Factorisation by inspection Suppose \[x^2+b\,x+c=(x+p)(x+q)\] for certain \(p\) and \(q\). Then finding the solution of the original equation is easy, namely \(x=-p\) or \(x=-q\). In this case, the following equality should be valid: \[x^2+b\,x+c=x^2+(p+q)x+p\times q\] Therefore, the task has become to find two numbers \(p\) and \(q\) such that \[p+q=b\quad\text{and}\quad p\times q=c\] The task actually has not become much easier, but sometimes you are lucky (in case of small integral coefficients) and you can see the solutions right in front of your eyes. But when it does not work, you do not really know if it is lack of inspiration or that there is indeed no solution possible in the real numbers. In that case, there is not much else to do than finding zeros via other methods and techniques.

Because of the task to find numbers \(p\) and \(q\) such that \(p+q=b\) and\(p\times q=c\), this method is also called the sum-product-method or product-sum-method. The last name indicates that the search for such numbers starts with finding pairs of integers with the prescribed product.