Factorise the expression f(x)=2 x^{3}-7 x^{2}-3 x+18. Hence, find all possible values of x for which f(x)= 0
Factorise the expression f(x)=2 x^{3}-7 x^{2}-3 x+18. Hence, find all possible values of x for which f(x)= 0

Given: f(x) = 2x3 – 7x2 – 3x + 18

By hit and trial method

For x = 2, the value of f(x) will be

f(2) = 2(2)3 – 7(2)2 – 3(2) + 18

= 16 – 28 – 6 + 18 = 0

As f(2) = 0, (x – 2) is a factor of f(x).

Now, performing long division we have

Therefore, factorized form of f(x) is,

f(x) = (x – 2) (2x2 – 3x – 9)

= (x – 2) (2x2 – 6x + 3x – 9)

= (x – 2) [2x(x – 3) + 3(x – 3)]

= (x – 2) (x – 3) (2x + 3)

Now, for f(x) = 0 we get roots as,

(x – 2) (x – 3) (2x + 3) = 0

Hence x = 2, 3 or -3/2