Question 214409This question is from textbook algebra 1 an integrated approach
: solve using long division, then check.
(-30+6y^3-28y+y^2) divided (2y-5)
PLEASE PLEASE PLEASE HELP ME WITH THIS PROBLEM I DONT UNDERSTAND HOW TO DO IT AT ALL THANK YOU SOOO MUCH.
This question is from textbook algebra 1 an integrated approach
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! (-30+6y^3-28y+y^2) divided (2y-5)
:
Arrange it as follows, it's alot like ordinary long division
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2y-5|6y^3 + y^2 - 28y - 30
:
Find the value that when multiply by 2y gives you 6y^3, that would be 3y^2
also multiply it by 5 and you have:
:
.......................3y^2
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2y-5|6y^3 + y^2 - 28y - 30
..........6y^3 - 15y^2; subtract
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...............0 + 16y^2 - 28y
:
Find the value that multiplied by 2y gives you 16y^2, that would be 8y
also mutiply by 5
:
.......................3y^2 + 8y
..........-----------------------
2y-5|6y^3 + y^2 - 28y - 30
..........6y^3 - 15y^2
..........--------------
...............0 + 16y^2 - 28y
.....................16y^2 - 40y; subtract
.....................-------------
..................................+12y - 30
:
Find the value that multiplied by 2y gives you 12y, that would be 6
also multiply 6 times 5
:
.......................3y^2 + 8y + 6
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2y-5|6y^3 + y^2 - 28y - 30
..........6y^3 - 15y^2
..........--------------
...............0 + 16y^2 - 28y
.....................16y^2 - 40y; subtract
.....................-------------
....................................+12y - 30
......................................12y - 30
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...............................................0 remainder
:
the answer is: 3y^2 + 8y + 6
;
;
Check the solution, multiply: 2y-5 * 3y^2 + 8y + 6
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