Answer:
\(\frac{(2z-3)}{(z-3)}\)
Step-by-step explanation:
\(\frac{4z^2-16z+15}{2z^2-11z+15}\)
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
\(4z^2 - 10z-6z+15\)
Take out GCf from first two terms and last two terms
\(2z(2z-5)-3(2z+5)\)
(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
\(2z^2 - 5z-6z+15\)
Take out GCf from first two terms and last two terms
\(z(2z-5)-3(2z-5)\)
(z-3)(2z-5)
Now we replace the factors
\(\frac{(2z-3)(2z-5)}{(z-3)(2z-5)}\)
Cancel out 2z-5 at the top and bottom
\(\frac{(2z-3)}{(z-3)}\)
