • one year ago
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    ok

  19. ashking1
    • one year ago
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    i need help understanding either the triangle or the theorem to answer this question Expand the following using either the Binomial Theorem or Pascal’s Triangle. You must show your work for credit. (x - 5)5

  20. ashking1
    • one year ago
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    like you dont have to tell me the answer more like help me understand how to get there

  21. phi
    • one year ago
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    please use the ^ otherwise it looks like (x-5) times 5 you mean \( (x-5)^5 \) first, write it like this: \( (x + (-5) )^5 \)

  22. ashking1
    • one year ago
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    sorry ok

  23. phi
    • one year ago
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    next, we start with this pattern \[ (a+b)^5 = a b + a b + a b + ...\] make the exponent of the a in the first term 5, and the exponent on b 0 (the exponents will *always* add up to 5) the next a will have an exponent of 4 (and the b will have 1) can you write out the a b pattern for (a+b)^5 ?

  24. ashking1
    • one year ago
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    so it will be (5+0)^5= (4)(1).....

  25. phi
    • one year ago
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    no. I am trying to say that \[ (a+b)^5 = a^5 b^0 + a^4 b^1 + ... \] (there are also numbers we have to put in, but first we need to learn how to write down this pattern) can you finish writing out the full pattern ?

  26. ashking1
    • one year ago
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    would the next one be like a^3b^2

  27. phi
    • one year ago
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    yes keep going

  28. ashking1
    • one year ago
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    a^2b^2

  29. phi
    • one year ago
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    you mean a^2 b^3 (remember the exponents add up to 5) or the a goes down, the one on b goes up

  30. ashking1
    • one year ago
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    ya sorry thats what i ment

  31. phi
    • one year ago
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    keep going, we need all the terms

  32. ashking1
    • one year ago
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    then a^1B^4

  33. ashking1
    • one year ago
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    a^0b^5

  34. phi
    • one year ago
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    yes. so we have (so far) \[ (a+b)^5 = a^5 b^0 + a^4 b^1 + a^3b^2+a^2b^3 + a^1 b^4 +a^0 b^5\] now we have to put in coefficients (numbers) in front of each term. we use Pascal's triangle to find these numbers.

  35. phi
    • one year ago
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    start by writing this pattern |dw:1437687651177:dw|

  36. phi
    • one year ago
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    next we add a new row by adding 1 on each side, and adding the two numbers in the row above it. Like this |dw:1437687710434:dw|

  37. ashking1
    • one year ago
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    ok

  38. phi
    • one year ago
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    now we keep adding rows until we get to a row with 6 entries... that will be the row we will use. |dw:1437687767682:dw|

  39. phi
    • one year ago
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    can you find the next row ?

  40. ashking1
    • one year ago
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    |dw:1437687781164:dw|

  41. ashking1
    • one year ago
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    i seen it before but how does it help for the problem

  42. phi
    • one year ago
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    and we add those to our pattern \[ (a+b)^5 = 1a^5 b^0 +5 a^4 b^1 + 10a^3b^2+10a^2b^3 + 5a^1 b^4 +1a^0 b^5 \]

  43. phi
    • one year ago
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    we can simplify that a little... a^0 and b^0 are both 1. And we can leave off a coefficient of 1. so the pattern can be \[ (a+b)^5 = a^5 +5 a^4 b + 10a^3b^2+10a^2b^3 + 5a b^4 +b^5 \]

  44. phi
    • one year ago
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    now we can tackle your problem \[ (x + (-5) )^5 \] every where you see a in the pattern , replace it with x and replace b with (-5) can you do that ?

  45. ashking1
    • one year ago
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    i changed everyhthing

  46. ashking1
    • one year ago
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    everything

  47. phi
    • one year ago
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    they probably want you to multiply out the (-5) to each exponent

  48. ashking1
    • one year ago
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    ok

  49. phi
    • one year ago
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    the first two terms should be \[ x^5 - 25x^4 ...\]

  50. ashking1
    • one year ago
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    sorry im writing everything on a piece of paper

  51. ashking1
    • one year ago
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    +250x^3 is the next one right

  52. phi
    • one year ago
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    yes

  53. ashking1
    • one year ago
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    -1250x^2 is next one

  54. phi
    • one year ago
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    yes

  55. ashking1
    • one year ago
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    +6250x2-3,125

  56. phi
    • one year ago
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    last -3125 is ok but redo 5*x*(-5)^4

  57. ashking1
    • one year ago
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    3125x

  58. phi
    • one year ago
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    yes. so put it together for the final answer.

  59. ashking1
    • one year ago
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    \[x^5-25x^4+250x^3-1250x^2+3125x-3125\]

  60. ashking1
    • one year ago
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    thats the answer right

  61. phi
    • one year ago
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    looks good

  62. ashking1
    • one year ago
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    thank you so much

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