What are the solutions to the equation (x^3-4x^2)=(16x-64) ?

The solutions to the equation (x^3-4x^2)=(16x-64) are x=4,x=-4

Find the zeros of (x^3-4x^2)=(16x-64)

The zeros of (x^3-4x^2)=(16x-64) are x=4,x=-4

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(x^{3} - 4 x^{2}) = (16 x - 64)

x = 4, x = - 4

Solution steps

(x^{3} - 4 x^{2}) = (16 x - 64)

Refine

x^{3} - 4 x^{2} = 16 x - 64

Move

64

to the left side

x^{3} - 4 x^{2} + 64 = 16 x

Move

16 x

to the left side

x^{3} - 4 x^{2} + 64 - 16 x = 0

Write in the standard form

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

x^{3} - 4 x^{2} - 16 x + 64 = 0

Factor

x^{3} - 4 x^{2} - 16 x + 64 : (x - 4)^{2} (x + 4)

(x - 4)^{2} (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 4 = 0 or x + 4 = 0

Solve

x - 4 = 0 : x = 4

Solve

x + 4 = 0 : x = - 4

The solutions are

x = 4, x = - 4

15 x^{4} + 15 x^{2} = 0

- x^{3} + 10 x^{2} - 31 x + 30 = 0

1000 + (3 - x)^{(10)} = 1978

0 = 6 x^{2} - 3 x^{3}

2 y^{3} - 5 y^{2} - 3 y = 0

What are the solutions to the equation (x^3-4x^2)=(16x-64) ?
The solutions to the equation (x^3-4x^2)=(16x-64) are x=4,x=-4
Find the zeros of (x^3-4x^2)=(16x-64)
The zeros of (x^3-4x^2)=(16x-64) are x=4,x=-4