##
Section3.3Factoring Trinomials with Leading Coefficients of 1

¶Factoring a trinomial of form \(x^2+bx+c\text{,}\) where \(b\) and \(c\) are integers, is essentially the reversal of a FOIL process. For this reason, we can develop a strategy by investigating a FOIL expansion. Let's expand \((x+4)(x+6)\text{.}\)

\begin{align*}
(x+4)(x+6)\amp=x^2+6x+4x+24\\
\amp=x^2+10x+24
\end{align*}

Let's observe that the linear coefficient, \(10\text{,}\) is the sum of \(4\) and \(6\) whereas the constant term, \(24\text{,}\) is the product of \(4\) and \(6\text{.}\) Since we expanded \((x+4)(x+6)\text{,}\) we can infer that finding the constants in a factorization of trinomials of form \(x^2+bx+c\) is dependent upon determining two numbers that sum to \(b\) and multiply to \(c\text{.}\)

From the factor pairs and sums shown in Table 3.3.1, we can infer the following factorizations.

\begin{align*}
x^2+7x+6\amp=(x+1)(x+6)\\
x^2+5x+6\amp=(x+2)(x+3)\\
x^2-7x+6\amp=(x-1)(x-6)\\
x^2-5x+6\amp=(x-2)(x-3)
\end{align*}

factors |
sum |

\(1,6\) |
\(7\) |

\(2,3\) |
\(5\) |

\(-1,-6\) |
\(-7\) |

\(-2,-3\) |
\(-5\) |

Table3.3.1Factor pairs of \(6\)

Just as noteworthy, since the factor pairs shown in Table 3.3.1 form an exhaustive list, we cannot factor any trinomial of form \(x^2+bx+6\) where b is not one of the four numbers shown in the sum column. Such trinomials are said to be prime. For example, \(x^2+3x+6\) is prime.

######
Example3.3.2

Factor \(x^2-8x+15\)

SolutionOur first task is to determine a factor pair that multiplies to \(15\) and adds to \(-8\text{.}\) Since the product is positive and the sum is negative, we are searching for two negative numbers. The pair that works is \(-3\) and \(-5\text{.}\) This gives us:

\begin{equation*}
x^2-8x+15=(x-3)(x-5)\text{.}
\end{equation*}

######
Example3.3.3

Factor \(w^2-4w-12\)

SolutionSince our factor pair needs to multiply to a negative value \((-12)\text{,}\) one the numbers in the pair must be positive and the other negative. Let's note that \(6\) and \(2\) multiply to 12 and have a difference of \(4\text{.}\) This can be helpful in determining that the factor pair that multiplies to \(-12\) and adds to \(-4\) is \(-6\) and \(2\text{.}\) This gives us:

\begin{equation*}
w^2-4w-12=(w-6)(w+2)\text{.}
\end{equation*}

###
Subsection3.3.1Exercises

Factor each trinomial. Check your answer by expanding the factorization. If the trinomial cannot be factored, state that it is prime.

###### 1

\(x^2+6x+5\)

###### 2

\(t^2-6t+8\)

###### 3

\(x^2+8x+4\)

###### 4

\(y^2-14y-32\)

###### 5

\(x^2+6x-20\)

###### 6

\(x^2+15x-100\)

###### 7

\(w^2-18w+45\)

###### 8

\(x^2+21x+80\)

###### 9

\(b^2-9b-400\)