What are the solutions to the equation (x-1)(2x^2-5x+2)=0 ?

The solutions to the equation (x-1)(2x^2-5x+2)=0 are x=1,x=2,x= 1/2

Find the zeros of (x-1)(2x^2-5x+2)=0

The zeros of (x-1)(2x^2-5x+2)=0 are x=1,x=2,x= 1/2

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(x - 1) (2 x^{2} - 5 x + 2) = 0

x = 1, x = 2, x = \frac{1}{2}

Decimal Notation

x = 1, x = 2, x = 0.5

Solution steps

(x - 1) (2 x^{2} - 5 x + 2) = 0

Factor

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

(x - 1) (x - 2) (2 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

x - 2 = 0 : x = 2

Solve

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

The solutions are

x = 1, x = 2, x = \frac{1}{2}

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

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

r^{5} + 5 r^{4} - 1 = 0

(a - 2) (a + 2) - (a - 2)^{2} a = 1

(x^{2} - 3 x + 3) (x^{2} - 2 x + 3) = 2 x^{2}

What are the solutions to the equation (x-1)(2x^2-5x+2)=0 ?
The solutions to the equation (x-1)(2x^2-5x+2)=0 are x=1,x=2,x= 1/2
Find the zeros of (x-1)(2x^2-5x+2)=0
The zeros of (x-1)(2x^2-5x+2)=0 are x=1,x=2,x= 1/2