What are the solutions to the equation (x^2-4)(x^2+1)(x-3)=0 ?

The solutions to the equation (x^2-4)(x^2+1)(x-3)=0 are x=-2,x=2,x=3,x=i,x=-i

Find the zeros of (x^2-4)(x^2+1)(x-3)=0

The zeros of (x^2-4)(x^2+1)(x-3)=0 are x=-2,x=2,x=3,x=i,x=-i

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(x^{2} - 4) (x^{2} + 1) (x - 3) = 0

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

Solution steps

(x^{2} - 4) (x^{2} + 1) (x - 3) = 0

Factor

(x^{2} - 4) (x^{2} + 1) (x - 3) : (x + 2) (x - 2) (x - 3) (x^{2} + 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

Solve

x - 3 = 0 : x = 3

Solve

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

The solutions are

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

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

x, x^{3} y - y^{2} = 3

x^{2} (x^{2} + 4) (x^{2} + 16) - 100 = 0

x^{4} - 2 x + 1 = 0

- t^{3} + 5 t^{2} - 3 t - 9 = 0

What are the solutions to the equation (x^2-4)(x^2+1)(x-3)=0 ?
The solutions to the equation (x^2-4)(x^2+1)(x-3)=0 are x=-2,x=2,x=3,x=i,x=-i
Find the zeros of (x^2-4)(x^2+1)(x-3)=0
The zeros of (x^2-4)(x^2+1)(x-3)=0 are x=-2,x=2,x=3,x=i,x=-i