What are the solutions to the equation n^3-3n^2+2n-336=0 ?

The solutions to the equation n^3-3n^2+2n-336=0 are n=8,n=-5/2+i(sqrt(143))/2 ,n=-5/2-i(sqrt(143))/2

Find the zeros of n^3-3n^2+2n-336=0

The zeros of n^3-3n^2+2n-336=0 are n=8,n=-5/2+i(sqrt(143))/2 ,n=-5/2-i(sqrt(143))/2

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n^{3} - 3 n^{2} + 2 n - 336 = 0

n = 8, n = - \frac{5}{2} + i \frac{143}{2}, n = - \frac{5}{2} - i \frac{143}{2}

Solution steps

n^{3} - 3 n^{2} + 2 n - 336 = 0

Factor

n^{3} - 3 n^{2} + 2 n - 336 : (n - 8) (n^{2} + 5 n + 42)

(n - 8) (n^{2} + 5 n + 42) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

n - 8 = 0 or n^{2} + 5 n + 42 = 0

Solve

n - 8 = 0 : n = 8

Solve

n^{2} + 5 n + 42 = 0 : n = - \frac{5}{2} + i \frac{143}{2}, n = - \frac{5}{2} - i \frac{143}{2}

The solutions are

n = 8, n = - \frac{5}{2} + i \frac{143}{2}, n = - \frac{5}{2} - i \frac{143}{2}

x, x^{3} - y^{3} + z^{3} = 4

m^{4} + 2 m^{3} + 3 m^{2} + 2 m + 1 = 0

x^{3} - 2 x + 1 = 0

3 x^{2} \cdot 2 x - 5 = 0

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

What are the solutions to the equation n^3-3n^2+2n-336=0 ?
The solutions to the equation n^3-3n^2+2n-336=0 are n=8,n=-5/2+i(sqrt(143))/2 ,n=-5/2-i(sqrt(143))/2
Find the zeros of n^3-3n^2+2n-336=0
The zeros of n^3-3n^2+2n-336=0 are n=8,n=-5/2+i(sqrt(143))/2 ,n=-5/2-i(sqrt(143))/2