Here, you will practice factoring trinomials of the form
$\,ax^2 + bx + c\,$, where
$\,a\,$, $\,b\,$, and $\,c\,$ are integers.
In this exercise, the coefficient of the $\,x^2\,$ term ($\,a\,$) is usually not going to be the number $\,1\,$,
which makes factoring more challenging.
For example, you might need to factor $\,6x^2-13x-5\,$.
Two methods are presented:
One common method for factoring when $\,a\ne 1\,$ is called Guess-and-Check
or Trial-and-Error.
In this method, we try to find four integers that work :
$$
\cssId{s15}{6x^2 -13x - 5}
\cssId{s16}{= (\text{__ }x + \text{__ })(\text{__ }x + \text{__ })}
$$
The unknown numbers in front of the $\,x\,$ variables must multiply together to give $\,6\,$.
What are the possibilities here?
Well, $\,2\cdot 3\,$ or $\,6\cdot 1\,$.
The unknown constants must multiply together to give $\,-5\,$.
What are the possibilities for this one?
Well, $\,1\cdot -5\,$ or $\,-1\cdot 5\,$.
Phew.
Not too many choices.
But even so, this gives lots of possible combinations:
$(2x+1)(3x-5)$ | $(6x+1)(x-5)$ |
$(2x-5)(3x+1)$ | $(6x-5)(x+1)$ |
$(2x-1)(3x+5)$ | $(6x-1)(x+5)$ |
$(2x+5)(3x-1)$ | $(6x+5)(x-1)$ |
Then, the inners and the outers have to work out right .
If you're not the particularly lucky type,
then it might take you a while to stumble on the one that works.
(Did you find it?)
And, for a random trinomial pulled out of the air, it's possible that nothing will work!
So, you could spend a long time looking, and find yourself wondering:
“Do they exist?
Have I just missed them?”
There must be a better way—and there is.
It's called the ‘Factor by Grouping’ method,
and it's usually much more efficient than trial-and-error.
Here's an example of the ‘factor by grouping’ method.
Details follow, for inquiring minds.
However, some of you may want to just take this example, and run with it.
$6x^2 - 13x - 5$ | (original trinomial) |
$= 6x^2\ \overset{=\ -13x}{\overbrace{- 15x + 2x}} - 5$ | (rename the middle term, using the two numbers you found) |
$= (6x^2 - 15x) + (2x - 5)$ | (group the first two terms, and the last two terms—hence the title of this technique!) |
$= 3x(2x - 5) + 1(2x - 5)$ | (factor the first part; put the $\,1\,$ in for clarity in the second part) |
$= 3x\overbrace{(2x-5)} + 1\overbrace{(2x-5)}$ | (Look carefully! There are two ‘big’ terms, with a common factor of $\,2x-5\,$) |
$= (2x-5)(3x + 1)$ | (factor out the common factor—done!) |
It may seem like a lot of work to you.
But, over the years, I've divided my classes in half:
told one side to use trial-and-error, and the other side to use factor by grouping.
The factor by grouping side always wins.
Let's do the same example one more time.
This will show you that the order you write the two middle terms doesn't matter.
Also, this version is much more compact, and is usually all you'll need to write down.
$6x^2 - 13x - 5$ | (original trinomial) |
$= 6x^2 \ \overset{\text{different order}}{\overbrace{+ 2x -15x}} - 5$ | (rename the middle term) |
$= (6x^2 + 2x) + (-15x - 5)$ | (group first two, last two) |
$= 2x(3x + 1) + (-5)(3x + 1)$ | (factor each group separately) |
$= (3x + 1)(2x - 5)$ | (factor again) |
Here's the motivation for the technique.
Start by pulling a random problem out of the air, and FOILing it out:
$$
\cssId{s116}{(3x - 1)(5x + 7)}
\cssId{s117}{= 15x^2 + 21x - 5x - 7}$$
Don't combine the middle terms.
Instead, look at the four numbers generated in the resulting sum:
$\,15\,$, $\,21\,$, $\,-5\,$, $\,-7$
The product of the first and last is:
$\,(15)(-7) = -105$
The product of the middle two is:
$\,(21)(-5) = -105$
Is this just a coincidence?
It's not a coincidence.
It's always true:
$$
\cssId{s128}{(ex + f)(gx + h)}
\cssId{s129}{= (eg)x^2 + (eh)x + (fg)x + (fh)}
$$
The product of the first and last is:
$\,(eg)(fh) = efgh\,$
The product of the middle two is:
$\,(eh)(fg) = efgh\,$
Same result.
Other than being a curious observation, what good is this?
Ends up that it's a LOT of good.
If we're trying to factor a trinomial by grouping,
and we're looking for the right way to rename the middle term,
then we know two things:
$$
\cssId{s140}{ax^2 + bx + c}
\cssId{s141}{= ax^2 + \text{__}\ x + \text{__}\ x + c}
$$
The rest of the argument is more advanced, and is primarily included for the sake of the teacher.
Click here, if you're interested in taking a look.
For more advanced students, a graph is displayed for each factoring problem.
For example, suppose you're asked to factor
$\,6x^2 - 13x - 5\,$.
Then, you'd see the graph of the equation
$\,y = 6x^2 - 13x - 5\,$.
(This graph is shown below when the web page is first loaded.)
Observe that the graph crosses the $\,x\,$-axis at $\,-\frac13\,$ and $\,\frac52\,$.
Look at the factorization:
$6x^2 - 13x - 5 = (3x + 1)(2x - 5) = 3(x + \frac 13)2(x - \frac52) = 6(x+\frac13)(x-\frac52)\,$
See the relationship between the factors and the $x$-axis intercepts (zeros)?
If $\,c\,$ is a zero, then $\,x - c\,$ is a factor!
(You're discovering the beautiful relationship between the zeroes of a polynomial, and its factors.)
Click the “show/hide graph” button if you prefer not to see the graph.