${2t^4 + 3t^3 – 2t^2 – 9t – 12}$
${t^2 – 3}$
$2t^2$
$+3t$
$+4$
${2t^4}$     ${-6t^2}$
($-$)
     ($+$)
$3t^3+4t^2-9t-12$
${3t^3}$      ${-9t}$
($-$)
       ($+$)  
${4t^2}$    ${-12}$
${4t^2}$    ${-12}$
($-$)    ($+$)
0
Therefore:${2t^4 + 3t^3 – 2t^2 – 9t – 12}$ = ${(t^2 – 3)}(2t^2+3t+4) + 0 $
Since the remainder is zero, hence first polynomial is a factor of second polynomial.