What are the solutions to the equation 12(x(x^2-4x+1)+6)=0 ?

The solutions to the equation 12(x(x^2-4x+1)+6)=0 are x=-1,x=2,x=3

Find the zeros of 12(x(x^2-4x+1)+6)=0

The zeros of 12(x(x^2-4x+1)+6)=0 are x=-1,x=2,x=3

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12 (x (x^{2} - 4 x + 1) + 6) = 0

x = - 1, x = 2, x = 3

Solution steps

12 (x (x^{2} - 4 x + 1) + 6) = 0

Factor

12 (x (x^{2} - 4 x + 1) + 6) : 12 (x + 1) (x - 2) (x - 3)

12 (x + 1) (x - 2) (x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 2 = 0 or x - 3 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

Solve

x - 3 = 0 : x = 3

The solutions are

x = - 1, x = 2, x = 3

x^{3} + 2 x^{2} + 8 = 0

(z^{2} - 1)^{2} - 10 (z^{2} - 1) + 21 = 0

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

x^{3} - 7 x^{2} + 10 x + 26 = 0

5 x^{4} - 6 x^{3} - 2 x^{2} = 0

What are the solutions to the equation 12(x(x^2-4x+1)+6)=0 ?
The solutions to the equation 12(x(x^2-4x+1)+6)=0 are x=-1,x=2,x=3
Find the zeros of 12(x(x^2-4x+1)+6)=0
The zeros of 12(x(x^2-4x+1)+6)=0 are x=-1,x=2,x=3