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anonymous

  • one year ago

pleaaase help.... fan and medals question in comments

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Michele_Laino

  3. Michele_Laino
    • one year ago
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    I call with F(x) the subsequent function: \[F\left( x \right) = {\left( {x - 2} \right)^2}g\left( x \right)\] then, applying the product rule, we can write the first derivative of F(x) as below: \[F'\left( x \right) = 2\left( {x - 2} \right)g\left( x \right) + {\left( {x - 2} \right)^2}g'\left( x \right)\] now please try to factor out x-2, what do you get?

  4. anonymous
    • one year ago
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    thank you.. got it.. it was so easy

  5. Michele_Laino
    • one year ago
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    I'm working on question #2, please wait...

  6. anonymous
    • one year ago
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    yes

  7. Michele_Laino
    • one year ago
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    here we can write this: \[{x^5} + a{x^4} + 3{x^3} + b{x^2} + a = P\left( x \right){\left( {x - 2} \right)^2}\] according to the hypothesis of your problem

  8. Michele_Laino
    • one year ago
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    where P(x) is another polynomial. Now if I replace x with 2, I substitute x=2 into that equation, what do you get?

  9. anonymous
    • one year ago
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    this will be equal to 0

  10. Michele_Laino
    • one year ago
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    hint: the right side is: \[P\left( 2 \right){\left( {2 - 2} \right)^2} = 0\]

  11. Michele_Laino
    • one year ago
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    please write the left side after that substitution

  12. Michele_Laino
    • one year ago
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    hint: we have: \[{2^5} + a \times {2^4} + 3 \times {2^3} + b \times {2^2} + a = 0\] please simplify

  13. anonymous
    • one year ago
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    32+16a+24+4b+a = 0

  14. Michele_Laino
    • one year ago
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    then you can simplify further: what is 32+24=... what is 16a+a=...

  15. anonymous
    • one year ago
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    56 and 17a

  16. anonymous
    • one year ago
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    i ve understood it now.. can work it/// thank you so much for your help

  17. Michele_Laino
    • one year ago
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    ok! so we got this condition or equation: \[56 + 17a + 4b = 0\]

  18. Michele_Laino
    • one year ago
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    :)

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