find the zeroes of g(x)=(x^2-5x-6)/(x^2-9)
I know i suppose to use synthetic division, but the denominator has x^2 in it. Does this matter or do I just divide by -9
to divide by a polynomial with power other than 1 you have to use long divison.
However for this problem that's not necessary... just factor
G(x) = (x-2)(x-3)/(x-3)(x+3)
thus your only zero is x = 2 since the function is not defined at x = 3
{1,2,3}
To find the zeroes of the function g(x)=(x^2-5x-6)/(x^2-9), we need to set the numerator equal to zero and solve for x.
First, let's factorize the numerator and the denominator separately:
Numerator: x^2 - 5x - 6 = (x - 6)(x + 1)
Denominator: x^2 - 9 = (x - 3)(x + 3)
Now, substitute the factored forms back into the function:
g(x) = [(x - 6)(x + 1)] / [(x - 3)(x + 3)]
To find the zeroes, we would set the numerator (x - 6)(x + 1) equal to zero:
(x - 6)(x + 1) = 0
Now solve for x:
x - 6 = 0 --> x = 6
x + 1 = 0 --> x = -1
Therefore, the zeroes of the function g(x) are x = 6 and x = -1.