What are the solutions to the equation 2x^3-13x^2+26x-24=0 ?

The solutions to the equation 2x^3-13x^2+26x-24=0 are x=4,x= 5/4+i(sqrt(23))/4 ,x= 5/4-i(sqrt(23))/4

Find the zeros of 2x^3-13x^2+26x-24=0

The zeros of 2x^3-13x^2+26x-24=0 are x=4,x= 5/4+i(sqrt(23))/4 ,x= 5/4-i(sqrt(23))/4

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2 x^{3} - 13 x^{2} + 26 x - 24 = 0

x = 4, x = \frac{5}{4} + i \frac{23}{4}, x = \frac{5}{4} - i \frac{23}{4}

Solution steps

2 x^{3} - 13 x^{2} + 26 x - 24 = 0

Factor

2 x^{3} - 13 x^{2} + 26 x - 24 : (x - 4) (2 x^{2} - 5 x + 6)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 4 = 0 : x = 4

Solve

2 x^{2} - 5 x + 6 = 0 : x = \frac{5}{4} + i \frac{23}{4}, x = \frac{5}{4} - i \frac{23}{4}

The solutions are

x = 4, x = \frac{5}{4} + i \frac{23}{4}, x = \frac{5}{4} - i \frac{23}{4}

a^{5} = 19

2 x^{4} - 3 x^{2} + 10 = 0

(x^{2} + 3 x + 2)^{2} - 8 (x^{2} + 3 x) - 4 = 0

x^{4} + 3 x^{3} - 5 x - 15 = 0

z, z^{2} (p^{2} z^{2} + q^{2}) = 1

What are the solutions to the equation 2x^3-13x^2+26x-24=0 ?
The solutions to the equation 2x^3-13x^2+26x-24=0 are x=4,x= 5/4+i(sqrt(23))/4 ,x= 5/4-i(sqrt(23))/4
Find the zeros of 2x^3-13x^2+26x-24=0
The zeros of 2x^3-13x^2+26x-24=0 are x=4,x= 5/4+i(sqrt(23))/4 ,x= 5/4-i(sqrt(23))/4