All potential rational zeros of a polynomial with only integer
coefficients are numbers of the form ±P/Q where P is a divisor
of the absolute value of the constant term and Q is a divisor of the
absolute value of the leading coefficient.
f(x) = x³-5x²-9x+45
The constant term is 45
The leading term is x³
The leading coefficient is 1.
In this case since the leading coefficient is 1, which has only
the divisor 1, we only need to consider as potential rational zeros
± the divisors of 45, which are
±1, ±3, ±5, ±9, ±15, and ±45.
We try 1
1 | 1 -5 -9 45
| 1 -4 -13
1 -4 -13 32
The remainder is 32, not 0, so 1 is not a zero of f(x)
We try -1
-1 | 1 -5 -9 45
| -1 6 4
1 -6 -4 48
The remainder is 48, not 0, so -1 is not a zero of f(x)
We try 3
3 | 1 -5 -9 45
| 3 -6 -45
1 -2 -15 0
The remainder is 0, so 3 is a zero of f(x).
Since the synthetic division was actually a division
of f(x) by x-3, we can use the numbers on the bottom
line of the synthetic division to factor f(x) as
f(x) = (x-3)(x²-2x-15)
We only need to find the zeros of x²-2x-15 to find
the other zeros, rational or otherwise of f(x).
We are able to do further factoring:
f(x) = (x-3)(x+3)(x-5)
Therefore all zeros of f(x) are found by setting each of
these expressions = 0 and solving:
x-3 = 0; x+3 = 0; x-5 = 0
x = 3 x = -3 x = 5
Thus there are three zeros, 3,-3, and -5.
In we graph of f(x) we see that these three zeros are the
x-intercepts:
Edwin
