What are the solutions to the equation (x^2+x)^2-8(x^2+x)+12=0 ?

The solutions to the equation (x^2+x)^2-8(x^2+x)+12=0 are x=1,x=-3,x=-2,x=2

Find the zeros of (x^2+x)^2-8(x^2+x)+12=0

The zeros of (x^2+x)^2-8(x^2+x)+12=0 are x=1,x=-3,x=-2,x=2

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(x^{2} + x)^{2} - 8 (x^{2} + x) + 12 = 0

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

Solution steps

(x^{2} + x)^{2} - 8 (x^{2} + x) + 12 = 0

Expand

(x^{2} + x)^{2} - 8 (x^{2} + x) + 12 : x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12

x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12 = 0

Factor

x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12 : (x - 1) (x + 3) (x + 2) (x - 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

x + 3 = 0 : x = - 3

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

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

725 = 5 x^{3}

27 x^{3} = 8

x^{3} - 7 x + 5 = 0

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

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

What are the solutions to the equation (x^2+x)^2-8(x^2+x)+12=0 ?
The solutions to the equation (x^2+x)^2-8(x^2+x)+12=0 are x=1,x=-3,x=-2,x=2
Find the zeros of (x^2+x)^2-8(x^2+x)+12=0
The zeros of (x^2+x)^2-8(x^2+x)+12=0 are x=1,x=-3,x=-2,x=2