Question 557526
i.
I suspect you mean
"When the expression C is divided by x-2, the remainder is R. When the expression is divided by x+1, the remainder is still R."
You have a quadratic trinomial {{{P(x)=ax^2+bx+c}}}
When you divide a polynomial P(x) by (x-a), the original polynomial, P(x), the quotient polynomial, Q(x), and the remainder R, are related through:
P(x)=Q(x)(x-a)+R (and P(a)=R).
So,
P(x)-R = Q(x)(x-a) is divisible by x-a.
In the case of your polynomial, P(x)-R is divisible by x-2 and by x+1.
So it is divisible by 
{{{(x-2)(x+1)=x^2-x-2}}}
Then {{{P(x)-R=ax^2+bx+c=a(x-2)(x+1)=a(x^2-x-2)=ax^2-ax-2a}}} and
{{{P(x)=a(x-2)(x+1)+R=ax^2-ax-2a+R}}}
So {{{b=-a}}}
ii.
P(4) = {{{a(4-2)(4+1)+R=10a+R=2R}}}
{{{10a+R=2R}}} --> {{{R=10a}}}
{{{c=-2a+R=-2a+10a=8a}}}
iii.
P(t) =  {{{at^2-at+8a}}} = {{{5R}}} = {{{50a}}}
{{{at^2-at+8a=50a}}} --> {{{t^2-t-42a=0}}}
Factoring, we get {{{t^2-t-42a=(t-7)(t+6)}}}
so {{{(t-7)(t+6)=0}}} has solutions {{{t-7}}} and {{{t=-6}}}