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Section 7.4 Optional Extension

Subsection A Property of Binomial Products

If the coefficients in a quadratic trinomial \(ax^2+bx+c\) are not prime numbers, the guess-and-check method may be time-consuming. In that case, we can use another technique that depends upon the following property of binomial products.

Example 7.28.
  1. Compute the product \((3t+2)(t+3)\) using the area of a rectangle.
  2. Verify that the products of the diagonal entries are equal.
  1. We construct a rectangle with sides \(3t+2\) and \(t+3\text{,}\) as shown below. We see that the product of the two binomials is



    \(3t\) \(3t^2\) \(9t\)
    \(2\) \(2t\) \(6\)

  2. Now let's compute the product of the expressions along each diagonal of the rectangle:

    \begin{equation*} 3t^2 \cdot 6 = 18t^2~~~~~~\text{and}~~~~~~9t \cdot 2t = 18t^2 \end{equation*}

    The two products are equal. This is not surprising when you think about it, because each diagonal product is the product of all four terms of the binomials, namely \(3t,~2,~t,\) and \(3\text{,}\) just multiplied in a different order. You can see where the diagonal entries came from in our example:

    \begin{align*} 18t^2 \amp = 3t^2 \cdot 6 =3t \cdot t \cdot 2 \cdot 3\\ 18t^2 \amp = 9t \cdot 2t =3t \cdot 3 \cdot 2 \cdot t \end{align*}
Product of Binomials.

When we represent the product of two binomials by the area of a rectangle, the products of the entries on the two diagonals are equal.

Reading Questions Reading Questions


In Example 7.28, why are the products on the two diagonals equal?

Look Ahead.

We can use rectangles to help us factor quadratic trinomials. Recall that factoring is the opposite or reverse of multiplying, so we must first understand how multiplication works.

Look Closer.

Look carefully at the rectangle for the product

\begin{equation*} (3x+4)(x+2)=3x^2+10x+8 \end{equation*}

Shown at right.


\(3x\) \(3x^2\) \(6x\)
\(4\) \(4x\) \(8\)
  • The quadratic term of the product, \(3x^2\text{,}\) appears in the upper left sub-rectangle.
  • The constant term of the product, 8, appears in the lower right sub-rectangle.
  • The linear term, \(10x\text{,}\) is the sum of the other two sub-rectangles.

Subsection Factoring Quadratic Trinomials by the Box Method

Now we'll factor the trinomial

\begin{equation*} 3x^2+10x+8 \end{equation*}

We'll try to reverse the steps for multiplication. Instead of starting with the factors on the outside of the rectangle, we begin by filling in the areas of the sub-rectangles.

Step 1 The quadratic term, goes in the upper left, and the constant term, 8, goes in the lower right, as shown in the figure.


\(\hphantom{0000}\) \(3x^2\) \(\hphantom{0000}\)
\(\hphantom{0000}\) \(\hphantom{0000}\) \(8\)

What about the other two entries? We know that their sum must be \(10x\text{,}\) but we don't know what expressions go in each! This is where we use our observation that the products on the two diagonals are equal.

Step 2 We compute the product of the entries on the first diagonal:

\begin{equation*} D=3x^2 \cdot 8 = 24x^2 \end{equation*}

The product of the entries on the other diagonal must also be \(24x^2\text{.}\) We now know two things about those entries:

\begin{align*} \amp \text {1. Their product is}~~24x^2~~\text{and}\\ \amp \text {2. Their sum is}~~10x \end{align*}

Step 3 To find the two unknown entries, we list all the ways to factor \(D=24x^2\text{,}\) then choose the factors whose sum is \(10x\text{.}\)

Factors of \(D=24x^2\) \(\hphantom{0000}\)Sum of Factors
\(x\) \(24x\) \(\hphantom{0000}x+24x=25x\)
\(2x\) \(12x\) \(\hphantom{0000}2x+12x=24x\)
\(3x\) \(8x\) \(\hphantom{0000}3x+8x=11x\)
\(\blert{4x}\) \(\blert{6x}\) \(\hphantom{0000}\blert{4x+6x=10x}\)

Step 4 We see that the last pair of factors, \(4x\) and \(6x\text{,}\) has a sum of \(10x\text{.}\) We enter these factors in the remaining sub-rectangles. (It doesn't matter which one goes in which spot.) We now have all the sub-rectangles filled in, as shown at right.


\(\hphantom{0000}\) \(\hphantom{00}3x^2\hphantom{00}\) \(\hphantom{00}4x\hphantom{00}\)
\(\hphantom{0000}\) \(\hphantom{00}6x\hphantom{00}\) \(\hphantom{00}8\hphantom{00}\)

Finally, we work backwards to discover what length and width produce the areas of the four subrectangles.

Step 5 We factor each row of the rectangle, and write the factors on the outside. Start with the top row, factoring out \(x\) and writing the result, \(3x+4\text{,}\) at the top, as shown at right. We get the same result when we factor \(2\) from the bottom row.

