What are the solutions to the equation 64y^3-7=y-448y^2 ?

The solutions to the equation 64y^3-7=y-448y^2 are y=-7,y=-1/8 ,y= 1/8

Find the zeros of 64y^3-7=y-448y^2

The zeros of 64y^3-7=y-448y^2 are y=-7,y=-1/8 ,y= 1/8

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64 y^{3} - 7 = y - 448 y^{2}

y = - 7, y = - \frac{1}{8}, y = \frac{1}{8}

Decimal Notation

y = - 7, y = - 0.125, y = 0.125

Solution steps

64 y^{3} - 7 = y - 448 y^{2}

Move

448 y^{2}

to the left side

64 y^{3} - 7 + 448 y^{2} = y

Move

y

to the left side

64 y^{3} - 7 + 448 y^{2} - y = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

64 y^{3} + 448 y^{2} - y - 7 = 0

Factor

64 y^{3} + 448 y^{2} - y - 7 : (y + 7) (8 y + 1) (8 y - 1)

(y + 7) (8 y + 1) (8 y - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y + 7 = 0 or 8 y + 1 = 0 or 8 y - 1 = 0

Solve

y + 7 = 0 : y = - 7

Solve

8 y + 1 = 0 : y = - \frac{1}{8}

Solve

8 y - 1 = 0 : y = \frac{1}{8}

The solutions are

y = - 7, y = - \frac{1}{8}, y = \frac{1}{8}

x^{4} - 8 x + 63 = 0

16 x^{4} - 9 = 0

x^{3} - 12 x + 7 = 0

- 2 x (x + 8) (x^{2} - 9) = 0

2^{24} = (2 y)^{3}

What are the solutions to the equation 64y^3-7=y-448y^2 ?
The solutions to the equation 64y^3-7=y-448y^2 are y=-7,y=-1/8 ,y= 1/8
Find the zeros of 64y^3-7=y-448y^2
The zeros of 64y^3-7=y-448y^2 are y=-7,y=-1/8 ,y= 1/8