To factor a trinomial with a leading coefficient of

1,

think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.

m^{2} + 18 m + 81

In this case, we have

81 .

This is a positive number, so for the product of the constant terms in the factors to be positive, these constants must have the same sign (both positive or both negative.)

Factor Constants	Product of Constants
$1$ and $81$	$81$
$- 1$ and $- 81$	$81$
$3$ and $27$	$81$
$- 3$ and $- 27$	$81$
$9$ and $9$	$81$
$- 9$ and $- 9$	$81$

Next, let's consider the coefficient of the linear term.

m^{2} + 18 m + 81

For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term,

18 .

Factors	Sum of Factors
$1$ and $81$	$82$
$- 1$ and $- 81$	$- 82$
$3$ and $27$	$30$
$- 3$ and $- 27$	$- 30$
$9$ and $9$	$18$
$- 9$ and $- 9$	$- 18$

We found the factors whose product is

81

and whose sum is

18 .

m^{2} + 18 m + 81 \Leftrightarrow (m + 9) (m + 9)

Both factors are the same, so we can write the expression as

(m + 9)^{2} .

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