3x²+5x+2 Multiply the 3 by the 2 ignoring signs. Get 6. Write down all the ways to have two positive integers which have product 6, starting with 6*1 6*1 3*2 Since the last sign in 3x²+5x+2 is +, ADD them, and place the SUM out beside that: 6*1 6+1=7 3*2 3+2=5 Now, again ignoring signs, we find in that list of sums the coefficient of the middle term in 3x²+5x+2 which is 5 So we replace the number 5 by 3+2 3x²+5x+2 3x²+(3+2)x+2 Then we distribute to remove the parentheses: 3x²+3x+2x+2 Factor the first two terms 3x²+3x by taking out the greatest common factor, getting 3x(x+1) Factor the last two terms +2x+2 by taking out the greatest common factor, +2, getting +2(x+1) So we have 3x(x+1)+2(x+1) Notice that there is a common factor, (x+1) 3x(x+1)+2(x+1) which we can factor out leaving the 3x and the +2 to put in parentheses: (x+1)(3x+2) ------------------------------ 5x²-14x-3 Multiply the 5 by the 3 ignoring signs. Get 15 Write down all the ways to have two positive integers which have product 15, starting with 15*1 15*1 5*3 Since the last sign in 5x²-14x-3 is -, SUBTRACT them, and place the DIFFERENCE out beside that: 15*1 15-1=14 5*3 5-3=2 Now, again ignoring signs, we find in that list of sums the coefficient of the middle term in 5x²-14x-3 So we replace the number 14 by 15-1 5x²-14x-3 5x²-(15-1)x-3 Then we distribute to remove the parentheses: 5x²-15x+1x-3 Factor the first two terms 5x²-15x by taking out the greatest common factor, 5x, getting 5x(x-3) Factor the last two terms +1x-3 by taking out the greatest common factor, +1, getting +1(x-3) So we have 5x(x-3)+1(x-3) Notice that there is a common factor, (x-3) 5x(x-3)+1(x-3) which we can factor out leaving the 5x and the +1 to put in parentheses: (x-3)(5x+1) --------------------------- The others are done just like the first one. I wrote a lesson on this method of factoring. Go here: http://www.algebra.com/algebra/homework/Expressions-with-variables/change-this-name32371.lesson Edwin