asked 22.5k views
0 votes
Find all the zeros of the function
g(x)=x^3-4x^2-x+22

asked
User Skamradt
by
8.1k points

1 Answer

5 votes

Answer:

x = -2, x = 3 − i√8, and x = 3 + i√8

Explanation:

g(x) = x³ − 4x² − x + 22

This is a cubic equation, so it must have either 1 or 3 real roots.

Using rational root theorem, we can check if any of those real roots are rational. Possible rational roots are ±1, ±2, ±11, and ±22.

g(-1) = 18

g(1) = 18

g(-2) = 0

g(2) = 12

g(-11) = 1782

g(11) = 858

g(-22) = -12540

g(22) = 8712

We know -2 is a root. The other two roots are irrational. To find them, we must find the other factor of g(x). We can do this using long division, or we can factor using grouping.

g(x) = x³ − 4x² − 12x + 11x + 22

g(x) = x (x² − 4x − 12) + 11 (x + 2)

g(x) = x (x − 6) (x + 2) + 11 (x + 2)

g(x) = (x (x − 6) + 11) (x + 2)

g(x) = (x² − 6x + 11) (x + 2)

x² − 6x + 11 = 0

Quadratic formula:

x = [ 6 ± √(36 − 4(1)(11)) ] / 2

x = (6 ± 2i√8) / 2

x = 3 ± i√8

The three roots are x = -2, x = 3 − i√8, and x = 3 + i√8.

answered
User Yan Bellavance
by
8.1k points

No related questions found

Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.