Factoring Trinomials of Type ax2 +bx +c, a la Shmoop
Farmer Jen Etticks <> produces donut-shaped watermelons.
Like Godzilla, they’re huge in Japan.
There’s just one problem: tossing the spoiled watermelons into the ocean has had some…
undesirable effects.
The guy in accounting says that the amount of spoilage, or “s”…
…for producing w number of watermelons per day can be approximated by the formula:
s equals 6w squared plus 13w plus 2.
What daily production level will give Jen zero spoilage?
This looks like a trinomial equation of the type ax squared plus bx plus c; here, we’re
using w’s instead of x’s.
To find the answer, we’re going to need to factor out the a and the c in the equation.
We’ll do this by using a table.
Our table has four columns.
The first involves our “a-term”, 6w squared.
There are two possible ways to factor 6w squared: 3w and 2w, and 6w and w.
The second column factors our “c-term”, 2.
The only possible factors are 2 and 1.
The third and fourth columns use the previous two columns to decipher possible “b-terms”.
Combinations of the a-term and c-term factors are combined in the “possible binomial factors”
column.
If we take a look at the first row of the first two columns, our possible binomial factors
come out to:
(3w + 2) times (2w + 1), and (3w+1) times (2w+2).
For the second row, we get (6w +2) times (w+1), and (6w+1) times (2+2).
Then, the sum of the inner and outer products are used to calculate what the b-term for
that product would be.
So for (3w +2) times (2w + 1), we take the outer product first, which is 3w times 1,
and add it to the inner products: 2w times 2.
3w times 1 is 3w, and 2w times 2 is 4w.
So the sum is 7w.
Not the right b-term...so we keep trying.
After doing all the calculations, we can see that
the factors (6w + 2) and (w + 2) give us our correct b-term, 13w, hiding out like a bunny
at a fox convention.
Our work can be checked using FOIL, multiplying out the first terms, outer terms, inner terms,
and last terms,.
We add them together…and check to see that they equal our original trinomial.
We passed the checkpoint, which means when spoilage equals zero, one of our factors must
also equal zero.
When we set each factor to zero and solve for w…
…we see there is zero spoilage when our daily watermelon production equals negative
one sixth and negative 2.
Uh-oh.
Since it’s impossible to have negative production…
…looks like Farmer Jen has no choice but to introduce spoiled watermelons into the
food supply.
Good luck, everyone.