For this case we have the following rational expression:
[tex] \frac{5x^2 + 26x + 24}{3x^2 + 10x - 8} [/tex]
What we must do for this case, is to factor the numerator and the denominator.
We have then:
[tex] \frac{(5x+6)(x+4)}{(3x-2)(x+4)} [/tex]
Then, canceling similar terms we have:
[tex] \frac{(5x+6)}{(3x-2)} [/tex]
Then, the function is not defined for the values that make the denominator zero.
We have then:
[tex]3x-2 = 0
3x = 2
[/tex]
[tex]x = \frac{2}{3} [/tex]
And on the other hand, a zero of the function is:
[tex]5x+6 = 0
5x = -6 [/tex]
[tex]x = \frac{-6}{5} [/tex]
Answer:
a zero of the function is:
[tex]x = \frac{-6}{5} [/tex]
the function is not defined for the value:
[tex]x = \frac{2}{3} [/tex]