Let
p(x)=2x3–3x2–17x+30
Constant term of p(x) = 30
∴ Factors of 30 are
±1,±2,±3,±5,±6,±10,±15,±30
By trial, we find that p(2) = 0, so (x – 2) is a factor of p(x).
[∵2(2)3−3(2)2−17(2)+30=16−12−34+30=0]
Now, we see that
2x3–3x2–17x+30
=(x−2)(2x2+x−15)
2x2+x−15=2x(x+3)–5(x+3) [By splitting the middle term]
=(x+3)(2x–5)
∴ 2x3–3x2–17x+30=(x−2)(x+3)(2x−5)