Question 347566: Could you please help me solve 2y^3-11y^2+12y=0?
I'm completely blank and very tired and can't think anymore (had a volleyball game today after school and am trying to do loads of homework tonight).
Here's what I tried:
2(y.y.y)-11(y.y)+12y=0
2(y)-11+12y=0
-11+14y=0
+11 = +11
14y=11
___ ____
14 14
y=11/14
Thank you so much for your time and help.
Darlyn
Answer by haileytucki(390)   (Show Source):
You can put this solution on YOUR website! 2y^(3)-11y^(2)+12y=0
Factor out the GCF of y from each term in the polynomial.
y(2y^(2))+y(-11y)+y(12)=0
Factor out the GCF of y from 2y^(3)-11y^(2)+12y.
y(2y^(2)-11y+12)=0
In this problem -(3)/(2)*-4=12 and -(3)/(2)-4=-11, so insert -(3)/(2) as the right hand term of one factor and -4 as the right-hand term of the other factor.
y(y-(3)/(2))(y-4)=0
Remove the fraction by multiplying the first term of the factor by the denominator of the second term.
y(2y-3)(y-4)=0
Set the single term factor on the left-hand side of the equation equal to 0.
y=0
Set each of the factors of the left-hand side of the equation equal to 0.
2y-3=0_y-4=0
Since -3 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 3 to both sides.
2y=3_y-4=0
Divide each term in the equation by 2.
(2y)/(2)=(3)/(2)_y-4=0
Simplify the left-hand side of the equation by canceling the common factors.
y=(3)/(2)_y-4=0
Set each of the factors of the left-hand side of the equation equal to 0.
y=(3)/(2)_y-4=0
Since -4 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 4 to both sides.
y=(3)/(2)_y=4
The complete solution is the set of the individual solutions.
y=0,(3)/(2),4
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