Solve the following polynomial equation by grouping and factoring. 4y^3−9y=0 y=?

Step 1
Given a polynomial equation,
${\displaystyle 4y^3-9y=0}$
Step 2
Now to solve the polynomial,
By using the factorization method,
${\displaystyle y(4y^2-9)=0}$
Now, by the formula of the square of difference,
${\displaystyle x^2-y^2=(x-y)(x+y)}$
Thus,
${\displaystyle \Rightarrow y(2y+3)(2y-3)=0}$
Step 3
Now, by zeros factorization principle,
If ab=0 then a=0 or b=0,
${\displaystyle \Rightarrow y=0}$ or (2y+3)=0 or (2y-3)=0
Thus,
${\displaystyle \Rightarrow 2y+3=0}$
${\displaystyle \Rightarrow y=-\frac{3}{2}}$
And
${\displaystyle \Rightarrow 2y-3=0}$
${\displaystyle \Rightarrow y=\frac{3}{2}}$
Step 4
Therefore, the solution of the polynomial is
y=0 or ${\displaystyle y=-\frac{3}{2}\text{or}y=\frac{3}{2}}$
Thus, the exact answer is
${\displaystyle y=0,-\frac{3}{2},\frac{3}{2}}$.