The final rectangle is shown at right, and the factors of \(3x^2+10x+8\) appear as the length and width of the rectangle. Our factorization is thus

\begin{equation*} 3x^2+10x+8=(x+2)(3x+4) \end{equation*}


\(x\) \(\hphantom{00}3x^2\hphantom{00}\) \(\hphantom{00}4x\hphantom{00}\)
\(2\) \(\hphantom{00}6x\hphantom{00}\) \(\hphantom{00}8\hphantom{00}\)

Reading Questions Reading Questions


Which terms of the quadratic trinomial go into the upper left and lower right sub-rectangles of the box?


Why do we list the possible factors of \(D\text{?}\)

Example 7.29.

Factor \(~~2x^2-11x+15\)


Step 1 Enter \(2x^2\) and \(15\) on the diagonal of the rectangle, as shown in the figure.


\(\hphantom{0000}\) \(2x^2\) \(\hphantom{0000}\)
\(\hphantom{0000}\) \(\hphantom{0000}\) \(15\)

Step 2 Compute the diagonal product:

\begin{equation*} D=2x^2 \cdot 15 = 30x^2 \end{equation*}

Step 3 List all possible factors of \(D\text{,}\) and compute the sum of each pair of factors. (Note that both factors must be negative.)

Factors of \(D=30x^2\) \(\hphantom{0000}\)Sum of Factors
\(-x\) \(-3024x\) \(\hphantom{0000}-x-30x=-31x\)
\(-2x\) \(-15x\) \(\hphantom{0000}-2x-15x=-17x\)
\(-3x\) \(-10x\) \(\hphantom{0000}-3x-10x=-13x\)
\(\blert{-5x}\) \(\blert{-6x}\) \(\hphantom{0000}\blert{-5x-6x=-11x}\)

The correct factors are \(-5x\) and \(-6x\text{.}\)

Step 4 Enter the factors \(-5x\) and \(-6x\) into the rectangle.


\(\hphantom{0000}\) \(\hphantom{00}2x^2\hphantom{00}\) \(\hphantom{00}-6x\hphantom{00}\)
\(\hphantom{0000}\) \(\hphantom{00}-5x\hphantom{00}\) \(\hphantom{00}15\hphantom{00}\)

Step 5 Factor \(2x\) from the top row of the rectangle, and write the result, \(x-3\text{,}\) at the top, as shown below.


\(2x\) \(\hphantom{00}2x^2\hphantom{00}\) \(\hphantom{00}-6x\hphantom{00}\)
\(\hphantom{00}\) \(\hphantom{00}-5x\hphantom{00}\) \(\hphantom{00}15\hphantom{00}\)


\(2x\) \(\hphantom{00}2x^2\hphantom{00}\) \(\hphantom{00}-6x\hphantom{00}\)
\(-5\) \(\hphantom{00}-5x\hphantom{00}\) \(\hphantom{00}15\hphantom{00}\)

Finally, factor \(x-3\) from the bottom row, and write the result, \(-5\text{,}\) on the left. The correct factorization is

\begin{equation*} 2x^2-11x+15=(2x-5)(x-3) \end{equation*}

Reading Questions Reading Questions


What do we do after we have filled in all the sub-rectangles of the box?


Where do the factors of the quadratic trinomial appear?

Here is a summary of our factoring method.

To Factor \(~ax^2+bx+c~\) Using the Box Method.
  1. Write the quadratic term \(ax^2\) in the upper left sub-rectangle, and the constant term \(c\) in the lower right.
  2. Multiply these two terms to find the diagonal product, \(D\text{.}\)
  3. List all possible factors \(px\) and \(qx\) of \(D\text{,}\) and choose the pair whose sum is the linear term, \(bx\text{,}\) of the quadratic trinomial.
  4. Write the factors \(px\) and \(qx\) in the remaining sub-rectangles.
  5. Factor each row of the rectangle, writing the factors on the outside. These are the factors of the quadratic trinomial.

Exercises Homework 7.3A

For Problems 1–4, Use the given areas to find the length and width of each rectangle.



\(\hphantom{0000}\) \(\hphantom{00}6x^2\hphantom{00}\) \(\hphantom{00}9x\hphantom{00}\)
\(\hphantom{0000}\) \(\hphantom{00}10x\hphantom{00}\) \(\hphantom{00}15\hphantom{00}\)


\(\hphantom{0000}\) \(\hphantom{00}8t^2\hphantom{00}\) \(\hphantom{00}-14t\hphantom{00}\)
\(\hphantom{0000}\) \(\hphantom{00}-12t\hphantom{00}\) \(\hphantom{00}21\hphantom{00}\)


\(\hphantom{0000}\) \(\hphantom{00}12m^2\hphantom{00}\) \(\hphantom{00}-10m\hphantom{00}\)
\(\hphantom{0000}\) \(\hphantom{00}30m\hphantom{00}\) \(\hphantom{00}-25\hphantom{00}\)


\(\hphantom{0000}\) \(\hphantom{00}9a^2\hphantom{00}\) \(\hphantom{00}21a\hphantom{00}\)
\(\hphantom{0000}\) \(\hphantom{00}-21a\hphantom{00}\) \(\hphantom{00}-49\hphantom{00}\)

For Problems 5–10, use the box method to factor the quadratic trinomials.