anonymous
  • anonymous
pleaaase help.... fan and medals question in comments
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
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?

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anonymous
  • anonymous
thank you.. got it.. it was so easy
Michele_Laino
  • Michele_Laino
I'm working on question #2, please wait...
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
where P(x) is another polynomial. Now if I replace x with 2, I substitute x=2 into that equation, what do you get?
anonymous
  • anonymous
this will be equal to 0
Michele_Laino
  • Michele_Laino
hint: the right side is: \[P\left( 2 \right){\left( {2 - 2} \right)^2} = 0\]
Michele_Laino
  • Michele_Laino
please write the left side after that substitution
Michele_Laino
  • Michele_Laino
hint: we have: \[{2^5} + a \times {2^4} + 3 \times {2^3} + b \times {2^2} + a = 0\] please simplify
anonymous
  • anonymous
32+16a+24+4b+a = 0
Michele_Laino
  • Michele_Laino
then you can simplify further: what is 32+24=... what is 16a+a=...
anonymous
  • anonymous
56 and 17a
anonymous
  • anonymous
i ve understood it now.. can work it/// thank you so much for your help
Michele_Laino
  • Michele_Laino
ok! so we got this condition or equation: \[56 + 17a + 4b = 0\]
Michele_Laino
  • Michele_Laino
:)

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