1.4 Factoring trinomials with leading coefficient 1

$x^2+5x+6$

1. Write two sets of parentheses: $(\begin{array}{cc}& \end{array})(\begin{array}{cc}& \end{array})$ .
2. Place the factors of $x^2$ into the first position of each set of parentheses:
   

   $\begin{array}{ccc}(x& )(x& )\end{array}$

3. The third term of the trinomial is 6. We seek two numbers whose
   (a) product is 6 and
   (b) sum is 5.
   The required numbers are 3 and 2. Place $+3\text{and}+2$ into the parentheses.
   

   $x^2+5x+6=(x+3)(x+2)$

   The factorization is complete. We’ll check to be sure.
   

   $\begin{array}{lll}(x+3)(x+2)\hfill & =\hfill & x^2+2x+3x+6\hfill \\ \hfill & =\hfill & x^2+5x+6\hfill \end{array}$

$y^2-2y-24$

1. Write two sets of parentheses: $\left(\begin{array}{cc}& \end{array}\right)\left(\begin{array}{cc}& \end{array}\right)$ .
2. Place the factors of $y^2$ into the first position of each set of parentheses:
   

   $\begin{array}{ccc}(y& )(y& )\end{array}$

3. The third term of the trinomial is $-24$ . We seek two numbers whose
   (a) product is $-24$ and
   (b) sum is $-2$ .
   The required numbers are $-6\text{and}4$ . Place $-6\text{and}+4$ into the parentheses.
   

   $y^2-2y-24=(y-6)(y+4)$

   The factorization is complete. We’ll check to be sure.
   

   $\begin{array}{lll}(y-6)(y+4)\hfill & =\hfill & y^2+4y-6y-24\hfill \\ \hfill & =\hfill & y^2-2y-24\hfill \end{array}$

Notice that the other combinations of the factors of $-24$ (some of which are $-2,12;3,-8;\text{and}-4,6$ ) do not work. For example,

$\begin{array}{ccc}(y-2)(y+12)& =& y^2+\begin{array}{||}\hline 10y\\ \hline\end{array}-24\\ (y+3)(y-8)& =& y^2-\begin{array}{||}\hline 5y\\ \hline\end{array}-24\\ (y-4)(y+6)& =& y^2+\begin{array}{||}\hline 2y\\ \hline\end{array}-24\end{array}$

In all of these equations, the middle terms are incorrect.

$a^2-11a+30$

1. Write two sets of parentheses: $(\begin{array}{cc}& \end{array})(\begin{array}{cc}& \end{array})$ .
2. Place the factors of $a^2$ into the first position of each set of parentheses:
   

   $\begin{array}{ccc}(a& )(a& )\end{array}$

3. The third term of the trinomial is $+30$ . We seek two numbers whose
   (a) product is 30 and
   (b) sum is $-11$ .
   The required numbers are $-5\text{and}-6$ . Place $-5\text{and}-6$ into the parentheses.
   

   $a^2-11a+30=(a-5)(a-6)$

   The factorization is complete. We’ll check to be sure.
   

   $\begin{array}{lll}(a-5)(a-6)\hfill & =\hfill & a^2-6a-5a+30\hfill \\ \hfill & =\hfill & a^2-11a+30\hfill \end{array}$

$3x^2-15x-42$

Before we begin, let’s recall the most basic rule of factoring: factor out common monomial factors first . Notice that 3 is the greatest common monomial factor of every term. Factor out 3.

$3x^2-15x-42=3(x^2-5x-14)$

Now we can continue.

1. Write two sets of parentheses: $3(\begin{array}{cc}& \end{array})(\begin{array}{cc}& \end{array})$ .
2. Place the factors of $x^2$ into the first position of each set of parentheses:
   

   $3\begin{array}{ccc}(x& )(x& )\end{array}$

3. The third term of the trinomial is $-14$ . We seek two numbers whose
   (a) product is $-14$ and
   (b) sum is $-5$ .
   The required numbers are $-7\text{and}2$ . Place $-7\text{and}+2$ into the parentheses.
   

   $3x^2-15x-42=3(x-7)(x+2)$

   The factorization is complete. We’ll check to be sure.
   

   $\begin{array}{lll}3(x-7)(x+2)\hfill & =\hfill & 3(x^2+2x-7x-14)\hfill \\ \hfill & =\hfill & 3(x^2-5x-14)\hfill \\ \hfill & =\hfill & 3x^2-15x-42\hfill \end{array}$

Factor $x^2-7x+12$ .

1. Write two sets of parentheses: $(\begin{array}{cc}& \end{array})(\begin{array}{cc}& \end{array})$ .
2. The third term of the trinomial is $+12$ . The sign is positive, so the two factors of 12 we are looking for must have the same sign. They will have the sign of the middle term. The sign of the middle term is negative, so both factors of 12 are negative. They are $-12\text{and}-1,-6\text{and}-2,\text{or}-4\text{and}-3$ . Only the factors $-4\text{and}-3\text{add}\text{to}-7,\text{so}-4\text{and}-3$ are the proper factors of 12 to be used.
   

   $x^2-7x+12=(x-4)(x-3)$