What are the solutions to the equation 5y(3-y)(4y+1)^2=0 ?

The solutions to the equation 5y(3-y)(4y+1)^2=0 are y=0,y=3,y=-1/4

Find the zeros of 5y(3-y)(4y+1)^2=0

The zeros of 5y(3-y)(4y+1)^2=0 are y=0,y=3,y=-1/4

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5 y (3 - y) (4 y + 1)^{2} = 0

y = 0, y = 3, y = - \frac{1}{4}

Solution steps

5 y (3 - y) (4 y + 1)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or 3 - y = 0 or 4 y + 1 = 0

Solve

3 - y = 0 : y = 3

Solve

4 y + 1 = 0 : y = - \frac{1}{4}

The solutions are

y = 0, y = 3, y = - \frac{1}{4}

x^{3} - x^{2} - x - 1 = 0

- x^{3} + 2 x^{2} + x - 2 = 0

x^{3} + x + 2 = 0

x^{4} = x^{2}

x^{3} + x^{2} + 1 = 0

What are the solutions to the equation 5y(3-y)(4y+1)^2=0 ?
The solutions to the equation 5y(3-y)(4y+1)^2=0 are y=0,y=3,y=-1/4
Find the zeros of 5y(3-y)(4y+1)^2=0
The zeros of 5y(3-y)(4y+1)^2=0 are y=0,y=3,y=-1/4