What are the solutions to the equation 4y^3-7y^2+28=16y ?

The solutions to the equation 4y^3-7y^2+28=16y are y= 7/4 ,y=-2,y=2

Find the zeros of 4y^3-7y^2+28=16y

The zeros of 4y^3-7y^2+28=16y are y= 7/4 ,y=-2,y=2

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4 y^{3} - 7 y^{2} + 28 = 16 y

y = \frac{7}{4}, y = - 2, y = 2

Decimal Notation

y = 1.75, y = - 2, y = 2

Solution steps

4 y^{3} - 7 y^{2} + 28 = 16 y

Move

16 y

to the left side

4 y^{3} - 7 y^{2} + 28 - 16 y = 0

Write in the standard form

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

4 y^{3} - 7 y^{2} - 16 y + 28 = 0

Factor

4 y^{3} - 7 y^{2} - 16 y + 28 : (4 y - 7) (y + 2) (y - 2)

(4 y - 7) (y + 2) (y - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

4 y - 7 = 0 or y + 2 = 0 or y - 2 = 0

Solve

4 y - 7 = 0 : y = \frac{7}{4}

Solve

y + 2 = 0 : y = - 2

Solve

y - 2 = 0 : y = 2

The solutions are

y = \frac{7}{4}, y = - 2, y = 2

x^{4} + 2 x^{3} + 6 x^{2} + 8 x + 8 = 0

w^{3} = 216

2 x^{5} - 5 x^{2} + 1 = 0

0 = - x^{3} + 15 x^{2} - 63 x + 49

3 n^{9} - 5 = 0

What are the solutions to the equation 4y^3-7y^2+28=16y ?
The solutions to the equation 4y^3-7y^2+28=16y are y= 7/4 ,y=-2,y=2
Find the zeros of 4y^3-7y^2+28=16y
The zeros of 4y^3-7y^2+28=16y are y= 7/4 ,y=-2,y=2