# Factoring Trinomial with “Box” Method

Factoring using the “box” or “grid” method is a great alternative to factoring trinomial by grouping method when the leading coefficient, [latex]a[/latex], is not equal to [latex]1[/latex] or [latex]- 1[/latex].

**TIP: **Before you can apply the general steps below, make sure to first take out common factors among the coefficients of the trinomial. Otherwise, the method will not work and therefore will give us a wrong answer.

In other words, for this to work, the Greatest Common Factor (GCF) of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] in [latex]a{x^2} + bx + c[/latex] must be** 1**.

## Steps to Factor a Trinomial using the “Box” Method

**Step 1**: Multiply the leading coefficient and the constant term (number without variable).

**Step 2**: Find **two numbers** such that the product is equal to *a*·*c* and the sum is equal to the middle coefficient, *b*. Let “*n*” and “*m*” be the two numbers satisfying the two conditions.

**Step 3**: Create a 2×2 grid and place the following terms in the right boxes:

- Place the first term in the upper left box.
- Place the constant term in the lower right box.
- Place the numbers you found in step 2 in the remaining empty boxes. This time, it doesn’t matter where you place them. Make sure that you attach a variable x to each number.

**Step 4**: Find the greatest common factor on each row and column. Place them outside the box. Take the sign of the term **closest** to it.

**Step 5**: The factors of the trinomial are coming from the outside terms.

[latex]a{x^2} + bx + c = \left( {qx + r} \right)\left( {sx + t} \right)[/latex]

### Examples of How to Factor a Trinomial using the “Box” Method

Let’s go over some examples!

**Example 1:** Factor the trinomial [latex]6{x^2} – 5x – 4[/latex] using the “box” method.

Start by multiplying the leading coefficient and the constant term.

[latex]\left( 6 \right)\left( { – 4} \right) = – 24[/latex]

Find the factor pair of [latex]–24[/latex] such that the sum equals the middle coefficient which is [latex]–5[/latex]. You can do some trial and error to figure this out. If you have done it correctly, you should have the two numbers [latex]−8[/latex] and [latex]3[/latex] because

[latex](–8)(3) = –24[/latex]

[latex](–8) + (3) = –5[/latex]

Next, we are going to fill in the box.

Find the greatest common factor (GCF) of each row and column. Its sign will depend on the term **closest** to it.

- Top Row

- Bottom Row

- First Column

- Second Column

We can read off the factors by looking at the outside terms on the sides of the box. The first factor comes from the sum of the terms found in the left column, while the second factor comes from the sum of the terms found on the top row.

So the final answer is

[latex]6{x^2} – 5x – 4 = \left( {3x – 4} \right)\left( {2x + 1} \right)[/latex]

**Example 2:** Factor the trinomial [latex]5{x^2} – 18x + 9[/latex] using the “box” method.

The product of the leading coefficient and the constant term is [latex](5)(9) = 45[/latex]. Can you find two numbers such that their product is [latex]45[/latex] and the sum would be the middle coefficient which is [latex]−18[/latex]?

If you think about it, the two numbers must have the same signs. That means they should be both positive or both negative. If you add two positive numbers, the sum will be positive. We don’t want this option since we want the sum to be negative.

This leaves us the second option that the two numbers must be both negative. After trial and error, the numbers that can satisfy the two conditions are [latex]−3[/latex] and [latex]−15[/latex]. Since,

[latex]\left( { – 3} \right)\left( { – 15} \right) = 45[/latex]

[latex]\left( { – 3} \right) + \left( { – 15} \right) = – 18[/latex]

Here’s our box with terms on the right places.

Determine the greatest common factor (GCF) of each row and column. Don’t forget to take the sign of the **closest **term in the box, which is either directly to its right or below it.

- Top Row

- Bottom Row

- First Column

- Second Column

The factors are obtained from the edges of the grid.

So our final answer is

[latex]5{x^2} – 18x + 9 = \left( {5x – 3} \right)\left( {x – 3} \right)[/latex]

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