What are the solutions to the equation λ^3-5λ^2+8λ-4=0 ?

The solutions to the equation λ^3-5λ^2+8λ-4=0 are λ=1,λ=2

Find the zeros of λ^3-5λ^2+8λ-4=0

The zeros of λ^3-5λ^2+8λ-4=0 are λ=1,λ=2

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λ^{3} - 5 λ^{2} + 8 λ - 4 = 0

λ = 1, λ = 2

Solution steps

λ^{3} - 5 λ^{2} + 8 λ - 4 = 0

Factor

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

(λ - 1) (λ - 2)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

λ - 1 = 0 or λ - 2 = 0

Solve

λ - 1 = 0 : λ = 1

Solve

λ - 2 = 0 : λ = 2

The solutions are

λ = 1, λ = 2

255 = 5 x^{3}

290 x^{3} + 54 x^{2} + 18 x - 221 = 0

x^{3} - 3 x^{2} - 3 x + 1 = 0

x (x + 2) (x + 4) = 0

x (6 x^{2} - x - 2) = 0

What are the solutions to the equation λ^3-5λ^2+8λ-4=0 ?
The solutions to the equation λ^3-5λ^2+8λ-4=0 are λ=1,λ=2
Find the zeros of λ^3-5λ^2+8λ-4=0
The zeros of λ^3-5λ^2+8λ-4=0 are λ=1,λ=2