Question

The function f(x)=x^3-7x^2+19x-18 has at least one integer root. Use the integer root theorem to find that root, then proceed to find all complex roots. (Note: roots may be integer, rational, irrational, and/or complex.)

Answer 1

The real root of the function f(x) = x³ - 7·x² + 19·x is; x = 2

The complex root of the function are; x = 2.5 + √(44)·i and x = 2.5 - √(44)i

The steps used to find the roots of the function, f are as follows;

The function can be presented as follows; f(x) = x³ - 7·x² + 19·x - 18

The rational roots theorem indicates that we get;

p/q = ±1, ±2, ±3, ±6, ±9, ±18

When x = 2, we get;

f(2) = 2³ - 7·2² + 19·2 - 18

2³ - 7·2² + 19·2 - 18 = 0

Therefore, x = 2 is a root of the equation, and we get;

(x - 2) is a factor of the function f(x) = x³ - 7·x² + 19·x - 18

Dividing x³ - 7·x² + 19·x - 18 by (x - 2), we get;

x² - 5·x + 9

(x - 2) |x³ - 7·x² + 19·x - 18

x³ - 2·x²

-5·x² + 19·x - 18

-5·x² + 10·x

9·x - 18

9·x - 18

0

The factors of f(x) = x³ - 7·x² + 19·x - 18 are therefore; (x - 2) and (x² - 5·x + 9)

The roots of the expression x² - 5·x + 9 are;

x = (-(-5) ± √((-5)² - 4× 1 × 9))/(2×1)

x = (5 ± √(-11))/2

The complex roots are; x = 2.5 + √(44)·i and x = 2.5 - √(44)·i