Factoring a Quadratic Trinomial

A quadratic trinomial in canonical form looks like this: $$ ax^2 + bx + c \qquad \text{with } a \ne 0 $$ The goal of factoring is to rewrite the trinomial as the product of two first-degree factors, whenever that’s possible.

There are several ways to factor a trinomial.

Note. Which method you choose depends on the specific form of the trinomial. The objective, however, is always the same: breaking the expression down into simpler factors whenever you can.

The Roots Method (Using the Discriminant)

This is the most general approach, and it always applies.

First, compute the discriminant $ \Delta $:

$$ \Delta = b^2 - 4ac $$

Then, depending on the value of $\Delta$, there are three scenarios:

• If $\Delta > 0$, the trinomial has two distinct real roots $x_1$ and $x_2$. In that case we can write: $$ ax^2 + bx + c = a(x - x_1)(x - x_2) $$
• If $\Delta = 0$, the trinomial has a repeated real root $x_0$. The factorization becomes: $$ ax^2 + bx + c = a(x - x_1)(x - x_2) = a(x - x_0)^2 $$
• If $\Delta < 0$, the trinomial has no real roots. It’s irreducible over the reals, though it can still be factored over the complex numbers.

Example

Take the trinomial

$$ x^2 - 5x + 6 $$

It’s in canonical form $ ax^2 + bx + c $, with coefficients:

• $a = 1$
• $b = -5$
• $c = 6$

The discriminant tells us whether the trinomial can be factored over the reals. Let’s compute it:

$$ \Delta = b^2 - 4ac = (-5)^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1 $$

Since $\Delta > 0$, we know there are two distinct real roots.

Now use the quadratic formula:

$$ x = \frac{-b \pm \sqrt{\Delta}}{2a} $$

Substitute the coefficients:

$$ x = \frac{-(-5) \pm \sqrt{1}}{2 \cdot 1} $$

$$ x = \frac{5 \pm 1}{2} $$

$$ x = \begin{cases} x_1 = \frac{5 - 1}{2} = 2 \\ \\ x_2 = \frac{5 + 1}{2} = 3 \end{cases} $$

So the roots are $x_1 = 2$ and $x_2 = 3$, and the trinomial factors as:

$$ x^2 - 5x + 6 = (x - x_1)(x - x_2) = (x - 2)(x - 3) $$

In other words, the equivalent factored form is:

$$ x^2 - 5x + 6 = (x - 2)(x - 3) $$

Check the result. Multiply it back out to be sure: $$ (x - 2)(x - 3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6 $$ It matches the original trinomial. The factorization is correct. Verifying isn’t strictly necessary, but it’s a good safeguard against small mistakes.

The "Sum and Product" Method (Monic Case)

This shortcut works only if the trinomial is monic, meaning the leading coefficient equals one ($a = 1$).

$$ x^2 + bx + c $$

Here, we look for two numbers $p$ and $q$ such that:

$$ \begin{cases} p + q = b \\ \\ pq = c \end{cases} $$

Of course, such numbers don’t always exist.

But if they do, the trinomial factors immediately as:

$$ x^2 + bx + c = (x + p)(x + q) $$

So this method is faster, but not universal. It only works when $a=1$.

Example

Consider the trinomial:

$$ x^2 + 5x + 6 $$

Here the coefficients are $ a = 1, \ b = 5, \ c = 6 $.

Since $a = 1$, we try to find two numbers $p$ and $q$ that satisfy:

$$ p + q = b \quad \text{and} \quad pq = c $$

Substituting the known values gives:

$$ p + q = 5, \quad pq = 6 $$

We now need integers that fit both conditions.

To check, list the factor pairs of $6$ (positive and negative) and see which pair sums to $5$.

$$ \begin{array}{|c|c|c|c|} \hline p & q & pq = 6 & p + q \\ \hline 1 & 6 & 6 & 7 \\ -1 & -6 & 6 & -7 \\ 2 & 3 & 6 & 5 \quad \text{OK} \\ -2 & -3 & 6 & -5 \\ \hline \end{array} $$

The pair $(2, 3)$ works, since $p + q = 5$ and $pq = 6$.

So the trinomial factors neatly as:

$$ x^2 + 5x + 6 = (x+2)(x+3) $$

The final result is:

$$ x^2 + 5x + 6 = (x+2)(x+3) $$

The "pq = ac" Method (factoring by grouping)

This method is especially useful when the trinomial $ax^2 + bx + c$ is not monic, that is, when $a \ne 1$.

$$ ax^2 + bx + c \qquad \text{with } a \ne 1 $$

Here we look for two integers $p$ and $q$ such that:

$$ \begin{cases} p + q = b \\ \\ pq = a \cdot c \end{cases} $$

Once you’ve found them, split the middle term $bx$ into $px + qx$ and then factor by grouping.

$$ ax^2 + px + qx + c $$

Example

Let’s factor the trinomial:

$$ 6x^2 + 7x + 2 $$

The coefficients are $a = 6, \ b = 7, \ c = 2$.

Since $a \ne 1$, the “$pq = ac$” method applies. Compute:

$$ ac = 6 \cdot 2 = 12 $$

We need integers $p$ and $q$ such that:

$$ p + q = b = 7, \quad pq = ac = 12 $$

The pair that works is:

$$ p = 3, \quad q = 4 $$

Now split the middle term $7x$ into $3x + 4x$:

$$ 6x^2 + 3x + 4x + 2 $$

Group the terms and factor each group:

$$ (6x^2 + 3x) + (4x + 2) $$

$$ 3x(2x + 1) + 2(2x + 1) $$

Since $(2x + 1)$ is common, factor it out:

$$ (3x + 2)(2x + 1) $$

So the trinomial factors neatly as:

$$ 6x^2 + 7x + 2 = (3x + 2)(2x + 1) $$

Special Products

Special products are standard identities that let you recognize a factorization pattern instantly. Mastering them saves a lot of time.

They do, however, require both sharp observation and a good memory of the main formulas.

• Square of a binomial
  $$ A^2 + 2AB + B^2 = (A + B)^2 \qquad A^2 - 2AB + B^2 = (A - B)^2 $$

  Example: $$ x^2 + 6x + 9 = (x + 3)^2 $$

• Cube of a binomial
  $$ A^3 + 3A^2B + 3AB^2 + B^3 = (A + B)^3 $$ $$ A^3 - 3A^2B + 3AB^2 - B^3 = (A - B)^3 $$

  Example: $$ x^3 + 6x^2 + 12x + 8 = (x + 2)^3 $$

• Difference of squares
  $$ A^2 - B^2 = (A - B)(A + B) $$

  Example: $$ x^2 - 9 = (x - 3)(x + 3) $$

• Sum and difference of cubes
  Less common in early practice, but still worth knowing: $$ A^3 + B^3 = (A + B)(A^2 - AB + B^2) $$ $$ A^3 - B^3 = (A - B)(A^2 + AB + B^2) $$

  Example: $$ x^3 - 8 = (x - 2)(x^2 + 2x + 4) $$

In most school settings, the ones you absolutely need to recognize right away are the square of a binomial and the difference of squares. But to really gain fluency in factoring, it’s wise to have at least five or six of these identities at your fingertips.

And beyond that, the list goes on.