Question

State the number of complex zeros, the possible number of real zeros,and the possible rational zeros for the function.Then find all rational zeros.

f(x)=x^3-5x^2-13x-7

Answer 1

First, try a few "simple" numbers, like 1, -1, 2, -2 etc.
You'll see that x = -1 is a zero.
This means that you can factor f(x) as (x+1)(ax²+bx+c).
How to find a, b and c? Just do a long division, or a synthetic division, which is much more efficient: (it is a really nice trick!)

-1 [ 1 -5 -13 -7 ]
-1 6 7
-------------------- +
1 -6 -7 0

So f(x) = (x+1)(x²-6x-7)

Explanation of synthetic division:

First, write down the zero you already have: -1
Then, between the brackets: the coefficients of the polynomial, in the right order:
1 -5 -13 -7

The first coefficient (1) goes straight under the dashed line.
Now multiply it by the zero -1 and put the result under the next coefficient (-5).
Add up to get -6. Again, multiply -1 and -6. Put the result 6 below next coeff. (-13).
Go on doing this until you have reached the end. The las sum is 0, which indicates that there is no remainder.
Now the numbers 1, -6 and -7 are the coefficients of the 2nd degree polynomial you were looking for.
This one can easily be factored again. This gives you (in this case) two more nice zeros, so no complex ones.