What are the solutions to the equation 0=(x+3)(x^2+4) ?

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

Find the zeros of 0=(x+3)(x^2+4)

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

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

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

Solution steps

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

Switch sides

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 3 = 0 : x = - 3

Solve

x^{2} + 4 = 0 : x = 2 i, x = - 2 i

The solutions are

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

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

- 8 m^{5} + 24 = 0

(\frac{1 + r}{100})^{10} = 1.5775

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

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

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