What are the solutions to the equation solvefor y,4(-32xy^4+12y^3)=0 ?

The solutions to the equation solvefor y,4(-32xy^4+12y^3)=0 are y=0,y= 3/(8x);\quad x\ne 0

Find the zeros of solvefor y,4(-32xy^4+12y^3)=0

The zeros of solvefor y,4(-32xy^4+12y^3)=0 are y=0,y= 3/(8x);\quad x\ne 0

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solve for

y, 4 (- 32 x y^{4} + 12 y^{3}) = 0

y = 0, y = \frac{3}{8 x}; x \neq = 0

Solution steps

4 (- 32 x y^{4} + 12 y^{3}) = 0

Factor

4 (- 32 x y^{4} + 12 y^{3}) : 16 y^{3} (- 8 y x + 3)

16 y^{3} (- 8 y x + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or - 8 y x + 3 = 0

Solve

- 8 y x + 3 = 0 : y = \frac{3}{8 x}; x \neq = 0

The solutions are

y = 0, y = \frac{3}{8 x}; x \neq = 0

y, 3 x + 7 y = - 21

x - 11 = 166 - 2 x

\frac{x}{3} - 2 = 25

\frac{1}{2} - \frac{7}{10} z = \frac{1}{2}

\frac{x}{2} + 8 = 10

What are the solutions to the equation solvefor y,4(-32xy^4+12y^3)=0 ?
The solutions to the equation solvefor y,4(-32xy^4+12y^3)=0 are y=0,y= 3/(8x);\quad x\ne 0
Find the zeros of solvefor y,4(-32xy^4+12y^3)=0
The zeros of solvefor y,4(-32xy^4+12y^3)=0 are y=0,y= 3/(8x);\quad x\ne 0