What are the solutions to the equation (x-3)(x^2+1)=0 ?

The solutions to the equation (x-3)(x^2+1)=0 are x=3,x=i,x=-i

Find the zeros of (x-3)(x^2+1)=0

The zeros of (x-3)(x^2+1)=0 are x=3,x=i,x=-i

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(x - 3) (x^{2} + 1) = 0

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

Solution steps

(x - 3) (x^{2} + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 3 = 0 or x^{2} + 1 = 0

Solve

x - 3 = 0 : x = 3

Solve

x^{2} + 1 = 0 : x = i, x = - i

The solutions are

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

c^{3} + 2 c = 2 c^{5}

n x^{3} + 8 = 11

0 = \frac{1}{6} x^{3} - x

x^{4} + 2 x^{3} - 3 x^{2} + 10 x + 25 = 0

12 x^{4} - 12 x^{3} + 1 = 0

What are the solutions to the equation (x-3)(x^2+1)=0 ?
The solutions to the equation (x-3)(x^2+1)=0 are x=3,x=i,x=-i
Find the zeros of (x-3)(x^2+1)=0
The zeros of (x-3)(x^2+1)=0 are x=3,x=i,x=-i