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].

in ax^2+bx+c, the leading coefficient cannot equal to 1 or -1

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).

in ax^2+bx+c, the product of the coefficient of the quadratic term and the constant is a times c

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.

n times m equals a times c, while the sum of n and m equals b

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

in a 2x2 table, the first row contains ax^2, mx; the second row contains nx, c
  • 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.

q is the GCF of a and m; r is the GCF of n and c; s is the GCF of a and n; t is the GCF of m and c

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) = â€“24[/latex]

[latex](–8) + (3) = â€“5[/latex]

Next, we are going to fill in the box.

in a 2x2 table, the first row contains 6x^2, 3x; the second row contains -8x, -4

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

  • Top Row
3x is the GCF of 6x^2 and 3x
  • Bottom Row
-4 is te GCF of -8x and -4
  • First Column
2x is the GCF of 6x^2 and -8x
  • Second Column
1 is the GCF of 3x and -4

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.

the left of the table generates the binomial 3x-4 while the top of the table generates the binomial 2x+1

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.

in a 2x2 table, the first row contains the elements 5x^2, -15x; the second row contains -3x, 9

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
5x is the GCF of 5x^2 and -15x
  • Bottom Row
-3 is the GCF of -3x and 9
  • First Column
x is the GCF of 5x^2 and -3x
  • Second Column
-3 is the GCF of -15x and 9

The factors are obtained from the edges of the grid.

(5x-3)(x-3)

So our final answer is

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


You may also be interested in these related math lessons or tutorials:

Factoring Trinomial where a=1 
Factoring Trinomial where a>1