Factoring Trinomials with a = 1

Factoring Trinomials with a = 1 exercise

• Explain how Johnny Redbeard can factor the trinomial $x^2+6x-27$.

  Hints

  To multiply two binomials like $(x + m)(x + n)$, we use:

  • First - multiply the first terms: $x$ times $x$
  • Outer - multiply the outer terms: $x$ times $n$
  • Inner - multiply the inner terms: $m$ times $x$
  • Last - multiply the last terms: $m$ times $n$

  In this example, $a=3$, $b=4$ and $c=5$.

  You are looking for two numbers, $m$ and $n$, that sum up to $b$ and whose product equals $ac$.

  Solution

  Captain Johnny Redbeard, is often referred to as the math expert of the seven seas.

  He realized he has to factorize the trinomial $x^2 + 6x - 27$ in order to open the treasure chest.

  $x^2 + 6x - 27$ is a special case of the standard form $ax^2 + bx + c$. Here we have $a=1$, $b=6$ and $c=-27$.

  The factorization of this polynomial will look like $(x + m)(x + n)$.

  Using the FOIL method we get the following:

  • First: $x\times x=x^2$
  • Outer: $x\times n=nx$
  • Inner: $m\times x=mx$
  • Last: $m\times n=mn$

  This leads us to $x^2 + mx + nx + mn = x^2 + (m + n)x + mn$.

  Comparing this with $x^2 + 6x - 27$, we have to find two numbers, $m$ and $n$, whose sum equals $6$ and whose product equals $-27$.

• Find all possible factors of $-27$ and add them together to identify $m$ and $n$.

  Hints

  The factors of $-27$ are $\pm 1$, $\pm 3$, $\pm9$ and $\pm 27$.

  Solution

  We've provided a table of the factors of $-27$ and the sum of the factors of $-27$.

  Captain Johnny Readbeard already knows he is looking for two numbers, $m$ and $n$ that, when added together, equal $6$ and whose product is $-27$. There can't be that many options.

  When given these kinds of problems, always start by figuring out the factors so that there are fewer possibilities. Then, write all combinations of the factors in a table.

  Next, calculate the sum of each of the factor pairs. For Johnny Redbeard's treasure, the sum has to be $6$.

  Taking a closer look at the table, there is only one possible combination which satisfies both requirements: $m=-3$ and $n=9$ added together equal $6$ and their product is $-27$.

  So, the trinomial $x^2+6x-27$ can be factored as $(x-3) (x+9)$.

• Assign each polynomial to its corresponding factorization.

  Hints

  Keep the reverse FOIL method in mind:

  $(x+m)(x+n)=x^2+(m+n)x+mn$

  The coefficient of $x$ is $m+n$ and the constant term is $mn$.

  First, look at the constant term and determine all possible factors, including the negative ones.

  Solution

  All given sets of binomials and trinomials look similar. Pay attention to their signs.

  First, we need the reverse FOIL method, $(x+m)(x+n)=x^2+(m+n)x+mn$.

  $~$
  $x^2-x-6$
  

  • Here we have $mn=-6$. The possible factors of $-6$ are: $\pm 1$, $\pm 2$, $\pm 3$, and $\pm 6$.
  • We also know that $m+n=-1$.
  • We then see that the only pair of factors that satisfy both requirements is $m=2$ and $n=-3$.
  • So $x^2-x-6=(x+2)(x-3)$

  $~$
  $x^2-5x+6$
  

  • Here, we have $mn=6$. The possible factors of $6$ are $\pm 1$, $\pm 2$, $\pm 3$, and $\pm 6$.
  • We also know that $m+n=-1$.
  • We then see that the only pair of factors that satisfy both requirements is $m=-2$ and $n=-3$.
  • So $x^2-5x+6=(x-2)(x-3)$

  $~$
  $x^2+5x+6$
  

  • Here, we have $mn=6$ and the factors are $\pm 1$, $\pm 2$, $\pm 3$, and $\pm 6$.
  • We also know that $m+n=5$.
  • We then see that the only pair that satisfies these requirements is $m=2$ and $n=3$.
  • So $x^2+5x+6=(x+2)(x+3)$

  $~$
  $x^2+x-6$
  

  • Here, we have $mn=-6$, so the possible factors of $-6$ are $\pm 1$, $\pm 2$, $\pm 3$, and $\pm 6$.
  • We also know that $m+n=1$.
  • We then see that the only pair that satisfies this is $m=-2$ and $n=3$.
  • So $x^2+x-6=(x-2)(x+3)$

• Factor the given polynomials in order to open the treasure chests.

