What are the solutions to the equation x^3-3x^2=(x-3)(x+20) ?

The solutions to the equation x^3-3x^2=(x-3)(x+20) are x=3,x=-4,x=5

Find the zeros of x^3-3x^2=(x-3)(x+20)

The zeros of x^3-3x^2=(x-3)(x+20) are x=3,x=-4,x=5

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x^{3} - 3 x^{2} = (x - 3) (x + 20)

x = 3, x = - 4, x = 5

Solution steps

x^{3} - 3 x^{2} = (x - 3) (x + 20)

Expand

(x - 3) (x + 20) : x^{2} + 17 x - 60

x^{3} - 3 x^{2} = x^{2} + 17 x - 60

Move

60

to the left side

x^{3} - 3 x^{2} + 60 = x^{2} + 17 x

Move

17 x

to the left side

x^{3} - 3 x^{2} + 60 - 17 x = x^{2}

Move

x^{2}

to the left side

x^{3} - 4 x^{2} - 17 x + 60 = 0

Factor

x^{3} - 4 x^{2} - 17 x + 60 : (x - 3) (x + 4) (x - 5)

(x - 3) (x + 4) (x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 3 = 0 or x + 4 = 0 or x - 5 = 0

Solve

x - 3 = 0 : x = 3

Solve

x + 4 = 0 : x = - 4

Solve

x - 5 = 0 : x = 5

The solutions are

x = 3, x = - 4, x = 5

t^{3} = t

x^{3} - 7 x - 4 = 0

- a^{3} - 2 a^{2} + 4 a + 8 = 0

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

a, y = 9 a^{4} - a^{2} b^{2}

What are the solutions to the equation x^3-3x^2=(x-3)(x+20) ?
The solutions to the equation x^3-3x^2=(x-3)(x+20) are x=3,x=-4,x=5
Find the zeros of x^3-3x^2=(x-3)(x+20)
The zeros of x^3-3x^2=(x-3)(x+20) are x=3,x=-4,x=5