Find a third-degree polynomial Q such that Q(1)=0, Q'(1)=0, Q''(1)=2, and Q'''(1)=12.

Another approach might be to work work the problem backwards. For a third degree polynomial, Q''' is just a constant. Set that to 12. Inegrate to get Q''

=12x + C Adjust the constant for the desired result.
Repeat for Q' and Q.

Q(x)=ax^3+bx^2+cx+d

Q(1)=a+b+c+d=0
Q'(x)=3ax^2+2bx+c
Q'(1)=3a+2b+c=0
Q''(x)=6ax+2b
Q''(1)=6a+2b=2
Q'''(x)=6a
Q'''(1)=6a=12
so u can solve and find the polynom

To find a third-degree polynomial Q that satisfies the given conditions, we can use the general form of a third-degree polynomial:

Q(x) = ax^3 + bx^2 + cx + d

Given Q(1) = 0, we can substitute the value of x=1 into the equation and solve for d:

Q(1) = a(1)^3 + b(1)^2 + c(1) + d
0 = a + b + c + d

Given Q'(1) = 0, we can differentiate Q(x) with respect to x and substitute x=1:

Q'(x) = 3ax^2 + 2bx + c
Q'(1) = 3a(1)^2 + 2b(1) + c
0 = 3a + 2b + c

Given Q''(1) = 2, we can differentiate Q'(x) with respect to x and substitute x=1:

Q''(x) = 6ax + 2b
Q''(1) = 6a(1) + 2b
2 = 6a + 2b

Given Q'''(1) = 12, we can differentiate Q''(x) with respect to x and substitute x=1:

Q'''(x) = 6a
Q'''(1) = 6a
12 = 6a

Now, we have a system of equations that we can solve to find the coefficients of the polynomial Q(a, b, c, d):

0 = a + b + c + d -- (1)
0 = 3a + 2b + c -- (2)
2 = 6a + 2b -- (3)
12 = 6a -- (4)

From equation (4), we have 6a = 12, which gives us a = 2.
Substituting a = 2 into equations (2) and (3), we get:

0 = 3(2) + 2b + c => 0 = 6 + 2b + c => -6 = 2b + c -- (5)
2 = 6(2) + 2b => 2 = 12 + 2b => -10 = 2b => b = -5.

Now substituting b = -5 into equation (5), we get:

-6 = 2(-5) + c
-6 = -10 + c
c = 4

Finally, substituting a = 2, b = -5, and c = 4 into equation (1), we can solve for d:

0 = 2 - 5 + 4 + d
0 = 1 + d
d = -1

Therefore, the third-degree polynomial that satisfies the given conditions is:

Q(x) = 2x^3 - 5x^2 + 4x - 1.

the second answer is better...