What are the solutions to the equation x^3+4x^2+x-26=0 ?

The solutions to the equation x^3+4x^2+x-26=0 are x=2,x=-3+2i,x=-3-2i

Find the zeros of x^3+4x^2+x-26=0

The zeros of x^3+4x^2+x-26=0 are x=2,x=-3+2i,x=-3-2i

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x^{3} + 4 x^{2} + x - 26 = 0

x = 2, x = - 3 + 2 i, x = - 3 - 2 i

Solution steps

x^{3} + 4 x^{2} + x - 26 = 0

Factor

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

(x - 2) (x^{2} + 6 x + 13) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x^{2} + 6 x + 13 = 0

Solve

x - 2 = 0 : x = 2

Solve

x^{2} + 6 x + 13 = 0 : x = - 3 + 2 i, x = - 3 - 2 i

The solutions are

x = 2, x = - 3 + 2 i, x = - 3 - 2 i

9 x^{3} - 25 x = 0

(1 + j)^{4} = 1.1

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

z^{5} = 4 - 4 i

(- x)^{4} = - (x)^{4}

What are the solutions to the equation x^3+4x^2+x-26=0 ?
The solutions to the equation x^3+4x^2+x-26=0 are x=2,x=-3+2i,x=-3-2i
Find the zeros of x^3+4x^2+x-26=0
The zeros of x^3+4x^2+x-26=0 are x=2,x=-3+2i,x=-3-2i