When factoring a quadratic expression $ax^2+bx+c$ using the grid method, we assume that $a, b$ and $c$ do not share any common factors (other than 1), and we assume that our expression does indeed factor, that is, there exist $m, n, r, t$ such that

$ax^2+bx+c=(mx+r)(nx+t)$

Given the values $a, b$ and $c$ , how does the grid method correctly calculate $m, n, r$ and $t$ ?

Since HCF $(a,b,c)=1$ , neither binomial will have a common factor: HCF $(m,r) = 1$ and HCF $(n, t) = 1$ .

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By expanding the right hand side, we have

$ax^2+bx+c=mnx^2+tmx+rnx+rt$

It is immediately apparent that $a=mn$ and $c=rt$ , however this does not help us calculate the values $m, n, r, t$ .

We also notice when we multiply two binomials of this kind that there are two $x$ terms.

Let us break the our given value of $b$ into two values, $p$ and $q$ , where $p+q=b$ :

$ax^2+px+qx+c=mnx^2+tmx+rnx+rt$

Calculating p, q given a, b, c

Let’s put this information into the grid format.

Given that $ax^2+px+qx+c=(mx+r)(nx+t)$ we have:

Given that $(mx+r)(nx+t)=mnx^2+tmx+rnx+rt$ we have:

In the second version, notice that when we multiply the two diagonals, we have the same expression:

$nrx \cdot mtx = mn\cdot rtx^2$

Therefore, from the first grid we have

$px \cdot qx = a\cdot cx^2$

Which leads us to conclude $p\times q=a\times c$ .

We already know that $p+q=b$ (that is how we defined $p$ and $q$ ).

This leads us to derive values for $p$ and $q$ directly from the coefficients $a, b$ and $c$ :

$\begin{align*}p\times q&=ac\\[10 pt]p+q&=b\end{align}$

If there are no integer solutions for $p$ and $q$ , then the values m, n, r, t are not integers. Our quadratic expression does not factor with integers.

By solving this system to calculate $p$ and $q$ we can fill in the two by two grid:

Calculating m, n, r, t

At this point in the process, values $m, n, r$ and $t$ are unknown, however the values in the grid are complete. In terms of $m, n, r, t$ , the two by two grid is:

Let’s consider the first row (any row or column will do)

Notice that $mx$ is common to both terms in this row. Also, as we noted in the beginning, HCF $(n,t)=1$ , therefore $mx$ is actually the highest common factor in this row.

Now let’s consider the first column. By a similar argument, we notice that $nx$ is the HCF of this column (because HCF $(m,r)=1$ ). We also notice that we need $nx$ to multiply with $mx$ to get the required product $mnx^2$ :