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## Section7.4Optional Extension

### SubsectionA 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.

###### Example7.28.
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.
Solution
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

$3t^2+9t+2t+6=3t^2+11t+6$

 $~~~~~t~~~~~$ $~~~~3~~~~~$ $3t$ $3t^2$ $9t$ $2$ $2t$ $6$

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 QuestionsReading Questions

###### 1.

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.

 $~~~~~x~~~~~$ $~~~~2~~~~~$ $3x$ $3x^2$ $6x$ $4$ $4x$ $8$
• 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.

### SubsectionFactoring 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.

 $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $3x^2$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $8$

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.

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

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*}
 $3x$ $4$ $x$ $\hphantom{00}3x^2\hphantom{00}$ $\hphantom{00}4x\hphantom{00}$ $2$ $\hphantom{00}6x\hphantom{00}$ $\hphantom{00}8\hphantom{00}$

#### Reading QuestionsReading Questions

###### 1.

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

###### 2.

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

###### Example7.29.

Factor $~~2x^2-11x+15$

Solution

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

 $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $2x^2$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $15$

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.

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

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

 $x$ $-3$ $2x$ $\hphantom{00}2x^2\hphantom{00}$ $\hphantom{00}-6x\hphantom{00}$ $\hphantom{00}$ $\hphantom{00}-5x\hphantom{00}$ $\hphantom{00}15\hphantom{00}$
 $x$ $-3$ $2x$ $\hphantom{00}2x^2\hphantom{00}$ $\hphantom{00}-6x\hphantom{00}$ $-5$ $\hphantom{00}-5x\hphantom{00}$ $\hphantom{00}15\hphantom{00}$

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 QuestionsReading Questions

###### 3.

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

###### 4.

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.

### ExercisesHomework 7.3A

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

###### 1.
 $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{00}6x^2\hphantom{00}$ $\hphantom{00}9x\hphantom{00}$ $\hphantom{0000}$ $\hphantom{00}10x\hphantom{00}$ $\hphantom{00}15\hphantom{00}$
###### 2.
 $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{00}8t^2\hphantom{00}$ $\hphantom{00}-14t\hphantom{00}$ $\hphantom{0000}$ $\hphantom{00}-12t\hphantom{00}$ $\hphantom{00}21\hphantom{00}$
###### 3.
 $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{00}12m^2\hphantom{00}$ $\hphantom{00}-10m\hphantom{00}$ $\hphantom{0000}$ $\hphantom{00}30m\hphantom{00}$ $\hphantom{00}-25\hphantom{00}$
###### 4.
 $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{00}9a^2\hphantom{00}$ $\hphantom{00}21a\hphantom{00}$ $\hphantom{0000}$ $\hphantom{00}-21a\hphantom{00}$ $\hphantom{00}-49\hphantom{00}$

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

###### 5.

$2x^2-13x+18$

###### 6.

$5x^2+16x-16$

###### 7.

$6h^2+7h+2$

###### 8.

$9n^2-8n-1$

###### 9.

$6t^2-5t-25$

###### 10.

$5x^2-14x-24$