Answer:
[tex]\frac{(2z-3)}{(z-3)}[/tex]
Step-by-step explanation:
[tex]\frac{4z^2-16z+15}{2z^2-11z+15}[/tex]
To simplify this we factor the numerator and denominator separately
factor 4z^2 - 16z+15
We apply 'ac' method, 4* 15 = 60
-10 * -6 = 60
-10 - 6 = -16
Now we split -16z using factors -10z -6z
[tex]4z^2 - 10z-6z+15[/tex]
Take out GCf from first two terms and last two terms
[tex]2z(2z-5)-3(2z+5)[/tex]
(2z-3)(2z-5)
factor 2z^2 - 11z+15
We apply 'ac' method, 2* 15 = 30
-5 * -6 = 30
-5 - 6 = -11
Now we split -11z using factors -5z -6z
[tex]2z^2 - 5z-6z+15[/tex]
Take out GCf from first two terms and last two terms
[tex]z(2z-5)-3(2z-5)[/tex]
(z-3)(2z-5)
Now we replace the factors
[tex]\frac{(2z-3)(2z-5)}{(z-3)(2z-5)}[/tex]
Cancel out 2z-5 at the top and bottom
[tex]\frac{(2z-3)}{(z-3)}[/tex]
