What are the solutions to the equation roots x^3-5x^2+8x-4 ?

The solutions to the equation roots x^3-5x^2+8x-4 are x=1,x=2

Find the zeros of roots x^3-5x^2+8x-4

The zeros of roots x^3-5x^2+8x-4 are x=1,x=2

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roots

x^{3} - 5 x^{2} + 8 x - 4

x = 1, x = 2

Solution steps

x^{3} - 5 x^{2} + 8 x - 4

The roots are the intercepts with the x-axis

(y = 0)

x^{3} - 5 x^{2} + 8 x - 4 = 0

Factor

x^{3} - 5 x^{2} + 8 x - 4 : (x - 1) (x - 2)^{2}

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 2 = 0 : x = 2

The solutions are

x = 1, x = 2

2 x^{4} + 9 x^{3} + 15 x^{2} + 11 x + 3 = 0

x^{3} + x^{2} - 4 x - 4

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

x^{3} - 5 x^{2} - 2 x + 6 = 0

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

What are the solutions to the equation roots x^3-5x^2+8x-4 ?
The solutions to the equation roots x^3-5x^2+8x-4 are x=1,x=2
Find the zeros of roots x^3-5x^2+8x-4
The zeros of roots x^3-5x^2+8x-4 are x=1,x=2