Given g(x) = (x + 5)(x2 - 6x + 13), find the one real zero of g(x):
I think it's -5
Find the two complex zeros of g(x):
*(x + 5)(x^2 - 6x + 13)
correct for real root x = -5
for complex:
x^2 - 6x + 13 = 0
x =(6 ± √-16)/2
= (6 ± 4i)/2
= 3 ± 2i
To find the real zero of g(x), we need to set g(x) equal to zero and solve for x:
g(x) = 0
(x + 5)(x^2 - 6x + 13) = 0
To find the real zero, we set each factor equal to zero:
x + 5 = 0 => x = -5
So, you are correct, the real zero of g(x) is -5.
Now, to find the complex zeros of g(x), we need to solve the quadratic equation x^2 - 6x + 13 = 0.
The discriminant (b^2 - 4ac) of this equation is (-6)^2 - 4(1)(13) = 36 - 52 = -16, which is negative. Since the discriminant is negative, the quadratic equation has no real roots, but instead has two complex roots.
The complex roots can be found using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our quadratic equation x^2 - 6x + 13 = 0, the values are:
a = 1, b = -6, c = 13
Substituting the values into the quadratic formula, we get:
x = (-(-6) ± √((-6)^2 - 4(1)(13))) / (2(1))
= (6 ± √(36 - 52)) / 2
= (6 ± √(-16)) / 2
Since √(-16) = 4i (where i is the imaginary unit), we can simplify further:
x = (6 ± 4i) / 2
= 3 ± 2i
Therefore, the two complex zeros of g(x) are 3 + 2i and 3 - 2i.