audio read-through Basic Concepts Involved in Factoring Trinomials

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Here, you will practice the basic concepts involved in factoring trinomials of the form $\,x^2 + bx + c\,.$ These trinomials have an $\,x^2\,$ term with a coefficient of $\,1\,,$ an $\,x\,$ term, and a constant term.

Recall that factoring is the process of taking a sum (things added) and rewriting it as a product (things multiplied).

Observe that for all real numbers $\,f\,,$ $\,g\,$ and $\,x\,$:

$$ \begin{align} &\cssId{s14}{(x+f)(x+g)}\cr\cr &\qquad \cssId{s15}{= \overset{\text{First}}{\overbrace{\strut\ x^2\ }}} \cssId{s16}{+ \overset{\text{Outer}}{\overbrace{\strut\ gx\ }}} \cssId{s17}{+ \overset{\text{Inner}}{\overbrace{\strut\ fx\ }}} \cssId{s18}{+ \overset{\text{Last}}{\overbrace{\strut\ fg\ }}}\cr\cr &\qquad \cssId{s19}{= x^2 + (f+g)x + fg} \end{align} $$

Now, think of going ‘backwards’:  from $\,x^2 + (f+g)x + fg\,$ back to the factored form $\,(x+f)(x+g)\,.$

We'd need two numbers that add together to give the coefficient of the $\,x\,$ term, and that multiply together to give the constant term.

This gives the following result, which is the primary tool used in factoring trinomials:

KEY TOOL FOR FACTORING TRINOMIALS (where the coefficient of the squared term is $\,1\,$)
To factor a trinomial of the form $\,x^2 + bx + c\,,$ start by finding two numbers, $\,f\,$ and $\,g\,,$ that
  • add together to give $\,b\,$ (the coefficient of the $\,x\,$ term); and
  • multiply together to give $\,c\,$ (the constant term).
Then: $$ \begin{align} &\cssId{s36}{x^2 + bx + c}\cr &\qquad \cssId{s37}{=\ \ x^2} \ \cssId{s38}{+ (\overset{= b}{\overbrace{f+g}})x} \cssId{s39}{+ \overset{= c}{\overbrace{\ fg\ }}}\cr\cr &\qquad \cssId{s40}{=\ \ (x + f)(x + g)} \end{align} $$

For example, to factor $\,x^2 + 5x + 6\,,$ we must find two numbers that add to $\,5\,$ and multiply to $\,6\,.$ The numbers $\,2\,$ and $\,3\,$ work, since $\,2+3 = 5\,$ and $2\cdot 3 = 6\,.$ Thus:

$$ \begin{align} &\cssId{s48}{x^2 + 5x + 6}\cr &\qquad \cssId{s49}{=\ \ x^2 + (2 + 3)x + (2\cdot 3)}\cr &\qquad \cssId{s50}{=\ \ (x+2)(x+3)} \end{align} $$

(FOIL it out to check!)

When everything in sight is positive and coefficients are small, then it may be easy to come up with the ‘numbers that work’. For example, it may not be too hard for you to find numbers that add to $\,5\,$ and multiply to $\,6\,.$

However, bring some negative numbers into the picture and make coefficients bigger, and things can get considerably trickier.

Fortunately, there are some key ideas that will help you find the ‘numbers that work’ (if they exist), and the purpose of this web exercise is to give you practice with these ideas.

Key Ideas for Finding the ‘Numbers that Work’

Examples

Question: Suppose two numbers multiply to $\,36\,.$ Then, the numbers have (choose one):
THE SAME SIGN DIFFERENT SIGNS
Answer: THE SAME SIGN
Question: Suppose two numbers multiply to $\,-36\,.$ Then, the numbers have (choose one):
THE SAME SIGN DIFFERENT SIGNS
Answer: DIFFERENT SIGNS
Question: Suppose two numbers have the same sign, and they add to $\,10\,.$ Then, the numbers are (choose one):
BOTH POSITIVE BOTH NEGATIVE
Answer: BOTH POSITIVE
Question: Suppose two numbers have the same sign, and they add to $\,-10\,.$ Then, the numbers are (choose one):
BOTH POSITIVE BOTH NEGATIVE
Answer: BOTH NEGATIVE
Question: When you add two numbers that have the same sign, then in your head you do a/an (choose one):
ADDITION PROBLEM SUBTRACTION PROBLEM
Answer: ADDITION PROBLEM
Question: When you add two numbers that have different signs, then in your head you do a/an (choose one):
ADDITION PROBLEM SUBTRACTION PROBLEM
Answer: SUBTRACTION PROBLEM
Question: Suppose two numbers have different signs, and they add to $\,10\,.$ Then, the bigger number is (choose one):
POSITIVE NEGATIVE
Answer: POSITIVE
Question: Suppose two numbers have different signs, and they add to $\,-10\,.$ Then, the bigger number is (choose one):
POSITIVE NEGATIVE
Answer: NEGATIVE

Concept Practice