3x² + 490x + 6000 Use the "ac" method: Multiply 3 by 6000 Get 18000 We write all the factor pairs of 18000 as well as the sum of the factor pairs, and see if any give you a sum of 490 18000 × 1; 18000 + 1 = 18001 9000 × 2; 9000 + 2 = 9002 6000 × 3; 6000 + 3 = 6003 4500 × 4; 4500 + 4 = 4504 3600 × 5; 3600 + 5 = 3605 3000 × 6; 3000 + 6 = 3006 2250 × 8; 2250 + 8 = 2258 2000 × 9; 2000 + 9 = 2009 1800 × 10; 1800 + 10 = 1810 1500 × 12; 1500 + 12 = 1512 1200 × 15; 1200 + 15 = 1215 1125 × 16; 1125 + 16 = 1141 1000 × 18; 1000 + 18 = 1018 900 × 20; 900 + 20 = 920 750 × 24; 750 + 24 = 774 720 × 25; 720 + 25 = 745 600 × 30; 600 + 30 = 630 500 × 36; 500 + 36 = 536 450 × 40; 450 + 40 = 490 400 × 45; 400 + 45 = 445 375 × 48; 375 + 48 = 423 360 × 50; 360 + 50 = 410 300 × 60; 300 + 60 = 360 250 × 72; 250 + 72 = 322 240 × 75; 240 + 75 = 315 225 × 80; 225 + 80 = 305 200 × 90; 200 + 90 = 290 180 × 100; 180 + 100 = 280 150 × 120; 150 + 120 = 270 144 × 125; 144 + 125 = 269 and we find that there is indeed one pair of factors, 450 and 40 with sum 490. So we write 3x² + 490x + 6000 as 3x² + 450x + 40x + 6000 Then we factor 3x out of the first two terms and 40 out of the last two terms: 3x(x + 150) + 40(x + 150) Next we factor out common factor of (x + 150) (x + 150)(3x + 40) What a horrible problem! Edwin