This is an interesting problem, as there are quite a few ways of solving this, depending on how you feel like going. I'll show 2 different approaches, although there actually are a few more.
You'll note that in either approach, a vital part is the Polynomial Remainder Theorem (PRT) , which basically states that the remainder when you divide any polynomial #f(x)# by a term #(x-a)# is equal to the value of #f(a)#. As a corollary to that, if #(x-a)# is a factor of #f(x)#, it must divide evenly into it with no remainder, which means #f(a) = 0# in that case.
Method 1 - Calculus Style Solution
In this method, we begin with the fact that #(x-1)# is a factor of #f(x)# as well as #f'(x)#. This means that #(x-1)# must be a repeated factor in #f(x)#. Why? If #(x-1)# is a factor of #f'(x)#, upon integration (the inverse process to differentiation) the exponent of that factor would be increased by 1.
Thus, the form of #f(x)# must be similar to the following:
#f(x) = (x-1)^2(ax^2 + bx + c) #
Since #f(0) = 3# (given), we can see through substitution that #c = 3#:
#f(0) = (0-1)^2(a*0^2 + b*0 + c) = c = 3#
We can now update #f(x)#: #f(x) = (x-1)^2(ax^2 + bx + 3)#
Next, we know #f'(0) = -5#. We find #f'(x)# and use substitution to find #b#:
#f'(x) = (x-1)^2(2ax + b) + (ax^2+bx+3)(2*(x-1)) #
#f'(0) = (0-1)^2(2a*0+b) + (a*0^2+b*0+3)(2*(0-1)) #
#" "=b + (3)(-2) = b - 6 #
#f'(0) = -5 -> b - 6 = -5 :. b = 1#
This now gives us:
#f(x) = (x-1)^2(ax^2 + x + 3)#
Finally, since we know the remainder when dividing #f(x)# by #(x-2)# is 13, the PRT tells us #f(2) = 13#:
#f(2) = (2-1)^2(a*2^2 + 2 + 3) #
#" "=4a + 5#
#f(2) = 13 -> 4a + 5 = 13 :. a = 2#
The final polynomial is:
#f(x) = (x-1)^2(2x^2 + x + 3)#
#" "=2x^4 - 3x^3 + 3x^2 - 5x + 3#
Method 2 - Simultaneous Equations Solution
In this method, we trade using the Product Rule of Derivatives for using simultaneous equations. We begin this time by writing out the general form of a 4th degree polynomial:
#f(x) = ax^4 + bx^3 + cx^2 + dx + e#
Again, we start with #f(0)=3# and substitution to immediately get the result that #e = 3#. (Why? All other terms include #x#.) We now have this:
#f(x) = ax^4 + bx^3 + cx^2 + dx + 3#
Next, we turn to the reference to #f'(0)=-5#. We can find #f'(x)#, evaluate at 0, and resolve another coefficient:
#f'(x) = 4ax^3 + 3bx^2 + 2cx + d #
Since #f'(0) = -5#, by simple substitution again we get the result that #d = -5# by the same login. Updating both #f(x)# and #f'(x)# now gives us:
#f(x) = ax^4 + bx^3 + cx^2 - 5x + 3#
#f'(x) = 4ax^3 + 3bx^2 + 2cx - 5 #
Time for some other facts and the PRT. Since we are told that #(x-1)# is a factor of both #f(x)# and #f'(x)#, we know that #f(1) = 0# and #f'(1) = 0#. This leads to the following:
#f(1) = a + b + c - 5 + 3 #
# \ \ \ \ \ \ \ \= a + b + c - 2 = 0 #
# :. a + b + c = 2" " [A]#
#f'(1) = 4a + 3b + 2c - 5 = 0#
# :. 4a + 3b + 2c = 5" "[B]#
We have two equations for three unknowns. We need one additional equation, which comes from the PRT and the fact that #f(2) = 13# as a result of the remainder of division by #(x-2)# being 13.
#f(2) = 16a + 8b + 4c - 10 + 3 #
# " "= 16a + 8b + 4c - 7 = 13#
# :. 16a + 8b + 4c = 20 or 4a + 2b + c = 5" "[C]#
By simultaneously saving all 3 equations [A], [B], and [C], you can determine the values of #a,b, and c#.
The answer is #f(x) = 2x^4 - 3x^3 + 3x^2 - 5x + 3#
