we can do the same as before, make it an equation =0 and use the LCD to get integers for the coefficients.
[tex]\cfrac{1}{3}x^2-1=0\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3\left( \cfrac{1}{3}x^2-1 \right)=3(0)} \\\\\\ x^2-3=0\implies 1x^2+0x-3=0\implies 3(1x^2+0x-3)\qquad \begin{cases} a = 1\\ b = 0\\ c = -3 \end{cases}[/tex]
notice that when writing a polynomial in descending order, any missing terms are not really missing per se, they are there but in disguise with our very good friend Mr. Zero, 0.
