Question

what must be subtracted from y^4+2y^2-3y+7 to get y^3+y^2+y-1?
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Answer 1

Answer:

y^4+y^3-y^2+4y-6

Step-by-step explanation:

y^3+y^2+y-1-(y^4+2y^2-3y+7)

=y^3+y^2+y+1-y^4-2y^2+3y-7

=-y^4+y^3+y^2-2y^2+y+3y+1-7

=-y^4+y^3-y^2+4y-6

Answer 2

Answer:

$y^4 + 3y^2 - 4y + 6 + y^3$

Step-by-step explanation:

 =>  $y^4+2y^2-3y+7$ + $y^3+y^2+y-1$

 =>  $y^4$ + $2y^2$ + $y^2$ - $3y$ + $y$ + $7$ - $1$ + $y^3$

 =>  $y^4 + 3y^2 - 4y + 6 + y^3$

Sorry if the answer is wrong.

If its right stars and thanks are appreciated :D