Let p(x)=2x4–5x3+2x2–x+2
Factorise x2–3x+2.
Now, x2–3x+2=x2–2x–x+2 [By splitting middle term]
=x(x–2)–1(x–2)=(x–1)(x–2)
Hence, zeros of x2–3x+2 are 1 and 2.
We have to prove that, 2x4–5x3+2x2–x+2 is divisible by x2–3x+2.
i.e. prove that p(1)=0 and p(2)=0
Now. p(1)=2(1)4–5(1)3+2(1)2–1+2
=2–5+2–1+2=6–6=0
p(2)=2(2)4–5(2)3+2(2)2–2+2
=2×16–5×8+2×4+0
=32−40+8=40–40=0
Hence, p(x) is divisible by x2–3x+2.