Polynomial function zeros:

   In math, a polynomial is an expression of finite length constructed from variables (also known as in determinates) and constants, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents. For example, x2 − 4x + 7 is a polynomial, but x2 − 4/x + 7x3/2 is not, because its second term involves division by the variable x and because its third term contains an exponent that is not a whole number.

   Polynomial function zeros are a set of theorems aiming to find (or determine the nature) of the complex zeros of a polynomial function.

Note: i = square root (-1)

 

Polynomial function zeros-Example 1:

 

 

2 + i is a zero of p(y) = x4 - 2.x3 - 6.x2 + 22.x - 15

Solution:

Step 1: The zero 2 + i is an imaginary element

Step 2: p(y) has real coefficients.

Step 3: The conjugate 2 - i is also zero of p(y).

Step 4: The factored form of p(y)
            p(y) = [x - (2 + i)] [x - (2 - i)]q(x)

Step 5: The expand the term of

          [x - (2 + i)][x - (2 - i)] in p(y)
          [x - (2 + i)][x - (2 - i)] = x2 -(2 + i)x -(2 - i)x + (2+i)(2-i)
           = x2 - 4·x + 5

Step 6: q(x) dividing by p(y)

            by x2 - 4·x + 5.
           (x4 - 2·x3 - 6·x2 + 22·x - 15) / (x2 - 4·x + 5)
           = x2 + 2·x - 3

Step 7: The factored form of p(y) is
            p(x) = [x - (2 + i)][x - (2 - i)](x2 + 2·x - 3)

Step 8: The remaining 2 zeros of p(y)

Step 9: The solutions to the quadratic equation are  
            x2 + 2·x - 3 = 0

Step 10:To factor the quadratic equation of x2 + 2·x - 3 = 0 .
           (x - 1)·(x + 3) = 0

Step 11: solutions
            x = 1, x = -3

Step 12: The zeros of p(y)

              2 + i , 2 - i, -3 and 1.

 

 

Polynomial function zeros practice problem:

 

Problem 1: To calculate the function of p(x) = x2 + 5x +6

Solution: The zero of the function is x = -2 and -3

Problem 2: To calculate the function of P (z) = 2z + 4

Solution: The zero of the function is z= -2