The step is to plug in some of integers to guess solutions. It’s found x=-2 is one of the solutions of the equation.

Then we factor the equation in order to produce the common factor of x+2.

Minus 4 from both sides

x^2(1+x)^2+x^2-4=8(1+x)^2-4

Move 4(1+x)^2 from RHS to LHS

x^2(1+x)^2-4(1+x)^2+x^2-4=4(1+x)^2-4

Then,

(1+x)^2(x^2-4)+(x^2-4)=4[(1+x)^2-1]

(x^2-4)[(1+x)^2+1]=4x(x+2)

(x+2)[(x-2)(x^2+2x+2)-4x]=0

(x+2)(x^3-6x-4)=0

Now we get one solution x=-2 and a cubic equation

x^3-6x-4=0

Similarly, it’s found x=-2 is also the solution of the cubic equation. So we aim to produce a common factor x+2.

x^3+8-6x-12=0

(x+2)(x^2-2x-2)=0

Solve the equation, we get three solutions

x=-2

or

x=1\pm \sqrt{3}

Steven Zheng posted 2 months ago

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