Question

y5+5y3-8y2+6y-16 ÷ y2+2​

Answer 1

Answer:

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Answer 2

$\frac{y^5+5y^3-8y^2+6y-16}{y^2+2}=y^3+3y-8$

Step-by-step explanation:

$\frac{y^5+5y^3-8y^2+6y-16}{y^2+2}$

Rearrange the terms of numerator of given fraction

$\frac{y^5+5y^3+6y-8y^2+16}{y^2+2}$

$\frac{y(y^4+5y^2+6)-8(y^2+2)}{y^2+2}$

$\frac{y(y^4+2y^2+3y^2+6)-8(y^2+2)}{y^2+2}$

$\frac{y((y^2(y^2+2)+3(y^2+2))-8(y^2+2)}{y^2+2}$

$\frac{y(y^2+2)(y^2+3)-8(y^2+2)}{y^2+2}$

$\frac{(y^2+2)(y(y^2+3)-8)}{y^2+2}$

$y(y^2+3)-8$

$y^3+3y-8$

Hence,$\frac{y^5+5y^3-8y^2+6y-16}{y^2+2}=y^3+3y-8$

#Learns more:

https://brainly.in/question/3323924:Answered by Khushi