4d+Factoring

A "quadratic" is a polynomial that looks like "  a//x//2 + b//x// + c   ", where "   a   ", "   b   ", and "   c   " are just numbers. For the easy case of factoring, you will find two numbers that will not only multiply to equal the constant term "  c   ", but also add up to equal "   b   ", the coefficient on the   //x//   -term. For instance: * ** Factor   //x//2 + 5//x// + 6. **  I need to find factors of   6   that add up to   5. Since  6   can be written as the product of   2   and   3  , and since   2 + 3 = 5   , then I'll use   2   and   3. I know from [|multiplying polynomials]  that this quadratic is formed from multiplying two factors of the form "   (//x// + //m//)(//x// + //n//)   ", for some numbers  //  m  //  and  //  n  //. So I'll draw my parentheses, with an " //  x  //  " in the front of each: (//x// )(//x// ) Then I'll write in the two numbers that I found above: (//x// + 2)(//x// + 3) This is the answer:  //x//2 + 5//x// + 6 =   **(//x// + 2)(//x// + 3)**  Your text or teacher may refer to factoring "by grouping", which is covered in the lesson on [|simple factoring]. In the "easy" case of factoring, using "grouping" just gives you some extra work. For instance, in the above problem, in addition to finding the factors of  6   that add to   5  , you would have had to do these additional steps: //x//2 + 5//x// + 6 =  //x//   2   + 3//x// + 2//x// + 6 = (//x//  2   + 3//x//) + (2//x// + 6) = //x//(//x// + 3) + 2(//x// + 3) = (//x// + 3)(//x// + 2) So far, "  c   " has always been positive. What if  c   is negative? Since I am multiplying to a negative six, I need factors of opposite signs; that is, one factor will be positive and the other will be negative. The larger factor (in [|absolute value]  ) will get the "plus" sign, because I am adding to a positive   1. Since these opposite-signed numbers will be adding to  1  , I need the two factors to be one unit apart. The factor pairs for six are  1   and   6  , and   2   and   3. The second pair are one apart, so I want to use  2   and   3  , with the   3   getting the "plus" sign (so the   2   gets the "minus" sign). //x//2 + //x// – 6  =   **(//x// – 2)(//x// + 3)**. There is one special case, by the way, for factoring. Back when you were [|factoring]  plain old numbers, there were some numbers that didn't factor, such as   5   or   13. Recall that they are called "prime" numbers. The terminology is the same for polynomials: Since the constant term is negative, I'll be needing a positive and a negative number such that, when I multiply them together, I get  6  , but when I add them, I get   7. The factor pairs for  6   are   1   and   6  , and   2   and   3. You may think that I should use  1   and   6  , but — One of the factors has to be negative in order to multiply to get a "minus" six! Trying the first factor pair of  1   and   6  , the sum would be either   (–1) + 6 = 5   or else   1 + (–6) = –5. And the other factor pair,  2   and   3  , won't work, either, because   (–2) + 3 = 1   and   2 + (–3) = –1. In other words, there is no pair of factors of  –6   that will add to   +7. And if something isn't factorable, it's prime. Then  //x//2 + 7//x// – 6   is "   **prime**   ", or "   **unfactorable over the integers**   " (because I couldn't find integers that would work). ||
 * ** Factor   //x//2 + //x// – 6.  **
 * ** Factor   //x//2 + 7//x// – 6  **.