Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$?

Question

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?

Answer 1

For $a=x^2+5x+4$ we get to factor $a(a+2)-48=a^2+2a-48=(a+8)(a-6)$.

Answer 2

$$(x^2 + 5x + 5-1)(x^2 + 5x + 5+1) - 48 = (x^2 + 5x + 5)^2-7^2$$

Answer 3

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.

Answer 4

For me it's just Vieta's formulas (or comparing coefficients as if $x^2+5x$ were a single variable):

$(x^2 + 5x + \color{red}4)(x^2 + 5x + \color{red}6) \color{red}{-48} = (x^2 + 5x + \color{red}{12})(x^2 + 5x \color{red}{-2})$?

$4+6=12+(-2)$, and $4\cdot 6 - 48 = 12 \cdot (-2)$, so yes.