What are the solutions to the equation (x^2+2)(x+5)=0 ?

The solutions to the equation (x^2+2)(x+5)=0 are x=sqrt(2)i,x=-sqrt(2)i,x=-5

Find the zeros of (x^2+2)(x+5)=0

The zeros of (x^2+2)(x+5)=0 are x=sqrt(2)i,x=-sqrt(2)i,x=-5

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(x^{2} + 2) (x + 5) = 0

x = 2 i, x = - 2 i, x = - 5

Solution steps

(x^{2} + 2) (x + 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x^{2} + 2 = 0 or x + 5 = 0

Solve

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

Solve

x + 5 = 0 : x = - 5

The solutions are

x = 2 i, x = - 2 i, x = - 5

3 x^{3} + 5 x^{2} = 28 x

x = 5 x - x^{3}

(x + 1) 4 - (x - 1) 4 = x^{2} (8 x + 1) + 15

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

V = \frac{4}{3} π r^{3}, r

What are the solutions to the equation (x^2+2)(x+5)=0 ?
The solutions to the equation (x^2+2)(x+5)=0 are x=sqrt(2)i,x=-sqrt(2)i,x=-5
Find the zeros of (x^2+2)(x+5)=0
The zeros of (x^2+2)(x+5)=0 are x=sqrt(2)i,x=-sqrt(2)i,x=-5