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

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

Mathematics
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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.

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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?

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

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