How do you find all the zeros of #f(x) = x^3 – 3x^2 – 16x + 48#?

1 Answer
May 11, 2016


#f(x)=x^3-3x^2-16x+48 = color(blue)((x+4)(x-3)(x-4))#
i.e. zeros are at #color(red)(x in {-4,+3,+4})#


Note that a polynomial of degree 3 (such as this one) should have 3 zeroes (although there is no way to be sure that they will all be unique or Real).

There are multiple ways to analyze this but I chose the way that required the least thinking on my part.

Hoping for at least one rational (in this case, integer) zero
I applied the Factor Theorem to all integer factors of #48#
and (using a spreadsheet for the donkey work) got:
enter image source here
This worked even better that I had hoped.
As you can see this gives me all 3 zeros!