Factoring Harder Quadratics

The leading coefficient splits across both factors

A trinomial like $2x^2 + 7x + 3$ is non-monic: its leading coefficient is $2$, not $1$. To factor a general non-monic trinomial $ax^2 + bx + c$ you are looking for two factors of the form $px + r$ and $qx + s$. In those two factors the coefficients $p$ and $q$ sit in front of the $x$ terms. The reason the old sum-and-product trick no longer applies directly is that $a$ is now the product of $p$ and $q$. So the leading coefficient is shared out between the two factors. To see exactly how every coefficient is built, expand the general product once and read it carefully.

The same rectangle that pictured expanding shows where each coefficient lives. Give a rectangle a width of $px + r$ and a height of $qx + s$. The two splits cut it into four pieces.

The two corner squares are the leading term $pq\,x^2$ and the constant $rs$. The two off-diagonal strips are $ps\,x$ and $qr\,x$, and together they make the middle term $(ps + qr)x$. That picture also hides a useful fact you will use in a moment: the two strips have areas whose coefficients multiply to $(ps)(qr) = pq \cdot rs = a \cdot c$.

Factoring by reverse FOIL

With no shortcut yet, the honest method is a guided trial. Choose a factor pair for $a$ and a factor pair for $c$, write the candidate $(px + r)(qx + s)$, and expand it to check the middle term. If the cross products $ps + qr$ come out to $b$, you are done; if not, swap the numbers around or try the next pair. Because expanding is quick, a handful of tries settles most trinomials. Start with $a > 0$ and $c > 0$, where every number in sight is positive.

Let the product of a and c point the way

Trying every arrangement works, but you can aim the search instead of guessing. The middle coefficient is the sum of the two cross products, $ps + qr = b$, and from the rectangle those same cross products multiply to $(ps)(qr) = a \cdot c$. So the two pieces the middle term splits into are just a pair of numbers with a known product and a known sum:

Finding that pair is exactly the sum-and-product search you already know, and it tells you what the two middle strips should be. Those strips point you straight at the factor pair to write down. The hint does not replace the trial; it tells you which candidate to build, and you still finish by expanding to be sure.

Pull out a common factor first

Before trying any factor pairs, check whether all three coefficients share a common factor. If they do, pull it out. It shrinks the leading coefficient you have to break apart and often turns a bristly trinomial into a gentler one.

Solving gives roots that can be fractions

Factoring pays off the same way it did before: once the trinomial is a product, the zero-product property from the earlier lesson turns it into roots. To solve $ax^2 + bx + c = 0$, factor the left side into $(px + r)(qx + s)$ and set the product to zero. A product is zero only when one factor is zero, so

which give $x = -\dfrac{r}{p}$ and $x = -\dfrac{s}{q}$. Here is the one genuinely new feature: because the leading numbers $p$ and $q$ can be bigger than $1$, dividing by them can produce a fraction. Non-monic quadratics routinely have fractional roots, and that is expected, not a mistake.

When a harder trinomial will not factor

As with monic trinomials, not every non-monic trinomial factors over the integers. The search can simply fail: no factor pair of $a$ and $c$ produces cross products that add to $b$. Equivalently, no integer pair has product $a \cdot c$ and sum $b$. Take $2x^2 + 2x + 1$. The middle would have to split into two integers with product $2 \cdot 1 = 2$ and sum $2$. But the only integer pairs multiplying to $2$ are $1$ and $2$ (sum $3$) or $-1$ and $-2$ (sum $-3$). Neither adds to $2$, so the trinomial is irreducible over the integers.

That does not say the equation $2x^2 + 2x + 1 = 0$ has no solutions, only that integer factoring is the wrong tool for it. Later lessons build methods that solve any quadratic, whether it factors or not. For now, if an honest run through the pairs turns up nothing, you can stop and report that the trinomial does not factor over the integers. Reporting the failure is better than forcing a factorization that is not there.

Common mistakes