If $p(x)=ax^3 -2x^2 +bx+c$, find $a, b$ and $c$ if $p(0)=12$, $p(-1)=3$ and $p(2)=36$ If $p(x)=ax^3 -2x^2 +bx+c$, find $a, b$ and $c$ if $p(0)=12$,  $p(-1)=3$ and $p(2)=36$
Can someone please teach me how to do this question thanks!

 A: Use $p(2)=36$ , you'll have
$8a - 8 + 2b + 12 = 36$.
Simplify it, you'll get $4a+b=16$.
Together with the equation you've written down,
$a=3$,$b=4$ 
A: This is really a problem in basic linear algebra.  If we simply plug in the three given values of the independent variable $x$, we obtain the three equations for $a$, $b$, $c$:
$p(0) = 12$ becomes
$c = 12$;
$p(-1) = 3$ becomes
$-a -b +c = 5$;
$p(2) = 36$ becomes
$8a + 2b + c = 44$;
looks like we get $c$ for free!  Plugging $c= 12$ into the second and third equations yields
$-a -b = -7$;
$8a + 2b = 32$;
looks like we get
$6a = 18$,
$a = 3$;
and finally,
$b= 4$.
This method actually generalizes to arbitrary polynomials; the powers of the different values of $x$ form a matrix and the coefficients form a vector; the rest is clearly linear algabra.  Glad to be of assistance in this matter.  Cheers.
A: So, you already found out that $c=12$. You also showed that $-a-b+10=3$, which we can simplify to $-a-b=-7$.
Doing a similar substitution,
$$a(2)^3-2(2)^2+b(2)+c=36.$$
The arithmetic is not too difficult:
$$8a-8+2b+c=36.$$
We can definitely move that $-8$ over, and remember how we already found out that $c=12$? Let's substitute that too:
$$8a+2b+12=44.$$
Now for the 12:
$$\color{blue}{8a+2b}=\color{blue}{32}.$$
This is good. Remember how we said that $-a-b=-7$? Let's just multiply that by $2$:
$$\color{red}{-2a-2b}=\color{red}{-14}.$$
Now check this out: if we add these two previous equations together, the $b$-terms will cancel (i.e., $2b-2b=0$):
$$\color{blue}{8a+2b}\color{red}{-2a-2b}=\color{blue}{32}\color{red}{-14}$$
or
$$6a=18.$$
From here, it is very easy to find $a$, and now that you have two of the three variables, finding $b$ should be very easy after a bit of substitution. I hope this helps.
