What are the solutions to the equation 60x^4-180x^3+120x^2=0 ?

The solutions to the equation 60x^4-180x^3+120x^2=0 are x=0,x=1,x=2

Find the zeros of 60x^4-180x^3+120x^2=0

The zeros of 60x^4-180x^3+120x^2=0 are x=0,x=1,x=2

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60 x^{4} - 180 x^{3} + 120 x^{2} = 0

x = 0, x = 1, x = 2

Solution steps

60 x^{4} - 180 x^{3} + 120 x^{2} = 0

Factor

60 x^{4} - 180 x^{3} + 120 x^{2} : 60 x^{2} (x - 1) (x - 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 1 = 0 or x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 2 = 0 : x = 2

The solutions are

x = 0, x = 1, x = 2

x (x + 2) (x - 3) + 3 = 0

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

16 x^{4} - 145 x^{2} + 324 = 0

x, x^{3} - 5 x + 1 = - 0.8

x^{7} = 168

What are the solutions to the equation 60x^4-180x^3+120x^2=0 ?
The solutions to the equation 60x^4-180x^3+120x^2=0 are x=0,x=1,x=2
Find the zeros of 60x^4-180x^3+120x^2=0
The zeros of 60x^4-180x^3+120x^2=0 are x=0,x=1,x=2