Question 3.4.9: Finding All Zeros of a Polynomial Function Find all zeros of...

We consider the number of positive zeros by observing the variations in signs for ƒ(x).

Because ƒ(x) has four sign changes, we can use Descartes’ rule of signs to determine that there are four, two, or zero positive real zeros. For negative zeros, we consider the variations in signs for ƒ(-x).

ƒ(-x) = (-x)^{4} – 3(-x)³ + 6(-x)² – 12(-x) + 8

ƒ(-x) = x^{4} + 3x³ + 6x² + 12x + 8

Because ƒ(-x) has no sign changes, there are no negative real zeros. The function has degree 4, so it has a maximum of four zeros with possibilities summarized in the table on the next page.

We can now use the rational zeros theorem to determine that the possible rational zeros are ±1, ±2, ±4, and ±8. Based on Descartes’ rule of signs, we discard the negative rational zeros from this list and try to find a positive rational zero. We start by using synthetic division to check 4.

Proposed zero →4)\overline{1    -3     6    -12       8}

            4       4     40      112

\overline{         1      1       10      28      120   }      ←ƒ(4) = 120

We find that 4 is not a zero. However, 4 > 0, and the numbers in the bottom row of the synthetic division are nonnegative. Thus, the boundedness theorem indicates that there are no zeros greater than 4. We can discard 8 as a possible rational zero and use synthetic division to show that 1 and 2 are zeros.

1)\overline{ 1  -3    6   -12  8}

   1   -2   4   -8

2)\overline{1    -2    4   -8   0}  ←ƒ(1)=0

          2      0        8

\overline{  1         0         4          0                 }    ←ƒ(2)=0

The polynomial now factors as

ƒ(x) = (x – 1)(x – 2)(x² + 4).

We find the remaining two zeros using algebra to solve for x in the quadratic factor of the following equation.

(x – 1)(x – 2)(x² + 4) = 0

x – 1 = 0    or     x – 2 = 0    or     x² + 4 = 0     Zero-factor property

x = 1         or     x = 2          or      x² = -4

                         x = ±2i       Square root property

The linear factored form of the polynomial is

ƒ(x) = (x – 1)(x – 2)(x – 2i)(x + 2i),

and the corresponding zeros are 1, 2, 2i, and -2i.