What are the solutions to the equation 16y^3-3=y-48y^2 ?

The solutions to the equation 16y^3-3=y-48y^2 are y=-3,y=-1/4 ,y= 1/4

Find the zeros of 16y^3-3=y-48y^2

The zeros of 16y^3-3=y-48y^2 are y=-3,y=-1/4 ,y= 1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

16 y^{3} - 3 = y - 48 y^{2}

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

Solution steps

16 y^{3} - 3 = y - 48 y^{2}

Move

48 y^{2}

to the left side

16 y^{3} - 3 + 48 y^{2} = y

Move

y

to the left side

16 y^{3} - 3 + 48 y^{2} - y = 0

Write in the standard form

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

16 y^{3} + 48 y^{2} - y - 3 = 0

Factor

16 y^{3} + 48 y^{2} - y - 3 : (y + 3) (4 y + 1) (4 y - 1)

(y + 3) (4 y + 1) (4 y - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

y + 3 = 0 : y = - 3

Solve

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

Solve

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

The solutions are

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

2 r^{3} - 4 r^{2} - r + 2 = 0

- 2 y^{4} - 7 y^{2} - 4 = 0

0 = (x^{2} + 1) (x - 2) (x + 1)

7 x^{4} - 21 x = 0

x, f = (x^{2} - 2)^{2}

What are the solutions to the equation 16y^3-3=y-48y^2 ?
The solutions to the equation 16y^3-3=y-48y^2 are y=-3,y=-1/4 ,y= 1/4
Find the zeros of 16y^3-3=y-48y^2
The zeros of 16y^3-3=y-48y^2 are y=-3,y=-1/4 ,y= 1/4