Asked by meme
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
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
Answers
Answered by
David
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
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
Answered by
nikkiroseledesma
{1,2,3}
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