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 \(-4\times 3=-12\).
\(a^2+4a + 3={}\)\((a+3)(a+1)\).
We try to find integers \(p\) and \(q\) such that \(a^2+4a + 3=(a+p)(a+q)\).
Expansion of brackets in the right-hand side then gives: \[a^2+4a + 3=a^2+(p+q)a+p\times q\] So we try to find integers \(p\) and \(q\) so that \(p+q=4\) and \(p \times q= 3\).
Because we may interchage \(p\) and \(q\) it suffices to choose \(p\) such that \(p^2\le 3\),
i.e., choosing \(p\) with \(|p|\le\sqrt{3}\approx 1.732\).
We make a table of possibilities integers: \[\begin{array}{||r|r|r||} \hline p & q & p+q\\ \hline 1 & 3 & 4\\ \hline -1 & -3 & -4\\ \hline \end{array}\] \(p=3\) and \(q=1\) meet the requirements.
The factorisation is: \[a^2+4a + 3=(a+3)(a+1)\]
We coupled the sum-product method to quadratic polynomials, but sometimes they are in disguise within the algebraic expressions. The examples below illustrate this.
\(a^{6}-4a^{5} + 3a^4={}\)\(a^4(a-3)(a-1)\).
Note first that all terms can be divided by \(a^4\) so that \[a^{6}-4a^{5} + 3a^4=a^4(a^2-4 a + 3)\] 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 \(a^2-4 a + 3=(a+p)(a+q)\).
Expansion of brackets in the right-hand side then gives: \[a^2-4a + 3=x^2+(p+q)a+p\times q\] So we try to find integers \(p\) and \(q\) so that \(p+q=-4\) and \(p \times q= 3\).
Because we may interchage \(p\) and \(q\) it suffices to choose \(p\) such that \(p^2\le 3\),
i.e., choosing \(p\) with \(|p|\le\sqrt{3}\approx 1.732\).
We make a table of possibilities integers: \[\begin{array}{||r|r|r||} \hline p & q & p+q\\ \hline 1 & 3 & 4\\ \hline -1 & -3 & -4\\ \hline \end{array}\] \(p=-3\) and \(q=-1\) meet the requirements.
The factorisation is: \[a^2-4a + 3=(a-3)(a-1)\] The final result is: \[a^{6}-4a^{5} + 3a^4=a^4(a-3)(a-1)\]
Mathcentre video
Factorization of a Quadratic Equation by Inspection (42:36)