Q. 134.3( 3 Votes )

# Using factor theorem, factorize each of the following polynomial:

2*y*^{3}-*5y*^{2}-19*y*+42

Answer :

Let, f (y) = 2*y*^{3}-*5y*^{2}-19*y*+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) = 2*y*^{3}-*5y*^{2}-19*y*+42 by (y - 2) to get the other factors of f (x)

Using long division method, we get

2*y*^{3}-*5y*^{2}-19*y*+42 = (y - 2) (2y^{2} – y – 21)

2y^{2} – y - 21 = (y + 3) (2y – 7)

Hence, 2*y*^{3}-*5y*^{2}-19*y*+42 = (y - 2) (2y - 7) (y + 3)

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