[tex]\displaystyle\bf\\\Big(a+1\Big)^{2} -4a^{2} =\Big(a+1\Big)^{2} -\Big(2a\Big)^{2} =\Big(a+1-2a\Big)\Big(a+1+2a\Big)=\\\\=\Big(1-a\Big)\Big(3a+1\Big)[/tex]
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[tex]\displaystyle\bf\\\Big(a+1\Big)^{2} -4a^{2} =\Big(a+1\Big)^{2} -\Big(2a\Big)^{2} =\Big(a+1-2a\Big)\Big(a+1+2a\Big)=\\\\=\Big(1-a\Big)\Big(3a+1\Big)[/tex]