What are the solutions to the equation n^2+n^3=-180 ?

The solutions to the equation n^2+n^3=-180 are n=-6,n= 5/2+i(sqrt(95))/2 ,n= 5/2-i(sqrt(95))/2

Find the zeros of n^2+n^3=-180

The zeros of n^2+n^3=-180 are n=-6,n= 5/2+i(sqrt(95))/2 ,n= 5/2-i(sqrt(95))/2

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n^{2} + n^{3} = - 180

n = - 6, n = \frac{5}{2} + i \frac{95}{2}, n = \frac{5}{2} - i \frac{95}{2}

Solution steps

n^{2} + n^{3} = - 180

Move

180

to the left side

n^{2} + n^{3} + 180 = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

n^{3} + n^{2} + 180 = 0

Factor

n^{3} + n^{2} + 180 : (n + 6) (n^{2} - 5 n + 30)

(n + 6) (n^{2} - 5 n + 30) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

n + 6 = 0 or n^{2} - 5 n + 30 = 0

Solve

n + 6 = 0 : n = - 6

Solve

n^{2} - 5 n + 30 = 0 : n = \frac{5}{2} + i \frac{95}{2}, n = \frac{5}{2} - i \frac{95}{2}

The solutions are

n = - 6, n = \frac{5}{2} + i \frac{95}{2}, n = \frac{5}{2} - i \frac{95}{2}

x^{3} - 27 x - 27 = 0

x^{3} + 5 x^{2} + 6 x = - 30

4 (x + 4)^{4} - 53 (x + 4)^{2} = - 49

0 = x^{3} - 2 x^{2} - 11 x + 12

a \cdot a \cdot a = 3 a

What are the solutions to the equation n^2+n^3=-180 ?
The solutions to the equation n^2+n^3=-180 are n=-6,n= 5/2+i(sqrt(95))/2 ,n= 5/2-i(sqrt(95))/2
Find the zeros of n^2+n^3=-180
The zeros of n^2+n^3=-180 are n=-6,n= 5/2+i(sqrt(95))/2 ,n= 5/2-i(sqrt(95))/2