text: how do you factor an algebraic expression? im in my first year algerbra and i am on chapter 10 section 5 in the prentice-hall book if that makes a diffrence! 1) x^2 - 4   this is a problem called the difference of squares. they are presented in the form of (a)^2 - (b)^2. a difference of squares problem always factors to (a+b)(a-b). in this case, a = x and b^2 = 4. sqrt(4) = 2, so b=2. plug it in to your algorithm, an you have (x+2)*(x-2)  2) 4x^2 - 9 another difference of squares. as a starter, sqrt(4x^2) = 2x, and sqrt(9) = 3.   3) x^2 + 8x + 16 now for something completely different. the following is a true statement. (ax + b)(cx + d) = (ac)x^2 + (bc)x + (da)x + (bd) so, if this will factor, you need to find a, b, c, and d so that ac = 1, bd = 16, and bc+da = 8. [the last one is true because of the distributive property of multiplication, in other words "xz + yz = (x+y)z" ]  the first assumption we are going to make is that because ac = 1; a = 1 and c = 1. this won't always be true, and if we can't find something to match the others, we can always go back and change that assumption. we do this because it's the easiest to work with. since a=1 and c=1, then we can rewrite "bc + da = 8" as "b+d=8". now, go back to when you learned to multiply signed numbers. the only way you can have a positive number as a product is when both multipliers are the same sign. now, go back farther to adding signed numbers and remember that a negative plus a negative is always going to be a negative. since "bd" is positive and "b+d" is positive, then both b and d are positive. in fact, the only two numbers that fulfill both statements are b=4, d=4 (if that jump's too great for you, factor 16 and you get 2*2*2*2. that means your choices for pairs are 2,8 (adds to 10, so wrong); 1,16 (17); and 4,4 (8).) so, when we plug our numbers back into that statement a few dozen lines ago, we get the factoring of  (x+4)*(x+4), or (x+4)^2  4) 4(x^2) + 4x + 1 ok, this is the same deal. (ax+b)(cx+d)=ac(x^2)+adx+bcx+bd so, "bd=1", "ad+bc=4", and "ac=4" since we have another product of 1; assume, for a minute, that "b=1", "d=1". that means "a+c=4" and "ac=4". again, both product and sum are positive, so both "a" and "c" are positive. factor 4, and you get 2*2. so your choices are 1,4 (sum of 5); and 2,2 (4). so, plug back in and you get (2x+1)(2x+1) or (2x+1)^2  5) 2d^2 - 18. this one is a tricky one. you can't do the difference of squares, because sqrt(2) and sqrt(18) are both irrational. but there is something you can factor out.... right, factor out a two and you have (2)(d^2-9) or (2)(d^2 - 3^2) now, you have a difference of squares, and can easily get the answer. (2)(d+3)(d-3)