Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$? As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?

If $x^2 + 5x+12 = 0$ then
$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (-8)(-6)-48 = 0.$$
If $x^2 + 5x-2 = 0$ then
$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = 6\cdot 8-48 = 0.$$
Therefore LHS and RHS have the same zeroes. Since RHS has no double roots and both sides are polynomials of degree $4$, we conclude LHS = RHS.
