Divide 57 into two parts whose product is 680.

Let the two parts be x and (57 − x).

According to the given condition,

$x(57-x)=680$

$\Rightarrow 57 x-x^{2}=680$

$\Rightarrow x^{2}-57 x+680=0$

$\Rightarrow x^{2}-40 x-17 x+680=0$

$\Rightarrow x(x-40)-17(x-40)=0$

$\Rightarrow(x-40)(x-17)=0$

$\Rightarrow x-40=0$ or $x-17=0$

$\Rightarrow x=40$ or $x=17$

When x = 40,
57 − x = 57 − 40 = 17

When x = 17,
57 − x = 57 − 17 = 40

Hence, the required parts are 17 and 40.