Solve the equation `(x+2)(x+3)(x+8)xx(x+12)=4x^2dot`

Explanation

Since `(-2)(-12)=(-3)(-8)` so we can we write given equation as
`(x+2)(x+12)(x+3)(x+8)=4x^(2)`
`implies(x^(2)+14x+24)(x^(2)+11x+24)=4x^(2)`……I
Now `x=0` is not a root of given equation on dividing by `x^(2)` in both sides of Eq. (i) we get
`(x+24/x+14)(x+24/x+11)=4` ........ii
Put `x+24/x=y` then Eq. (ii) can be reduced in the form
`(y+14)(y+11)=4` or `y^(2)+25y+150=0`
`:.y_(1)=-15` and `y_(2)=-10`
Thus, the original equation is equivalent to the collection of equatiosn
`[(x+24/x=-15),(x+24/x=-10):]`
i.e. `[(x^(2)+15x+24=0),(x^(2)+10x+24=0):]`
On solving these collection we get
`x_(1)=(-15-sqrt(129))/2,x_(2)=(-15+sqrt(129))/2,x_(3)=-6,x_(4)=-4`