Consider the trinomial
\begin{equation*}
x^2 + 10x + 16
\end{equation*}
Can we find two binomial factors,
\begin{equation*}
(x + a)(x + b)
\end{equation*}
whose product is the given trinomial? The product of the binomials is
\begin{equation*}
(x + a)(x + b) = x^2 + (a + b)x + ab
\end{equation*}
Thus, we are looking for two numbers, \(a\) and \(b\text{,}\) that satisfy
\begin{equation*}
(x + a)(x + b) = x^2 + (\alert{a + b})x + \blert{ab}= x^2 + \alert{10}x + \blert{16}
\end{equation*}
By comparing the coefficients of the terms in the two trinomials, we see that \(\alert{a + b = 10}\) and \(\blert{ab = 16}\text{.}\) That is, the sum of the two numbers is the coefficient of the linear term, \(10\text{,}\) and their product is the constant term, \(16\text{.}\)
To find the numbers, we list all the possible integer factorizations of \(16\text{:}\)
\begin{equation*}
1 \cdot 16, ~~~2 \cdot 8, ~~~\text{ and }~~~ 4 \cdot 4
\end{equation*}
We see that only one combination gives the correct linear term: \(8\) and \(2\text{.}\) These are the numbers \(a\) and \(b\text{,}\) so
\begin{equation*}
x^2 + 10x + 16 = (x + 8)(x + 2)
\end{equation*}
In Example A.57 we factor quadratic trinomials in which one or more of the coefficients is negative.
Example A.57
Factor.
\(x^27x+12\)
\(x^2x12\)
Solution

Find two numbers whose product is \(12\) and whose sum is \(7\text{.}\) Because the product is positive and the sum is negative, the two numbers must both be negative. The possible factors of \(12\) are \(1\) and \(12\text{,}\) \(2\) and \(6\text{,}\) or \(3\) and \(4\text{.}\) Only \(4\) and \(3\) have the correct sum, \(7\text{.}\) Hence,
\begin{equation*}
x^2  7x + 12 = (x  4) (x  3)
\end{equation*}

Find two numbers whose product is \(12\) and whose sum is \(1\text{.}\) Because the product is negative, the two numbers must be of opposite sign and their sum must be \(1\text{.}\) By listing the possible factors of \(12\text{,}\) we find that the two numbers are \(4\) and \(3\text{.}\) Hence,
\begin{equation*}
x^2  x  12 = (x  4) (x + 3)
\end{equation*}
If the coefficient of the quadratic term is not \(1\text{,}\) we must also consider its factors.
Example A.58
Factor \(~8x^2  9  21x\)
Solution

Write the trinomial in decreasing powers of \(x\text{.}\)
\begin{equation*}
8x^221x9
\end{equation*}

List the possible factors for the quadratic term.
\begin{align*}
(8x\hphantom{00000})\amp(x\hphantom{000000})\\
(4x\hphantom{00000})\amp(2x\hphantom{00000})
\end{align*}

Consider possible factors for the constant term: \(9\) may be factored as \(9\cdot 1\) or as \(3\cdot 3\text{.}\) Form all possible pairs of binomial factor using these factorizations.

Select the combinations of the products ① and ② whose sum or difference could be the linear term, \(21x\text{.}\)
\begin{equation*}
(8x \hphantom{000} 3) (x \hphantom{000} 3)
\end{equation*}

Insert the proper signs:
\begin{equation*}
(8x + 3) (x  3)
\end{equation*}
With practice, you can usually factor trinomials of the form \(Ax^2 + Bx + C\) mentally. The following observations may help.

If \(A\text{,}\) \(B\) and \(C\) are all positive, both signs in the factored form are positive. For example, as a first step in factoring \(6x^2 + 11x + 4\text{,}\) we could write
\begin{equation*}
(\hphantom{000} + \hphantom{000} ) (\hphantom{000} + \hphantom{000} )
\end{equation*}

If \(A\) and \(C\) are positive and \(B\) is negative, both signs in the factored form are negative. Thus as a first step in factoring \(6x^2  11x + 4\text{,}\) we could write
\begin{equation*}
(\hphantom{000}  \hphantom{000} ) (\hphantom{000}  \hphantom{000} )
\end{equation*}

