Factoring Trinomials: Leading Coefficient ≠ 1
Factor trinomials ax² + bx + c where a ≠ 1 using the AC method (or trial and error). Understand why multiple methods exist and choose efficiently based on the coefficients.
A lesson of Algebra 1, in Factoring Polynomials, about 35 minutes.
Factoring Trinomials: Leading Coefficient Not 1
By the end of this lesson, you will be able to factor $ax^2 + bx + c$ using the AC method, know why the two-number trick needs adjusting, split the middle term reliably, and choose between AC and trial-and-error sensibly.
Why the Old Trick Needs Adjusting
The two-number method worked because the $x$ terms were plain.
$$(x + p)(x + q) = x^2 + (p+q)x + pq$$
Put a coefficient in front and that clean structure breaks:
$$(2x + 3)(x + 1) = 2x^2 + 5x + 3$$
The $3$ is still $3 \times 1$, but the $5$ is not $3 + 1$. It is $2 \cdot 1 + 3 \cdot 1$, because the $2$ multiplied into one of the middle products and not the other. The leading coefficient interferes with the middle term, so "two numbers adding to $b$" no longer describes anything.
What survives is a modified version, and it is exact rather than approximate.
$$\textbf{find two numbers multiplying to } a \cdot c \textbf{ and adding to } b$$
The AC Method
Five steps, and each one is mechanical.
- Multiply $a \cdot c$
- Find two numbers multiplying to $ac$ and adding to $b$
- Split the middle term into those two pieces
- Group the four terms in pairs and factor each pair
- The shared bracket comes out
$$2x^2 + 7x + 3$$
$ac = 2 \times 3 = 6$. Two numbers multiplying to $6$ and adding to $7$: that is $1$ and $6$.
$$2x^2 + x + 6x + 3$$
$$x(2x + 1) + 3(2x + 1)$$
$$(2x + 1)(x + 3)$$
The shared bracket is the whole mechanism. If the two pairs do not produce identical brackets, something earlier went wrong, and that makes step 4 a built-in check rather than just a step.
The order of the split does not matter. Writing $6x$ first gives $2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3)$, the same answer by a different route.
Example 1: The Whole Method
Here is every step on one trinomial, worked all the way through. $$3x^2 + 10x + 8$$
- What is a times c? 24
- Two positive numbers multiply to 24 and add to 10. What is the smaller of them? 4
- Split the middle term. Write the four-term expression with 4x first 3x^2 + 4x + 6x + 8
- Factor the first pair, 3x^2 + 4x x(3x + 4)