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