What are the solutions to the equation (x-3)(x-1)^2(x^2+4x+13)=0 ?

The solutions to the equation (x-3)(x-1)^2(x^2+4x+13)=0 are x=3,x=1,x=-2+3i,x=-2-3i

Find the zeros of (x-3)(x-1)^2(x^2+4x+13)=0

The zeros of (x-3)(x-1)^2(x^2+4x+13)=0 are x=3,x=1,x=-2+3i,x=-2-3i

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(x - 3) (x - 1)^{2} (x^{2} + 4 x + 13) = 0

x = 3, x = 1, x = - 2 + 3 i, x = - 2 - 3 i

Solution steps

(x - 3) (x - 1)^{2} (x^{2} + 4 x + 13) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 3 = 0 or x - 1 = 0 or x^{2} + 4 x + 13 = 0

Solve

x - 3 = 0 : x = 3

Solve

x - 1 = 0 : x = 1

Solve

x^{2} + 4 x + 13 = 0 : x = - 2 + 3 i, x = - 2 - 3 i

The solutions are

x = 3, x = 1, x = - 2 + 3 i, x = - 2 - 3 i

- 2 x^{3} + x^{2} + 13 x + 6 = 0

x x^{2} = 4

x^{2} (4 x + 8)^{2} = 14 0^{2}

x^{8} = 81 x

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

What are the solutions to the equation (x-3)(x-1)^2(x^2+4x+13)=0 ?
The solutions to the equation (x-3)(x-1)^2(x^2+4x+13)=0 are x=3,x=1,x=-2+3i,x=-2-3i
Find the zeros of (x-3)(x-1)^2(x^2+4x+13)=0
The zeros of (x-3)(x-1)^2(x^2+4x+13)=0 are x=3,x=1,x=-2+3i,x=-2-3i