Using factor theorem, factorize of the polynomials:

Solution:

Given, $f(x)=2 y^{3}-5 y^{2}-19 y+42$

The constant in f(x) is + 42

The factors of 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42

Let, y – 2 = 0

=> y = 2

$f(2)=2(2)^{3}-5(2)^{2}-19(2)+42$

= 16 – 20 – 38 + 42

= 0

So, (y – 2) is the factor of f(y)

Now, divide f(y) with (y – 2) to get other factors

By, long division method

$2 y^{2}-y-21$

$y-22 y^{3}-5 y^{2}-19 y+42$

$2 y^{3}-4 y^{2}$

$(-) \quad(+)$

$-y^{2}-19 y$

 

$-y^{2}+2 y$

(+)      (-)

- 21y + 42

- 21y + 42

(+)       (-)

0

$=>2 y^{3}-5 y^{2}-19 y+42=(y-2)\left(2 y^{2}-y-21\right)$

Now,

$2 y^{2}-y-21$

The factors are (y + 3) (2y – 7)

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