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

The solutions to the equation 2r^3-4r^2-r+2=0 are r=2,r=-1/(sqrt(2)),r= 1/(sqrt(2))

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

The zeros of 2r^3-4r^2-r+2=0 are r=2,r=-1/(sqrt(2)),r= 1/(sqrt(2))

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

r = 2, r = - \frac{1}{2}, r = \frac{1}{2}

Solution steps

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

Factor

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

(r - 2) (2 r + 1) (2 r - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

r - 2 = 0 or 2 r + 1 = 0 or 2 r - 1 = 0

Solve

r - 2 = 0 : r = 2

Solve

2 r + 1 = 0 : r = - \frac{1}{2}

Solve

2 r - 1 = 0 : r = \frac{1}{2}

The solutions are

r = 2, r = - \frac{1}{2}, r = \frac{1}{2}

- 2 y^{4} - 7 y^{2} - 4 = 0

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

7 x^{4} - 21 x = 0

x, f = (x^{2} - 2)^{2}

x, x^{8} + 15 x^{4} = 16

What are the solutions to the equation 2r^3-4r^2-r+2=0 ?
The solutions to the equation 2r^3-4r^2-r+2=0 are r=2,r=-1/(sqrt(2)),r= 1/(sqrt(2))
Find the zeros of 2r^3-4r^2-r+2=0
The zeros of 2r^3-4r^2-r+2=0 are r=2,r=-1/(sqrt(2)),r= 1/(sqrt(2))