<s>[INST] 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!

 [/INST] 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.
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