(A) 9x²-13x-10 Multiply the 9 by the 10 ignoring signs. Get 90 Write down all the ways to have two positive integers which have product 90, starting with 90*1 90*1 45*2 30*3 18*5 15*6 10*9 Since the last sign in 9x²-13x-10 is -, SUBTRACT them, and place the DIFFERENCE out beside that: 90*1 90-1=89 45*2 45-2=44 30*3 30-3=27 18*5 18-5=13 15*6 15-6= 9 10*9 10-9= 1 Now, again ignoring signs, we find in that list of sums the coefficient of the middle term in 9x²-13x-10 So we replace the number 13 by 18-5 9x²-13x-10 9x²-(18-5)x-10 Then we distribute to remove the parentheses: 9x²-18x+5x-10 Factor the first two terms 9x²-18 by taking out the greatest common factor, getting 9x(x-2) Factor the last two terms +5x-10 by taking out the greatest common factor, getting +5(x-2) So we have 9x(x-2)+5(x-2) Notice that there is a common factor, (x-2) 9x(x-2)+5(x-2) which we can factor out leaving the 9x and the +5 to put in parentheses: (x-2)(9x+5) --------------------------- (B) 12x²-32x+21 Multiply the 12 by the 21 ignoring signs. Get 252 Write down all the ways to have two positive integers which have product 252, starting with 252*1 252*1 126*2 84*3 63*4 42*6 36*7 28*9 21*12 18*14 Since the last sign in 12x²-32x+21 is +, ADD them, and place the SUM out beside that: 252*1 252+1=253 126*2 126+2=128 84*3 84+3-87 63*4 63+4=67 42*6 42+6=48 36*7 36+7=43 28*9 28+9=37 21*12 21+12=33 18*14 18+14=32 Now, again ignoring signs, we find in that list of sums the coefficient of the middle term in 12x²-32x+21 So we replace the number 32 by 18+14 12x²-32x+21 12x²-(18+14)x+21 Then we distribute to remove the parentheses: 12x²-18x-14x+21 Factor the first two terms 12x²-18x by taking out the greatest common factor, getting 6x(2x-3) Factor the last two terms -14x+21 by taking out the greatest common factor, getting -7(2x-3) So we have 6x(2x-3)-7(2x-3) Notice that there is a common factor, (2x-3) 6x(2x-3)-7(2x-3) which we can factor out leaving the 6x and the -7 to put in parentheses: (2x-3)(6x-7) Edwin