What are the solutions to the equation roots 2x^3+2x^2-19x+20=0 ?

The solutions to the equation roots 2x^3+2x^2-19x+20=0 are x=-4,x= 3/2+i 1/2 ,x= 3/2-i 1/2

Find the zeros of roots 2x^3+2x^2-19x+20=0

The zeros of roots 2x^3+2x^2-19x+20=0 are x=-4,x= 3/2+i 1/2 ,x= 3/2-i 1/2

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roots

2 x^{3} + 2 x^{2} - 19 x + 20 = 0

x = - 4, x = \frac{3}{2} + i \frac{1}{2}, x = \frac{3}{2} - i \frac{1}{2}

Solution steps

2 x^{3} + 2 x^{2} - 19 x + 20 = 0

Factor

2 x^{3} + 2 x^{2} - 19 x + 20 : (x + 4) (2 x^{2} - 6 x + 5)

(x + 4) (2 x^{2} - 6 x + 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 4 = 0 or 2 x^{2} - 6 x + 5 = 0

Solve

x + 4 = 0 : x = - 4

Solve

2 x^{2} - 6 x + 5 = 0 : x = \frac{3}{2} + i \frac{1}{2}, x = \frac{3}{2} - i \frac{1}{2}

The solutions are

x = - 4, x = \frac{3}{2} + i \frac{1}{2}, x = \frac{3}{2} - i \frac{1}{2}

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

y, x = y^{3} - 1

4 x^{3} + 12 x^{2} - 36 x = 0

x^{5} + x^{15} + x^{25} = 780

9 x^{4} - 42 x^{3} + 64 x^{2} - 35 x + 4 = 0

What are the solutions to the equation roots 2x^3+2x^2-19x+20=0 ?
The solutions to the equation roots 2x^3+2x^2-19x+20=0 are x=-4,x= 3/2+i 1/2 ,x= 3/2-i 1/2
Find the zeros of roots 2x^3+2x^2-19x+20=0
The zeros of roots 2x^3+2x^2-19x+20=0 are x=-4,x= 3/2+i 1/2 ,x= 3/2-i 1/2