Find zeros of polynomial functions. Problems with detailled solutions are presented.
Review
Note: In what follows the imaginary unit i is defined as i = square root (-1)
Let p(x) be a polynomial function with real coefficients. If a + ib is an imaginary zero of p(x), the conjuagte a - bi is also a zero of p(x).
TUTORIAL
Example - Problem 1: 2 + i is a zero of polynomial p(x) given below, find all the other zeros.
p(x) = x^{4} - 2·x^{3} - 6·x^{2} + 22·x - 15
Solution to Problem 1:
The zero 2 + i is an imaginary number and p(x) has real coefficients. It follows that the conjugate 2 - i is also a zero of p(x). p(x) may be written in factored form as follows
p(x) = [x - (2 + i)][x - (2 - i)]q(x)
Let us expand the term [x - (2 + i)][x - (2 - i)] in p(x)
[x - (2 + i)][x - (2 - i)] = x^{2} -(2 + i)x -(2 - i)x + (2+i)(2-i)
= x^{2} - 4·x + 5
q(x) can be found by dividing p(x) by x^{2} - 4·x + 5.
(x^{4} - 2·x^{3} - 6·x^{2} + 22·x - 15) / (x^{2} - 4·x + 5)
= x^{2} + 2·x - 3
We now write p(x) in factored form
p(x) = [x - (2 + i)][x - (2 - i)](x^{2} + 2·x - 3)
The remaining 2 zeros of p(x) are the solutions to the quadratic equation.
x^{2} + 2·x - 3 = 0
Factor the above quadratic equation and solve.
(x - 1)·(x + 3) = 0
solutions
x = 1
x = -3
p(x) has the following zeros.
2 + i , 2 - i, -3 and 1.
Matched Problem 1: -3 - i is a zero of polynomial p(x) given below, find all the other zeros.