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McGraw Hill Glencoe Algebra 1, 2012 View details

5. Using the Distributive Property

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Exercise 51 Page 499

a Let's start by writing the factors of

- 6,

and calculate their sum.

Factors	Sum of Factors
$1$ and $- 6$	$- 5$
$- 1$ and $6$	$5$
$2$ and $- 3$	$- 1$
$- 2$ and $3$	$1$

We see that $- 2$ and $3$ have a product of $- 6$ and a sum of $1 .$

b We will now write each factor in an empty square. We will also include the positive or negative sign and variable.

c For this part, we will find the factor for each row and column of the box.

The factors of

x^{2} + x - 6

are

(x + 3)

and

(x - 2) .

x^{2} + x - 6 = (x + 3) (x - 2)

d When we use the box method to factor

x^{2} - 3 x - 40,

we need to follow four steps.

Place the first term, $x^{2},$ in the top left-hand corner and the last term, $- 40,$ in the lower right-hand corner.
Determine which two factors have a product of $- 40$ and a sum of $- 3 .$
Place the variable and the factors with their signs in the box.
Find the factor for each row and column.

Extra

Use Box Method to Factor It

We will start by placing the first term, $x^{2},$ in the top left-hand corner and the last term, $- 40,$ in the lower right-hand corner.

Now, we will determine the factor pair with a product of

- 40

and a sum of

- 3 .

- 40 ↙ ↘ - 8 + 5 = - 3

Next, we will place the variable and the factors with their signs in the box.

Finally, we find the factor for each row and column.

The factors of

x^{2} - 3 x - 40

are

(x + 5)

and

(x - 8) .

x^{2} - 3 x - 40 = (x + 5) (x - 8)

Subchapter links

Check Your Understanding p.497

Practice and Problem Solving p.498-499

H.O.T. Problems p.499

Standardized Test Practice p.500

Skills Review p.500