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p-x-2x-2-y-3-3xy-10-a-0-a-1-a-2-a-10-




Question Number 190982 by sciencestudentW last updated on 15/Apr/23
p(x)=2x^2 y^3 −3xy+10  a_0 =?  a_1 =?     a_2 =?  a_(10) =?
$${p}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{3}} −\mathrm{3}{xy}+\mathrm{10} \\ $$$${a}_{\mathrm{0}} =?\:\:{a}_{\mathrm{1}} =?\:\:\:\:\:{a}_{\mathrm{2}} =?\:\:{a}_{\mathrm{10}} =? \\ $$
Commented by mr W last updated on 16/Apr/23
what do you mean with a_0 , a_1  etc?
$${what}\:{do}\:{you}\:{mean}\:{with}\:{a}_{\mathrm{0}} ,\:{a}_{\mathrm{1}} \:{etc}? \\ $$
Answered by a.lgnaoui last updated on 16/Apr/23
posons  a_0 =p(0);  a_1 =p(1);  a_2 =p(2);a_(10) =p(10)  ⇒ a_0 =10            a_1 =y(2y^2 −3)+10           a_2 =2y(4y^2 −3)+10            a_(10) =10y(20y^2 −3)+10  suite  a_1 =y(2y^2 −3)+a_0   a_2 =2[y(2y^2 )−3y]+a_0 =2a_1 −a_0   a_5 =5a_1 −4a_0   ......  a_n =na_1 −(n−1)a_0 =n(a_1 −1)+a_0     ?
$$\mathrm{posons}\:\:\mathrm{a}_{\mathrm{0}} =\mathrm{p}\left(\mathrm{0}\right);\:\:\mathrm{a}_{\mathrm{1}} =\mathrm{p}\left(\mathrm{1}\right);\:\:\mathrm{a}_{\mathrm{2}} =\mathrm{p}\left(\mathrm{2}\right);\mathrm{a}_{\mathrm{10}} =\mathrm{p}\left(\mathrm{10}\right) \\ $$$$\Rightarrow\:\mathrm{a}_{\mathrm{0}} =\mathrm{10}\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{a}_{\mathrm{1}} =\mathrm{y}\left(\mathrm{2y}^{\mathrm{2}} −\mathrm{3}\right)+\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{a}_{\mathrm{2}} =\mathrm{2y}\left(\mathrm{4y}^{\mathrm{2}} −\mathrm{3}\right)+\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{a}_{\mathrm{10}} =\mathrm{10y}\left(\mathrm{20y}^{\mathrm{2}} −\mathrm{3}\right)+\mathrm{10} \\ $$$$\mathrm{suite} \\ $$$$\mathrm{a}_{\mathrm{1}} =\mathrm{y}\left(\mathrm{2y}^{\mathrm{2}} −\mathrm{3}\right)+\mathrm{a}_{\mathrm{0}} \\ $$$$\mathrm{a}_{\mathrm{2}} =\mathrm{2}\left[\mathrm{y}\left(\mathrm{2y}^{\mathrm{2}} \right)−\mathrm{3y}\right]+\mathrm{a}_{\mathrm{0}} =\mathrm{2a}_{\mathrm{1}} −\mathrm{a}_{\mathrm{0}} \\ $$$$\mathrm{a}_{\mathrm{5}} =\mathrm{5a}_{\mathrm{1}} −\mathrm{4a}_{\mathrm{0}} \\ $$$$…… \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{na}_{\mathrm{1}} −\left(\mathrm{n}−\mathrm{1}\right)\mathrm{a}_{\mathrm{0}} =\mathrm{n}\left(\mathrm{a}_{\mathrm{1}} −\mathrm{1}\right)+\mathrm{a}_{\mathrm{0}} \:\:\:\:? \\ $$$$ \\ $$

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