Factorisation of a quadratic polynomial via the sum-product method

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

In order to factorise \(x^2+b\, x+c\) as \((x+p)(x+q)\) we set ourselves the task to find two numbers such that their sum equals the given \(b\) and their product equals the given \(c\). Therefore, the task has become to find two numbers \(p\) and \(q\) such that \[p+q=b\quad\mathrm{and}\quad p\times q=c\] 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 factorisation possible within the set of the real numbers.

In the example of \(x^2-x-12\), for example, you try to find two numbers of which the product equals\(-12\) and the sum equals \(-1\), because the term \(-x\) can be read as \(-1x\). Because the product is negative one of the numbers searched for must be negative too, say \(p\). On a piece of srap paper or in your head you inspect the sequence \[-12=-12\times 1=-6\times 2=-4\times 3 = -3\times 4 = -2\times 6=-1\times 12\] It is more handy to create a table of all possibilities and calculate the sum of the two numbers: \[\begin{array}{||c|c|c||} \hline p & q & p+q\\ \hline -12 & 1 & -11\\ \hline -6 & 2 & -4\\ \hline -4 & 3 & -1\\ \hline -3 & 4 & 1\\ \hline -2 & 6 & 4\\ \hline -1 & 12 & 11\\ \hline\end{array}\] From the table you read off that the numbers \(-4\) and \(3\) have the sum \(-1\).
Thus: \[x^2-x-12=(x-4)(x+3)\] You can verify this results by expanding the brackets in the right-hand side.

Decompose \(\;a^2+3a + 2\;\) into factors with integer coefficients via the sum-product method.

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

We try to find integers \(p\) and \(q\) such that \(a^2+3a + 2=(a+p)(a+q)\).
Expansion of brackets in the right-hand side then gives: \[a^2+3a + 2=a^2+(p+q)a+p\times q\] 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: \[\begin{array}{||r|r|r||} \hline p & q & p+q\\ \hline 1 & 2 & 3\\ \hline -1 & -2 & -3\\ \hline \end{array}\] \(p=2\) and \(q=1\) meet the requirements.
The factorisation is: \[a^2+3a + 2=(a+2)(a+1)\]

New example

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

Decompose\(\;x^{5}-x^{4} -6x^3\;\) into factors with integer coefficients.

\(x^{5}-x^{4} -6x^3={}\)\(x^3(x-3)(x+2)\).

Note first that all terms can be divided by \(x^3\) so that \[x^{5}-x^{4} -6x^3=x^3(x^2-x -6)\] 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 \(x^2-x -6=(x+p)(x+q)\).
Expansion of brackets in the right-hand side then gives: \[x^2-x -6=x^2+(p+q)x+p\times q\] 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: \[\begin{array}{||r|r|r||} \hline p & q & p+q\\ \hline 1 & -6 & -5\\ \hline -1 & 6 & 5\\ \hline 2 & -3 & -1\\ \hline -2 & 3 & 1\\ \hline \end{array}\] \(p=-3\) and \(q=2\) meet the requirements.
The factorisation is: \[x^2-x -6=(x-3)(x+2)\] The final result is: \[x^{5}-x^{4} -6x^3=x^3(x-3)(x+2)\]

New example

Mathcentre video

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