$(6x+2)(x+5)(x+15)(x+18)+360=0$ [duplicate]

Solution 1:

$f(x) = (x+4)(x+7)(x+8)(x+11)+20\ $ is symmetric about the line $x = -7.5$

$y = x + 7.5$

$(y-3.5)(y-0.5)(y+0.5)(y+3.5)+20=0$

$(y^2 - 0.5^2)(y^2 -3.5^2)+20=0$

$y^4 - 12.5 y^2 + 23.0625 = 0$

Use the quadratic formula to solve for $y^2.$ The square roots of that give $y.$ And, subtract $7.5$ to get $x.$

Solution 2:

If we're looking for integer solutions, we can see that $x + 4$, $x + 7$, $x + 8$, and $x + 11$ must be four integers whose product is $-20$, and so each must be $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$. The difference between the smallest and largest factor is $(x + 11) - (x + 4) = 7$, so those two factors must be either $-5$ and $2$ or $-2$ and $5$. These values correspond to the possibilities $x = -9, -6$, and substituting gives that these are indeed solutions.

With these in hand, one can expand the quartic into standard form and divide by the linear factors $x + 6$ and $x + 9$ we now know it has, leaving a quadratic to which we may apply the quadratic formula to find the remaining roots.

Of course, most quartic polynomials are irreducible over $\Bbb Q$, and so most quartic equations cannot be solved this way.

Solution 3:

Rearranging the LHS, $$(x+4)(x+11)(x+7)(x+8)+20=0$$ to have $$(x^2+15x+44)(x^2+15x+56)+20=0$$ Now setting $x^2+15x=t$ gives $$(t+44)(t+56)+20=0,$$ i.e. $$t^2+100t+2484=0\iff (t+54)(t+46)=0$$

Now solve $$x^2+15x=-54,-46$$