[tex]p(x)=4x^2-5x+6[/tex]
[tex]c=1\implies p(c)=4(1)^2-5(1)+6=5[/tex]
By the remainder theorem, the remainder upon dividing a polynomial [tex]p(x)[/tex] by a linear binomial [tex]x-c[/tex] is equal to [tex]p(c)[/tex]. Via synthetic division, we get
1 | 4 -5 6
. | 4 -1
- - - - - - - - - - -
. | 4 -1 5
which translates to
[tex]\dfrac{4x^2-5x+6}{x-1}=4x-1+\dfrac5{x-1}[/tex]
or
[tex]4x^2-5x+6=(4x-1)(x-1)+5[/tex]
Indeed, when [tex]x=1[/tex], the first term on the right hand side vanished and we're left with 5.
