How do you multiply $\frac{4 z^{2} + 24 z}{3 z^{2} - 9 z - 12} \cdot \frac{z^{2} - 4 z - 5}{z - 4}$ $\frac { 4z ^ { 2} + 24z } { 3z ^ { 2} - 9z - 12} \cdot \frac { z ^ { 2} - 4z - 5} { z - 4}$ ?

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Answer 1 · 2016-12-22T13:46:49

Perhaps some one else can spot how to simplify this further.

$\frac{4 z (z^{2} + z - 30)}{3 (z - 4) (z - 4)}$ $(4z(z^2+z-30))/(3(z-4)(z-4))$

Just playing with formats to start with. Looking for things to cancel out.

$\frac{4 z (z + 6)}{3 (z^{2} - 3 z - 4)} \times \frac{(z + 1) (z - 5)}{z - 4}$ $(4z(z+6))/(3(z^2-3z-4))xx((z+1)(z-5))/(z-4)$

$\frac{4 z (z + 6)}{3 (z + 1) (z - 4)} \times \frac{(z + 1) (z - 5)}{z - 4}$ $(4z(z+6))/(3(z+1)(z-4))xx((z+1)(z-5))/(z-4)$

$(4z(z+6))/(3cancel((z+1))(z-4))xx(cancel((z+1))(z-5))/(z-4)$

$(4z)/3xx((z+6)(z-5))/((z-4)(z-4)$

$(4z)/3xx(z^2+z-30)/((z-4)(z-4)$

$(4z(z^2+z-30))/(3(z-4)(z-4)$

Answer 2 · 2016-12-22T18:17:26

$(4z(z+6)(z-5))/(3(z-4)(z-4))$

$(4z^2+24z)/(3z^2-9z-12)*(z^2-4z-5)/(z-4)$

$(4z(z+6))/(3(z^2-3z-4))*((z+1)(z-5))/(z-4)$

$(4z)/3*((z+6))/((z+1)(z-4))*((z+1)(z-5))/((z-4))$

$(4z(z+6)(z-5))/(3(z-4)(z-4))$

How do you multiply \frac { 4z ^ { 2} + 24z } { 3z ^ { 2} - 9z - 12} \cdot \frac { z ^ { 2} - 4z - 5} { z - 4} 4z2+24z3z2−9z−12⋅z2−4z−5z−4\frac { 4z ^ { 2} + 24z } { 3z ^ { 2} - 9z - 12} \cdot \frac { z ^ { 2} - 4z - 5} { z - 4} ?