What are the solutions to the equation 4x^3+12x^2+16x=0 ?

The solutions to the equation 4x^3+12x^2+16x=0 are x=0,x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2

Find the zeros of 4x^3+12x^2+16x=0

The zeros of 4x^3+12x^2+16x=0 are x=0,x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2

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4 x^{3} + 12 x^{2} + 16 x = 0

x = 0, x = - \frac{3}{2} + i \frac{7}{2}, x = - \frac{3}{2} - i \frac{7}{2}

Solution steps

4 x^{3} + 12 x^{2} + 16 x = 0

Factor

4 x^{3} + 12 x^{2} + 16 x : 4 x (x^{2} + 3 x + 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x^{2} + 3 x + 4 = 0 : x = - \frac{3}{2} + i \frac{7}{2}, x = - \frac{3}{2} - i \frac{7}{2}

The solutions are

x = 0, x = - \frac{3}{2} + i \frac{7}{2}, x = - \frac{3}{2} - i \frac{7}{2}

x^{4} - 6 x^{3} + 6 x^{2} - 6 x + 5 = 0

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

z, (a) z^{4} = 24 i - 7

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

2 x^{2} - 4 x^{4} = 5

What are the solutions to the equation 4x^3+12x^2+16x=0 ?
The solutions to the equation 4x^3+12x^2+16x=0 are x=0,x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2
Find the zeros of 4x^3+12x^2+16x=0
The zeros of 4x^3+12x^2+16x=0 are x=0,x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2