A Polynomial that Passes through the following four points? I'm trying to do this for practice but I'm just going nowhere with it, I'd love to see some work and answers on it.
Thanks :)
Find a polynomial that passes through the points (-2,-1), (-1,7), (2,-5), (3,-1). Present the answer in standard form.
What I've tried:


 A: Hints: 
Way 1: Consider the polynomial
$$\small A(x+1)(x-2)(x-3)+B(x+2)(x-2)(x-3)+C(x+2)(x+1)(x-3)+D(x+2)(x+1)(x-2).$$
You can find constants $A,B,C,D$ such that the above polynomial will do the job. For example, to make the polynomial be equal to $-1$ at $-2$, all we need to do is to make $A(-2+1)(-2+-2)(-2+-3)=1$. 
The hard thing about this procedure is presenting the answer in standard form. That is a routine but very unpleasant calculation.   
Way 2: Let our polynomial be $ax^3+bx^2+cx+d$.  Because the curve $y=ax^3+bx^2+cx+d$  passes through $(-2,-1)$ we have
$$-8a+4b-2c+d=-1.$$
Similarly, we obtain three other linear equations in $4$ unknowns. It akes a fair amount of routine work, but you can solve the resulting system of $4$ linear equations in $4$  unknowns, and then you are finished. 
A: $$f(x) = a x^3 + b x^2 + c x + d$$
\begin{eqnarray*} 
-1 &=& a (-2)^3 + b(-2)^2 + c(-2) + d \\
 7 &=& a (-1)^3 + b(-1)^2 + c(-1) + d \\
-5 &=& a(2)^3 + b (2)^2 + c (2) + d \\
-1 &=& a(3)^3 + b(3)^2 + c (3) + d
\end{eqnarray*}
or 
\begin{eqnarray*} 
-8 a  + 4b -2c + d &=&  -1\\
-{\ }a{\ } +{\ } b -{\ } c + d &=&{\ } 7 \\
 8a + 4b  + 2c  + d&=&-5 \\
27a + 9b + 3c + d&=&-1
\end{eqnarray*}
with solution
\begin{eqnarray*} 
a&=&{\ }{\ }1\\
b&=&-2\\
c&=&-5\\
d&=&{\ }{\ }5
\end{eqnarray*}
and so $f(x) = x^3 -2x^2-5x+5$
A: Step 1 : List the equations that define the function according to ax3 + bx2 + cx + d:
-8a+ 4b - 2c + d = -1 (equation1)
-1a + 1b + -1c + d = 7 (equation2) 
8a + 4b + 2c + d = -5 (equation3) 
27a + 9b + 3c + d = -1 (equation4)
Step 2 : Use matrix operations to simplify the system of equations
for example: add/subtract multiples of equation1 from/to each of the equations below it to remove the a's from them.
Then add/subtract multiples of the new equation2 from/to the equations below it to remove the b's from them. .
Then add/subtract multiples of the new equation3 from/to the equations below it (which is only equation 4) to remove the c's from it.
After completing this process, the system of equations will be easy to solve since equation4 will end up having 1 variable (which will be d), which you can calculate and substitute in equation3 to get c and then substitute d and c into equation2 and so on
