Question 1049759
P(x)=9x^3+12x^2+hx+k. at x=1/3


2=9(1/3)^3+12(1/3)^2+h(1/3)+k.
2=9(1/27)+12(1/9)+h(1/3)+k.
2=(1/3)+(4/3)+(h/3)+k.
6=(1)+(4)+(h)+3k.

{{{1}}}={{{h+3k}}}

at x=-2
-5=9(-2)^3+12(-2)^2+h(-2)+k.

-5=-72+12(4)-2h+k.

-5=-72+(48)-2h+k.
-5=(-24)-2h+k.

{{{19}}}={{{-2h+k}}}

solve
{{{1}}}={{{h+3k}}}
{{{19}}}={{{-2h+k}}}

*[invoke linear_substitution "h", "k", 1, 3, 1, -2, 1, 19 ]