Q. 104.2( 17 Votes )

Let us factorise

Answer :

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)


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