What are the solutions to the equation 3y(3y-5)(2y+7)=0 ?

The solutions to the equation 3y(3y-5)(2y+7)=0 are y=0,y= 5/3 ,y=-7/2

Find the zeros of 3y(3y-5)(2y+7)=0

The zeros of 3y(3y-5)(2y+7)=0 are y=0,y= 5/3 ,y=-7/2

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3 y (3 y - 5) (2 y + 7) = 0

y = 0, y = \frac{5}{3}, y = - \frac{7}{2}

Solution steps

3 y (3 y - 5) (2 y + 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or 3 y - 5 = 0 or 2 y + 7 = 0

Solve

3 y - 5 = 0 : y = \frac{5}{3}

Solve

2 y + 7 = 0 : y = - \frac{7}{2}

The solutions are

y = 0, y = \frac{5}{3}, y = - \frac{7}{2}

x^{4} + 2 x^{2} - 2 x + 1 = 0

12 x^{3} - 84 x^{2} + 120 x = 0

- x^{3} + 12 x^{2} - 29 x + 18 = 0

715 = 5 x^{3}

4 x^{4} - x + 1 = 0

What are the solutions to the equation 3y(3y-5)(2y+7)=0 ?
The solutions to the equation 3y(3y-5)(2y+7)=0 are y=0,y= 5/3 ,y=-7/2
Find the zeros of 3y(3y-5)(2y+7)=0
The zeros of 3y(3y-5)(2y+7)=0 are y=0,y= 5/3 ,y=-7/2