Calculating with letters: Computing with letters
Factorisation of a quadratic polynomial via the sum-product method
A quadratic equation in the variable \(x\) is an expression of the form \[ax^2+bx+c=0\] for certain numbers \(a\), \(b\), and \(c\) with \(a\neq0\). The basic form of the sum-product method assumes \(a=1\).
The sum-product method
In the sum-product method, also called product-sum-method or factorisation by inspection, we try to factorise \(x^2+b\,x+c\) as \((x+p)(x+q)\) for certain \(p\) en \(q\). If you expand brackets in the factored form, then we get \[x^2+b\,x+c=x^2+(p+q)x+p\times q\text.\] Therefore, the task has become to find two numbers \(p\) and \(q\) such that \[p+q=b\quad\mathrm{and}\quad p\times q=c\]
Examples
\[x^2+3x+2=(x+1)(x+2)\] because \(1+2=3\) and \(1\times 2=2\).
\[x^2-x-12=(x-4)(x+3)\] because \(-4+3=-1\) and \(-4\times 3=-12\).
\(t^2+t -6={}\)\((t-2)(t+3)\).
We try to find integers \(p\) and \(q\) such that \(t^2+t -6=(t+p)(t+q)\).
Expansion of brackets in the right-hand side then gives: \[t^2+t -6=t^2+(p+q)t+p\times q\] So we try to find integers \(p\) and \(q\) so that \(p+q=1\) and \(p \times q= -6\).
Because we may interchage \(p\) and \(q\) it suffices to choose \(p\) such that \(p^2\le 6\),
i.e., choosing \(p\) with \(|p|\le\sqrt{6}\approx 2.449\).
We make a table of possibilities integers: \[\begin{array}{||r|r|r||} \hline p & q & p+q\\ \hline 1 & -6 & -5\\ \hline -1 & 6 & 5\\ \hline 2 & -3 & -1\\ \hline -2 & 3 & 1\\ \hline \end{array}\] \(p=-2\) and \(q=3\) meet the requirements.
The factorisation is: \[t^2+t -6=(t-2)(t+3)\]
We coupled the sum-product method to quadratic polynomials, but sometimes they are in disguise within the algebraic expressions. The examples below illustrate this.
\(a^{4}+a^{3} -12a^2={}\)\(a^2(a-3)(a+4)\).
Note first that all terms can be divided by \(a^2\) so that \[a^{4}+a^{3} -12a^2=a^2(a^2+a -12)\] and that the quadratic polynomial between the brackets is in the form in which the product-sum method with integer coefficients is applicable.
Next we try to find integers \(p\) and \(q\) such that \(a^2+a -12=(a+p)(a+q)\).
Expansion of brackets in the right-hand side then gives: \[a^2+a -12=x^2+(p+q)a+p\times q\] So we try to find integers \(p\) and \(q\) so that \(p+q=1\) and \(p \times q= -12\).
Because we may interchage \(p\) and \(q\) it suffices to choose \(p\) such that \(p^2\le 12\),
i.e., choosing \(p\) with \(|p|\le\sqrt{12}\approx 3.464\).
We make a table of possibilities integers: \[\begin{array}{||r|r|r||} \hline p & q & p+q\\ \hline 1 & -12 & -11\\ \hline -1 & 12 & 11\\ \hline 2 & -6 & -4\\ \hline -2 & 6 & 4\\ \hline 3 & -4 & -1\\ \hline -3 & 4 & 1\\ \hline \end{array}\] \(p=-3\) and \(q=4\) meet the requirements.
The factorisation is: \[a^2+a -12=(a-3)(a+4)\] The final result is: \[a^{4}+a^{3} -12a^2=a^2(a-3)(a+4)\]
Mathcentre video
Factorization of a Quadratic Equation by Inspection (42:36)