C++I need really appreciate your help with the two questions, professor labeled them as a and a by accident I believe. Please please help, i'd really appreciate it. Thanks so so much! In this lab you will explore the conventional means to evaluate a polynomial and compare its efficiencyto Horner's method (see textbook, page 242).1) Implement each method as a C++ function2) Time each method against a set of test polynomials to collect some empirical data.Recommend using C++'s high resolution clock, for an example see time point:https://www.geeksforgeeks.org/chrono-in-c/3) Submit your code as a .cpp file and your findings as a PDF file.The conventional algorithm for evaluating a polynomial a„x" + a,-17-1 + ... + a,x + a, at x = c can be expressed in pseudocode byprocedure polynomial(c, aɔ, a,, .., an: real numbers)power := 1Y := dofor i := 1 to npower := power * cY:= Y + a; * powerreturn y {y = aņc" + an-1c²-1+ ... + a,c + ao}where the final value of y is the value of the polynomial at x = c.a) Evaluate 3x² + x + 1 at x = 2 by working through each step of the algorithm showing the values assigned at each assignment step.pusexercise. It is called Horner's method. This pseudocode shows how to use this method to find the value of a," + an-pr-1 + .. + a,x+ a, at x = c.Paprocedure Horner(c, Ag, aq. ɑz. .., qn : real numbers)Y = anfor i=1 to ny= y*c+ an-ireturn y {y = a,c" + an-1c"-1 + . + a;c + a,}a) Evaluate 3x² + x + 1 at x = 2 by working through each step of the algorithm showing the values assigned at each assignment step.

C++ program to solve the given problem is below.Answer a) CODE: #include <bits/stdc++.h> #include <chrono> #include <ctime> using namespace std; int main() { chrono::time_point<chrono::system_clock> starTime, endTime; int n; cin >> n; int arr[n + 1]; for (int i = 0; i < n + 1; i++) { cin >> arr[i]; } int c; cin >> c; int output = 0; starTime = chrono::system_clock::now(); for (int i = n; i >= 0; i--) { output += (arr[i] * pow(c, i)); } endTime = chrono::system_clock::now(); cout << "output: " << output << '\n'; chrono::duration<double> timeReq = endTime - starTime; cout << "Time required: " << timeReq.count() << "s\n"; } OUTPUT: SCREENSHOT OF CODE FOR REFERENCE: Answer b) 3x2 + x + 1, at x = 23×22 + 2 + 1 = 15 for (int i = n; i >= 0; i--) { output += (arr[i] * pow(c, i)); } In the above for loop the vale of array in this example is arr = {1,1,3} and c = 2, n = 2 we are iterating from i = 2 to i = 0 At each iteration i = 2, output += arr[i] * pow(c,i) = 3 * pow(2,2) = 12 i = 1, output += arr[i] * pow(c,i) = 1 * pow(2,1) = 2 i = 0, output += arr[i] * pow(c,i) = 1 * pow(2,0) = 1