Factor Quadratic by Grid Method Explained

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


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.

By expanding the right hand side, we have


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:


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:

We continue in this way, drawing out the required factor in each row and column:

Beginning with the values a, p, q and c, we have calculated m, n, r and t.

The example once again:

This grid above shows the values a, p, q and c. By drawing out the HCF of each row/column, we get m, n, r and t:

Here, m=2; n=1, r=-3 and t=4.