[tex]p(x-1)[/tex] is a quadratic in the form
[tex]a(x-1)^2 + b(x-1) + c[/tex]
We want to write [tex]p(x-1)[/tex] instead in the form
[tex]a' x^2 + b' x + c[/tex]
To do this, simply expand the first form completely.
[tex]a(x-1)^2 + b(x-1) + c = a(x^2 - 2x + 1) + b(x-1) + c \\\\ ~~~~~~~~ = ax^2 + (-2a + b)x + (a - b + c)[/tex]
Then [tex]a'=a[/tex], [tex]b'=-2a+b[/tex], and [tex]c'=a-b+c[/tex].
Given
[tex]p(x-1) = 2(x-1)^2 + 3(x-1) + 4[/tex]
(so [tex]a=2[/tex], [tex]b=3[/tex], and [tex]c=4[/tex]) it follows that
[tex]a' = a = 2[/tex]
[tex]b' = -2a + b = -2^2 + 3 = -1[/tex]
[tex]c' = a - b + c = 2 - 3 + 4 = 3[/tex]
[tex]\implies p(x-1) = \boxed{2}\,x^2 - \boxed{1}\,x + \boxed{3}[/tex]
