Let us factorise

Given, f(y)= 2y³ – 5y² –19y+ 42

In f(y) putting y=±1, ±2, ±3, we see for which value of y, f(y)=0

f(2)=2.(2)³−5.(2)²−19.2+42=0

We observe that f(2) = 0

From factor theorem, we can say, (y−2) is a factor of f(y)

2y³ – 5y² –19y+ 42= 2y³ – 5y² –19y+ 42

= 2y³−4y²−y²+2y−21y+42

= 2y²(y−2)−y(y−2)−21(y−2)

= (y−2)(2y²−y−21)