### 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 \(-4\times 3=-12\).

\(a^2+a -6={}\)\((a-2)(a+3)\).

We try to find integers \(p\) and \(q\) such that \(a^2+a -6=(a+p)(a+q)\).

Expansion of brackets in the right-hand side then gives: \[a^2+a -6=a^2+(p+q)a+p\times q\] So we try to find integers \(p\) and \(q\) so that \(p+q=1\) and \(p \times q= -6\).

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: \[a^2+a -6=(a-2)(a+3)\]

We coupled the sum-product method to quadratic polynomials, but sometimes they are in disguise within the algebraic expressions. The examples below illustrate this.

\(a^{5}+3a^{4} + 2a^3={}\)\(a^3(a+1)(a+2)\).

Note first that all terms can be divided by \(a^3\) so that \[a^{5}+3a^{4} + 2a^3=a^3(a^2+3 a + 2)\] 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 \(a^2+3 a + 2=(a+p)(a+q)\).

Expansion of brackets in the right-hand side then gives: \[a^2+3a + 2=x^2+(p+q)a+p\times q\] So we try to find integers \(p\) and \(q\) so that \(p+q=3\) and \(p \times q= 2\).

Because we may interchage \(p\) and \(q\) it suffices to choose \(p\) such that \(p^2\le 2\),

i.e., choosing \(p\) with \(|p|\le\sqrt{2}\approx 1.414\).

We make a table of possibilities integers: \[\begin{array}{||r|r|r||} \hline p & q & p+q\\ \hline 1 & 2 & 3\\ \hline -1 & -2 & -3\\ \hline \end{array}\] \(p=1\) and \(q=2\) meet the requirements.

The factorisation is: \[a^2+3a + 2=(a+1)(a+2)\] The final result is: \[a^{5}+3a^{4} + 2a^3=a^3(a+1)(a+2)\]

Mathcentre video clips

Factorization of a Quadratic Equation by Inspection (42:36)