(x+1)(x+2)(x+4)(x+8)=4x2

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Solution for (x+1)(x+2)(x+4)(x+8)=4x2 equation: 



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

We move all terms to the left:

(x+1)(x+2)(x+4)(x+8)-(4x^2)=0

determiningTheFunctionDomain
-4x^2+(x+1)(x+2)(x+4)(x+8)=0

We multiply parentheses ..

-4x^2+(+x^2+2x+x+2)(x+4)(x+8)=0

We multiply parentheses ..

-4x^2+(+x^2+2x+x+2)(+x^2+8x+4x+32)=0

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