Case VII.

115. When a trinomial has the form $x^2 + ax + b$.

Where $a$ is the algebraic sum of two numbers, and is either positive or negative; and $b$ is the product of these two numbers, and is either positive or negative.

the factors of $x^{2} + 8 x + 15$ are $x + 5$ and $x + 3$ .

the factors of $x^{2} + 2 x - 15$ are $(x + 5)$ and $(x - 3)$ .

Hence, if a trinomial of the form $x^{2} + a x + b$ is such an expression that it can be resolved into two binomial factors, it is obvious that the first term of each factor will be $x$ , and that the second terms of the factors will be two numbers whose product is $b$ , the last term of the trinomial, and whose algebraic sum is $a$ , the coefficient of $x$ in the middle term of the trinomial.

Resolve into factors $x^{2} + 11 x + 30$ .

We are required to find two numbers whose product is $30$ and whose sum is $11$ .

Two numbers whose product is $30$ are $1$ and $30$ , $2$ and $15$ , $3$ and $10$ , $5$ and $6$ , and the sum of the last two numbers is $11$ . Hence, $x^{2} + 11 x + 30 = (x + 5) (x + 6) .$

Resolve into factors $x^{2} - 7 x + 12$ .

We are required to find two numbers whose product is $12$ and whose algebraic sum is $- 7$ .

Since the product is $+ 12$ , the two numbers are both positive or both negative, and since their sum is $- 7$ , they must both be negative.

Two negative numbers whose product is $12$ are $- 12$ and $- 1$ , $- 6$ and $- 2$ , $- 4$ and $- 3$ , and the sum of the last two numbers is $- 7$ . Hence, $x^{2} - 7 x + 12 = (x - 4) (x - 3) .$

Resolve into factors $x^{2} + 2 x - 24$ .

We are required to find two numbers whose product is $- 24$ and whose algebraic sum is $2$ .

Since the product is $- 24$ , one of the numbers is positive and the other negative, and since their sum is $+ 2$ , the larger number is positive.

Two numbers whose product is $- 24$ , and the larger number positive, are $24$ and $- 1$ , $12$ and $- 2$ , $8$ and $- 3$ , $6$ and $- 4$ , and the sum of the last two numbers is $+ 2$ . Hence, $x^{2} + 2 x - 24 = (x + 6) (x - 4) .$

Resolve into factors $x^{2} - 3 x - 18$ .

Since the product is $- 18$ , one of the numbers is positive and the other negative, and since their sum is $- 3$ , the larger number is negative.