[tex]\dfrac{k+3}{4k-2}\cdot(12k^2-3)=\dfrac{(k+3)(12k^2-3)}{(2)(2k)-(2)(1)}\\\\=\dfrac{(k+3)[(3)(4k^2)-(3)(1)]}{2(2k-1)}=\dfrac{(k+3)(3)(4k^2-1)}{2(2k-1)}\\\\=\dfrac{3(k+3)[(2k)^2-1^2]}{2(2k-1)}=\dfrac{3(k+3)(2k-1)(2k+1)}{2(2k-1)}=\dfrac{3(k+3)(2k+1)}{2}[/tex]
[tex]\text{Used}\\\\a(b\pm c)=ab\pm ac\\\\a^2-b^2=(a-b)(a+b)[/tex]
