What is the correct method for squaring a binomial?

Answer 1

First, memorize this equation: (a-b)(a-b), which, factored out will be: a2-2ab+b2 or, if your bionomial looks like this... (a+b)(a+b), leave everything positive and you'll get a2+2ab+b2 If you don't know how to get to those aswers, use the FOIL method, which stands for Firsts, Outer, Inner, Last! For this example use (a+b)(a+b) First - multiply the first variable in the first paranthesis with the first in the second parenthesis. (a+b)(a+b) a x a = a2 Outer - multiply the outer 2 variables. Use the first variable of the first parenthesis and the last of the second parenthesis. (a+b)(a+b) a x b = ab Inner - multiply the inner variables. Use the second variable of the first parenthisis and the first variable of the second parenthesis. (a+b)(a+b) b x a = ba or also written as ab Last - multiply the last variables. (a+b)(a+b) b x b = b2 Now when you add all that together, you have a2+ab+ba+b2 which, in its smallest version is a2+2ab+b2 A generalized way of doing this is to use the formula for the binomial coefficient: C(n,k) = n!/(k!(n-k)!). This formula is actually a nice concise way of writing the number of unique size-k combinations (each of which is called a k-combination) which can be made from n different objects. It turns out from inspection that the coefficient in from of a term in the multiplication (a+b)(a+b), say the term ab, is equal to C(n,k), where n = 2 (the order of the term - ab is an order 2 term), and k = 1 (the power on one of the terms - will be explained shortly). This gives C(2,1) = 2!/(1!1!) = 2, meaning that the product ab shows up 2 times in the process of doing the multiplication shown above. To give a more complex example and to better illustrate how this works, consider doing the product (a+b)(a+b)(a+b). What if I wanted to know what the coefficient would be on the a2 b term? Identify n and k first. Here, n = 3 since the order of the term is 3. In this case, the powers on the individual terms a and b are different, but see what happens if you pick k = 2 (for a) and k = 1 (for b) in separate calculations. First, C(3,2) = 3!/(2!(3-2)!) = 3. If we try it with k = 1, we get C(3,1) = 3!/(1!(3-1)!) = 3 as well! So regardless of which variable of which we choose to use the power for k, we get that the coefficient is equal to 3. In short, you can choose either and you will get the right answer. So finally, how does this allow you to calculate a polynomial product? All you have to do is think of all of the combinations of terms that might come from the multiplication. (a+b)(a+b)(a+b) gives the following: a3, a2 b, ab2, and b3. Once you write these down, you need only sum them with their appropriate coefficients. (a+b)(a+b)(a+b) = C(3,3) a3 + C(3,2)a2 b + C(3,1)ab2 + C(3,0)b3 = a3 + 3a2 b + 3ab2 + b3 Just remember that 0! = 1 (for calculating C(3,0)). This is obviously true because we need C(3,3) = C(3,0) since on this last term (b3), we can choose the power on either variable (the power on a being 0 since a0 = 1) and in both cases the coefficient must be the same. This method may also be used for polynomials with more than 2 terms each. An example is (a+b+c)(a+b+c). This topic falls under the area of combinatorics, which is quite interesting in its own right. Enjoy.