More on Zeros of Polynomials
F(x) =
is a rational function where P(x) and Q(x) are polynomials
If deg P(x) < deg Q(x), then F(x) is a proper rational function (fraction).
If deg P(x) ≥ deg Q(x), then F(x) is improper fraction and we can divide polynomials in one of
two ways:
1) Long division of polynomials
2) Synthetic division of polynomials
Example 1: Divide F(x) = x3 + 2x2 – 1 by x – 2.
x3  2x2  1
x2
Solution:
1) Long division:
x2 + 4x + 8
x–2
x 3  2x 2  0x  1
-(x3 – 2x2)
Note: we fill in missing degrees with coefficient 0 (x here)
x2 (x – 2x)
4x2 + 0x
-(4x2 – 8x)
4x(x – 2)
8x – 1
-(8x – 16)
15
8(x – 2)
Remainder
Answer: x2 + 4x + 8 with remainder 15.
2) Synthetic division:
First we need to find the root of x – 2 (divisor) by setting x – 2 = 0. This gives us x = 2.
Then, we write the root (2) first, then all coefficients of (x3 + 2x2 + 0x – 1). It’s same process.
2| 1 2 0 1
2
8 16
1 4 8 |15
1) We copy the first coefficient (1)
2) We multiply each coefficient in bottom row by root (2)
to write the results in second row
3) Then we add coefficients in first and second rows
4) The last result is the remainder
Now, we work backwards to write the answer: Remainder is 15, then 8 + 4x + x2 (ascending
order), or x2 + 4x + 8 (descending order).
The Remainder Theorem:
If F(x) is divided by x – a, then the remainder is F(a).
This theorem is useful to check to see if you performed division correctly (or vice versa).
For the example above, by Remainder Theorem, F(2) should equal 15 (remainder). Let’s check if
it’s true.
F(2) = 23 + 2(22) – 1 = 8 + 8 – 1 = 15
Yep, it works!
Factor Theorem: (follows from Remainder Theorem)
x – a is a factor of F(x) if and only if F(a) = 0.
The Rational Zeros Test:
If F(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0
and is a zero of F(x) (reduced fraction) then
1) p is a factor of a0
2) q is a factor of an
Example 2: Find the set of all possible zeros.
h(x) = 4x3 – 9x2 + x + 6
a0 = 6, a3 = 4
p (factors of 6): ±1, ±2, ±3, ±6
q(factors of 4): ±1, ±2, ±4
1 2 3 6 1 2 3 6 1 2 3 6
Then, all possible zeros are ( ):  , , , , , , , , , , ,
1 1 1 1 2 2 2 2 4 4 4 4
Note: we simply listed all possibilities of p/q. Now, we must reduce fractions (to eliminate
redundant choices).
Possible zeros (reduced): {±1, ±2, ±3, ±6, 
1
1
3
3
, , , }
2
4
2
4
Note: The rational zeros test only gives us possible rational zeros (not all of these possibilities
are actual zeros), so we must check these possibilities with synthetic division or by plugging
them into the polynomial. Also, the rational zeros test does not tell us complex (imaginary)
zeros (if any).
It may be time-consuming the check all possibilities, so sometimes we find the first zero (by
checking a few possibilities), then use long (or synthetic) division to find the rest of factors
(zeros).
Example 3: First find all rational zeros, then use the remaining (depressed) equation to find the
rest of zeros of f(x).
f(x) = x3 – 7x2 – 5x + 3
p = ± 1, ± 3
(how?)
q = ±1
(how?)
possible zeros (p/q): ±
f(1) = -8
1 3
, ± , or ± 1, ± 3
1 1
1 is not a zero
f(-1) = 0
Bingo!
Since -1 is a zero of f(x), we perform synthetic division on f(x) to find the (depressed) equation.
-1| 1 -7 -5 3
1
-1
8 -3
-8
3 |0
The depressed equation (answer) is x2 – 8x + 3.
Since -1 is a zero, then (x + 1) is a factor of f(x), then
f(x) = x3 – 7x2 – 5x + 3 = (x + 1) (x2 – 8x + 3)
To find the remaining zeros, we can use Quadratic Formula on x2 – 8x + 3.
x=
8  64  4  1  3 8  52 8  2 13


 4  13
2 1
2
2
So, the roots are {-1, 4 ±
}.