What are the solutions to the equation 3x^3+5x^2=28x ?

The solutions to the equation 3x^3+5x^2=28x are x=0,x= 7/3 ,x=-4

Find the zeros of 3x^3+5x^2=28x

The zeros of 3x^3+5x^2=28x are x=0,x= 7/3 ,x=-4

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3 x^{3} + 5 x^{2} = 28 x

x = 0, x = \frac{7}{3}, x = - 4

Decimal Notation

x = 0, x = 2.33333 \dots, x = - 4

Solution steps

3 x^{3} + 5 x^{2} = 28 x

Move

28 x

to the left side

3 x^{3} + 5 x^{2} - 28 x = 0

Factor

3 x^{3} + 5 x^{2} - 28 x : x (3 x - 7) (x + 4)

x (3 x - 7) (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 3 x - 7 = 0 or x + 4 = 0

Solve

3 x - 7 = 0 : x = \frac{7}{3}

Solve

x + 4 = 0 : x = - 4

The solutions are

x = 0, x = \frac{7}{3}, x = - 4

x = 5 x - x^{3}

(x + 1) 4 - (x - 1) 4 = x^{2} (8 x + 1) + 15

x^{4} - 5 x^{2} 4 = 0

V = \frac{4}{3} π r^{3}, r

(x^{3} - 1) (x^{5} - 1) = 0

What are the solutions to the equation 3x^3+5x^2=28x ?
The solutions to the equation 3x^3+5x^2=28x are x=0,x= 7/3 ,x=-4
Find the zeros of 3x^3+5x^2=28x
The zeros of 3x^3+5x^2=28x are x=0,x= 7/3 ,x=-4