  Hints

  Write each term in this form:

  $(x+m)(x+n)$

  Let's look at an example:

  $x^2-5x-6=(x-6)(x+1)=(x+1)(x-6)$

  The product of $m$ and $n$ is the constant of the polynomial and the sum of $m$ and $n$ the coefficient of $x$.

  Solution

  For each given polynomial, we look at the constant term first. Keep in mind that $mn=c$, where $c$ is the constant term. So you have to find all possible factors.

  Then, we check for each possible combination of factors if $m$ and $n$ also satisfy the second requirement, $m+n$ is equal to the coefficient of $x$.

  $~$
  $x^2+8x+12$
  

  • Here, $mn=12$
  • The possible factors of $12$ are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, and $\pm 12$
  • To get the sum $m+n=8$, we have $m=2$ and $n=6$ or, according to the Commutative Property, $m=6$ and $n=2$
  • So $x^2+8x+12=(x+2)(x+6)=(x+6)(x+2)$

  $~$
  $x^2-x-12$
  

  • $mn=-12$
  • The possible factors of $-12$ are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, and $\pm 12$
  • To get the sum $m+n=-1$, we have $m=3$ and $n=-4$. It could be the other way around, but here the $+$ is already given for the first binomial
  • So $x^2-x+12=(x+3)(x-4)$

  $~$
  $x^2-11x+24$
  

  • $mn=24$
  • The possible factors are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 8$, $\pm 12$, and $\pm 24$
  • To get the sum $m+n=-11$, we have $m=-8$ and $n=-3$ or, according to the Commutative Property, $m=-3$ and $n=-8$
  • So $x^2-11x+24=(x-8)(x-3)=(x-3)(x-8)$

• Describe the FOIL method for multiplying binomials.

  Hints

  Pay attention to the signs.

  FOIL is the mnemonic used to remember how to multiply two binomials.

  Keep in mind that the result of the multiplication above is

  $x^2+6x-27$

  Solution

  FOIL multiplication stands for

  • First - multiply the both first terms $x\times x=x^2$
  • Outer - multiply the outer terms $x\times 9=9x$
  • Inner - multiply the inner terms $-3\times x=-3x$
  • Last - multiply the last terms $-3\times 9=-27$

  So we get:

  $\begin{array}{rcl} (x-3)(x+9) &=& \\ x^2 + 9x - 3x - 27 &=& \\ &=& x^2+6x-27 \end{array}$

• Write each trinomial as a product of two binomials by factoring.

  Hints

  The reverse FOIL method tells us that

  $m+n=b$ and $mn=c$

  First, determine all combinations $m$ and $n$ that satisfy $mn=c$.

  Products are commutative, so instead of $(x-3)(x+2)$, you can also write $(x+2)(x-3)$.

  Solution

  When factoring polynomials, we can use the reverse FOIL method to help us more easily solve the equation, $x^2+bx+c=(x+m)(x+n)$:

  1. First, look at the constant $c$. We must find all possible factors, $m$ and $n$, of $c$.
  2. Check each combination of $m$ and $n$ that satisfies $mn=c$. Don't forget the negative factors.
  3. $m$ and $n$ also have to satisfy $m+n=b$.

  $~$
  $x^2+17x+70$
  

  • Looking at the constant term, we know $mn=70$.
  • We also know that $m+n$ has to equal $b$. For example, $2\times 35=70$, but $2+35=37\neq 17$.
  • Another example is $7\times 10=70$; adding these factors gives us $7+10=17$.
  • We can conclude that $x^2+17x+70=(x+7)(x+10)=(x+10)(x+7)$

  $~$
  $x^2-19x+84$
  

  • Looking at the constant term, we know $mn=84$.
  • We know that $m+n$ has to equal $b$. For example, $12\times 7=84$, but $12+7=19\neq -19$.
  • Let's check. $-12\times -7=84$. Now we get $-12+-7=-19$.
  • We can conclude that $x^2-19x+84=(x-12)(x-7)=(x-7)(x-12)$

  $~$
  $x^2+4x-96$
  

  • Here, we have $c=-96$ and so $mn=-96$. So we have either $m=\pm$ and $n=\mp 96$ or $m=\pm 2$ and $n=\mp 48$ ...
  • Because $m+n=4$, we have to look at a special combination of $m$ and $n$. In this combination, the difference between the absolute values of $m$ and $n$ must be $4$.
  • Applying these rules leads us to $m=\pm12$ and $n=\mp8$.
  • $-12\times 8=-96$ and $-12+8=-4$... that can't be the solution.
  • $12\times -8=-96$ and $12+-8=4$ $\surd$
  • So $x^2+4x-96=(x+12)(x-8)=(x-8)(x+12)$