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

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

Find the zeros of r^3-5r^2+8r-4=0

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

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

r = 1, r = 2

Solution steps

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

r - 1 = 0 or r - 2 = 0

Solve

r - 1 = 0 : r = 1

Solve

r - 2 = 0 : r = 2

The solutions are

r = 1, r = 2

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

x^{3} - 6 x + 9 = 0

x^{3} + 2 x^{2} - 24 x = 0

12 x^{3} - 5 x + 132 = 0

2 x^{3} - 2 x^{2} - 4 x + 1 = 0

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