What are the solutions to the equation (x^2-49)(x^2+16)=0 ?

The solutions to the equation (x^2-49)(x^2+16)=0 are x=-7,x=7,x=4i,x=-4i

Find the zeros of (x^2-49)(x^2+16)=0

The zeros of (x^2-49)(x^2+16)=0 are x=-7,x=7,x=4i,x=-4i

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(x^{2} - 49) (x^{2} + 16) = 0

x = - 7, x = 7, x = 4 i, x = - 4 i

Solution steps

(x^{2} - 49) (x^{2} + 16) = 0

Factor

(x^{2} - 49) (x^{2} + 16) : (x + 7) (x - 7) (x^{2} + 16)

(x + 7) (x - 7) (x^{2} + 16) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 7 = 0 or x - 7 = 0 or x^{2} + 16 = 0

Solve

x + 7 = 0 : x = - 7

Solve

x - 7 = 0 : x = 7

Solve

x^{2} + 16 = 0 : x = 4 i, x = - 4 i

The solutions are

x = - 7, x = 7, x = 4 i, x = - 4 i

x^{3} - 4 x^{2} - 12 x = 0

x^{3} - 9 x + 3 = 0

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

7 x^{2} (3 x - 5) = 0

48 x^{5} = 27 x^{3}

What are the solutions to the equation (x^2-49)(x^2+16)=0 ?
The solutions to the equation (x^2-49)(x^2+16)=0 are x=-7,x=7,x=4i,x=-4i
Find the zeros of (x^2-49)(x^2+16)=0
The zeros of (x^2-49)(x^2+16)=0 are x=-7,x=7,x=4i,x=-4i