Q. 134.3( 3 Votes )

Using factor theo

Let, f (y) = 2y3-5y2-19y+42

The factors of the constant term + 42 are

Putting y = 2, we have

f (2) = 2 (2)3 – 5 (2)2 – 19 (2) + 42

= 16 – 20 - 38 + 42

= 0

So, (y - 2) is a factor of f (y)

Let us now divide

f (y) = 2y3-5y2-19y+42 by (y - 2) to get the other factors of f (x)

Using long division method, we get

2y3-5y2-19y+42 = (y - 2) (2y2 – y – 21)

2y2 – y - 21 = (y + 3) (2y – 7)

Hence, 2y3-5y2-19y+42 = (y - 2) (2y - 7) (y + 3)

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