[tex]\frac{4z^2-16z+15}{2z^2-11z+15}[/tex] Factor the numerator and the denominator
[ax² + bx + c]
4z² - 16z + 15 Since a > 1, multiply a and c together, then find the factors of (a·c), that adds or subtracts to = b.
(a·c) --> (4 · 15) = 60
Factors of 60:
1 · 60, 2 · 30, 3 · 30, 4 · 15, 5 · 12, 6 · 10
[The only factor is 6 and 10, (-6) + (-10) = -16] So you substitute -6z - 10z for -16z
4z² - 6z - 10z + 15 Factor out 2z from 4z² - 6z, and factor out -5 from -10z + 15
2z(2z - 3) - 5(2z - 3) Factor out (2z - 3) and your left with:
(2z - 3)(2z - 5)
Do the same for the denominator, and you should get (z - 3)(2z - 5)
Now you have:
[tex]\frac{(2z - 3)(2z-5)}{(z-3)(2z - 5)}[/tex] You can cancel out (2z - 5)
[tex]\frac{2z - 3}{z-3}[/tex]
ANSWER
[tex]\frac{(2z-3)}{(z-3)} [/tex]
EXPLANATION
The given expression is:
[tex] \frac{4z^2-16z+15}{2z^2-11z+15} [/tex]
Let us split the middle terms to get,
[tex]\frac{4z^2-6z - 10z+15}{2z^2-6z - 5z+15} [/tex]
Factor by grouping:
[tex]\frac{2z(2z-3) - 5(2z - 3)}{2z(z-3) - 5(z - 3)} [/tex]
Factor further:
[tex]\frac{(2z - 5)(2z-3)}{(2z - 5)(z-3)} [/tex]
We cancel the common factors,
[tex]\frac{(2z-3)}{(z-3)} [/tex]
where
[tex]z \ne \frac{5}{2} \: or \: z \ne3[/tex]
