Question

divide the following
$12z {}^{2} + 3z {}^{2} + 4z + 1 \div 4z + 1$
​

Answer 1

$4z + 1 = 0 \\ \\ 4z = - 1 \\ \\ z = \frac{ - 1}{4}$

$= 12 {z}^{2} + 3 {z}^{2} + 4z + 1 \\ \\ = 12 {( \frac{ - 1}{4}) }^{2} + 3 {( \frac{ - 1}{4} )}^{2} + 4( \frac{ - 1}{4} )+ 1 \\ \\ = 12( \frac{1}{16} ) + 3( \frac{1}{16} ) - 1 + 1 \\ \\ = \frac{12}{16} + \frac{3}{16} \\ \\ = \frac{15}{16}$