Without actual division, prove that `2x^4-5x^3+2x^2-x+2` is exactly divisible by `x^2-3x+2.`

Explanation

Let `p(x) = 2x^4 – 5x^3 + 2x^2 – x+ 2` firstly, factorise `x^2-3x+2`.
Now, `x^2-3x+2 = x^2-2x-x+2` [by splitting middle term]
`= x(x-2)-1 (x-2)= (x-1)(x-2)`
Hence, zeroes of `x^2-3x+2` are `1` and `2`.
We have to prove that, `2x^4 – 5x^3 + 2x^2 – x+ 2` is divisible by` x^2-3x+2` i.e.,
to 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`
and `p(2) = 2(2)^4 – 5(2)^3 + 2(2)^2 – 2 + 2 = 2x^16-5x^8+2x^4+ 0 = 32 – 40 + 8 = 40 – 40 =0`
Hence, `p(x)` is divisible by `x^2-3x+2`.