Question

Someone answer this
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Answer 1

             

Answer: 17y - 17

     

$\boxed{p(y)=6y^5+5y^4+11y^3-3y^2+y+5}$

$\boxed{q(y)=3y^2-2y+4}$

Okay, p(y) has to be divided by q(y).

Splitting middle terms...

$\boxed{6y^5+5y^4+11y^3-3y^2+y+5} \\ \\ \boxed{6y^5-4y^4+9y^4+8y^3-6y^3+9y^3+12y^2-6y^2-9y^2+12y+6y-17y-12+17} \\ \\ \boxed{6y^5-4y^4+8y^3+9y^4-6y^3+12y^2+9y^3-6y^2+12y-9y^2+6y-12-17y+17} \\ \\ \boxed{2y^3(3y^2-2y+4)+3y^2(3y^2-2y+4)+3y(3y^2-2y+4)-3(3y^2-2y+4)-(17y-17)} \\ \\ \boxed{(3y^2-2y+4)(2y^3+3y^2+3y-3)-(17y-17)}$

Here, the remainder got is - (17y - 17).

When the negative form of the remainder is added to p(y), the remainder will be vanished, and then p(y) can exactly be divided by q(y).

∴ The answer is  - ( - (17y - 17)) = 17y - 17.

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Thank you. :-))