# 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?

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

• Great catch! Upvoted. Commented Nov 22, 2020 at 17:02
• This is what was intended. Wonder what makes the admin not accept it till now @EmperorConcerto? Commented Nov 22, 2020 at 17:10

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

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.

• You also need to see that they both have leading coefficient one Commented Nov 22, 2020 at 17:08
• @HagenvonEitzen Sure, that is evident. Commented Nov 22, 2020 at 17:08

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.