College Algebra 7th Edition

(a) $249999$ (b) $4819$ (c) $3999951$
We use the special product formula: (a) $501*499=(500+1)(500-1)=500^{2}-1=250000-1=249999$ (b) $79*61=(70+9)(70-9)=70^{2}-9^{2}=4900-81=4819$ (c) $2007*1993=(2000+7)(2000-7)=2000^{2}-7^{2}=4000000-49=3999951$