Question

If 4x^3-6x^2+1=(x-2)Q(x)+ax-3 for all values of x ,where Q(x) is a polynomial in x ,find
(i)the value of constant a
(ii)Q(x)

Answer 1

the statement must be true for all values of x
suppose we let x = 2 , (can you see why ?)

then
4x^3-6x^2+1=(x-2)Q(x)+ax-3
32 - 24 + 1 = (0)(Q(x) + 2a - 3
9 = 2a - 3
2a = 12
a = 6

so 4x^3-6x^2+1=(x-2)Q(x)+ax-3
becomes
4x^3-6x^2+1=(x-2)Q(x)+6x-3
4x^3 - 6x^2 - 6x + 3 = (x-2)Q(x)
Q(x) = (4x^3 - 6x^2 - 6x + 3)/(x-2)

I noticed your other post is just about the same type.