What are the solutions to the equation 2x^3-7x^2-5x+4=0 ?

The solutions to the equation 2x^3-7x^2-5x+4=0 are x=-1,x= 1/2 ,x=4

Find the zeros of 2x^3-7x^2-5x+4=0

The zeros of 2x^3-7x^2-5x+4=0 are x=-1,x= 1/2 ,x=4

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2 x^{3} - 7 x^{2} - 5 x + 4 = 0

x = - 1, x = \frac{1}{2}, x = 4

Decimal

x = - 1, x = 0.5, x = 4

Solution steps

2 x^{3} - 7 x^{2} - 5 x + 4 = 0

Factor

2 x^{3} - 7 x^{2} - 5 x + 4 : (x + 1) (2 x - 1) (x - 4)

(x + 1) (2 x - 1) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or 2 x - 1 = 0 or x - 4 = 0

Solve

x + 1 = 0 : x = - 1

Solve

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

Solve

x - 4 = 0 : x = 4

The solutions are

x = - 1, x = \frac{1}{2}, x = 4

7 x^{3} - 2 x^{2} - 35 x + 10 = 0

5.5 x^{2} + 5.5 x + 105.5 = 107.5 x^{3}

x^{4} - 16 x^{2} - 720 = 0

x^{4} = (5 x - 66)^{2}

x^{4} + 30 x^{2} + 225 = 0

What are the solutions to the equation 2x^3-7x^2-5x+4=0 ?
The solutions to the equation 2x^3-7x^2-5x+4=0 are x=-1,x= 1/2 ,x=4
Find the zeros of 2x^3-7x^2-5x+4=0
The zeros of 2x^3-7x^2-5x+4=0 are x=-1,x= 1/2 ,x=4