Question

What is the simplified form of the quantity 4 z squared minus 16z plus 15 over the quantity 2 z squared minus 11 z plus 15?

Answer 1

Factor the numerator and denominator to get
(2z-3)(2z-5) / (z-3)(2z-5)
Since there is a common factor of (2z -5) on the numerator and denominator this factor will cancel leaving (2z-3)/(z-3)

Answer 2

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

$\frac{(2z-3)}{(z-3)}$