If \(C\) is negative, the signs in the factored form are opposite. Thus as a first step in factoring \(6x^2  5x  4\text{,}\) we could write
\begin{equation*}
(\hphantom{000} + \hphantom{000} ) (\hphantom{000}  \hphantom{000} )
~~\text{ or }~~ (\hphantom{000}  \hphantom{000} ) (\hphantom{000} + \hphantom{000} )
\end{equation*}
Example A.59
\(\begin{aligned}[t]
6x^2 + 5x + 1 \amp = (\hphantom{00} + \hphantom{00} ) (\hphantom{00} + \hphantom{00} )\\
\amp = (3x + 1) (2x + 1)
\end{aligned}\)
\(\begin{aligned}[t]
6x^2  5x + 1 \amp = (\hphantom{00}  \hphantom{00} ) (\hphantom{00}  \hphantom{00} )\\
\amp = (3x  1) (2x  1)
\end{aligned}\)
\(\begin{aligned}[t]
6x^2  x  1 \amp = (\hphantom{00} + \hphantom{00} ) (\hphantom{00}  \hphantom{00} )\\
\amp = (3x + 1) (2x  1)
\end{aligned}\)
\(\begin{aligned}[t]
6x^2  xy  y^2 \amp = (\hphantom{00} + \hphantom{00} ) (\hphantom{00}  \hphantom{00} )\\
\amp = (3x + y) (2x  y)
\end{aligned}\)
Subsection Special Products and Factors
The products below are special cases of the multiplication of binomials. They occur so often that you should learn to recognize them on sight.
Special Products
\((a + b)^2 = (a + b) (a + b) = a^2 + 2ab + b^2\)
\((a  b)^2 = (a  b) (a  b) = a^2  2ab + b^2\)
\((a + b)(ab) = a^2  b^2\)
Example A.61
\(\begin{aligned}[t]
3(x + 4)^2 \amp = 3(x^2 + 2 \cdot 4x + 4^2) \\
\amp = 3x^2 + 24x + 48
\end{aligned}\)
\(\begin{aligned}[t]
(y + 5) (y  5) \amp = y^25^2 \\
\amp = y^225
\end{aligned}\)
\(\begin{aligned}[t]
(3x  2y)^2 \amp = (3x)^2  2(3x)(2y) + (2y)^2 \\
\amp = 9x^212xy+4y^2
\end{aligned}\)
Each of the formulas for special products, when viewed from right to left, also represents a special case of factoring quadratic polynomials.
Special Factorizations
\(a^2 + 2ab + b^2=(a + b)^2\)
\(a^2  2ab + b^2=(a  b)^2 \)
\(a^2  b^2=(a + b)(ab)\)
\(a^2+b^2 ~~\text{ cannot be factored}\)
The trinomials in (I) and (II) are sometimes called perfectsquare trinomials because they are squares of binomials. Note that the sum of two squares, \(a^2 + b^2\text{,}\) cannot be factored.
Example A.62
Factor.
\(x^2 + 8x + 16\)
\(y^2 10y + 25\)
\(4a^2  12ab + 9b^2\)
\(25m^2n^2 + 20mn + 4\)
Solution

Because \(16\) is equal to \(4^2\) and \(8\) is equal to \(2\cdot 4\text{,}\)
\begin{align*}
x^2 + 8x + 16 \amp = x^2  2 \cdot 4x + 4^2\\
\amp = (x + 4)^2
\end{align*}

Because \(25=5^2\) and \(10=2\cdot 5\text{,}\)
\begin{align*}
y^2  10y + 25 \amp = y^2  2 \cdot 5y + 5^2\\
\amp = (y5)^2
\end{align*}

Because \(4a^2=(2a)^2\text{,}\) \(9b^2=(3b)^2\text{,}\) and \(2ab=2(2a)(3b)\text{,}\)
\begin{align*}
4a^2  12ab + 9b^2 \amp = (2a)^2  2(2a)(3b) + (3b)^2\\
\amp = (2a3b)^2
\end{align*}

Because \(25m^2n^2=(5mn)^2\text{,}\) \(4=2^2\text{,}\) and \(20mn=2(5mn)(2)\text{,}\)
\begin{align*}
25m^2n^2 + 20mn + 4 \amp = (5mn)^2 + 2(5mn)(2) + 2^2\\
\amp = (5mn+2)^2
\end{align*}
Binomials of the form \(a^2  b^2\) are often called the difference of two squares.
Example A.63
Factor if possible.
\(x^2  81\)
\(4x^2  9y^2\)
\(x^2 + 81\)
Solution

The expression \(x^2  81\) is the difference of two squares, \(x^2  9^2\text{,}\) and thus can be factored according to Special Factorization (III) above.
\begin{align*}
x^281 \amp = x^29^2\\
\amp = (x+9)(x9)
\end{align*}

Because \(4x^2  9y^2\) can be written as \((2x)^2 (3y)^2\text{,}\)
\begin{align*}
4x^2  9y^2 \amp =(2x^2)(3y)^2 \\
\amp =(2x+3y)(2x3y)
\end{align*}
The expression \(x^2 + 81\text{,}\) or \(x^2 + 0x + 81\text{,}\) is not factorable, because no two real numbers have a product of \(81\) and a sum of \(0\text{.}\)
The factors \(x + 9\) and \(x  9\) in Example A.63a are called conjugates of each other. In general, any binomials of the form \(a+b\) and \(ab\) are called a conjugate pair.