Question

Find the remainder when
${4y}^{3} + {4y}^{2} - y - 1$
is divided by 2y+1

Answer 1

Here is your answer it may help...

Answer 2

Hola..!!

For zeros,
$2y + 1 = 0 \\ = > 2y = - 1 \\ = > y = \frac{ - 1}{2}$
Let
$p(y) = 4 {y}^{3} + 4 {x}^{2} - y - 1$
Now,
$p( \frac{ - 1}{2} ) = 4 \times { (\frac{ - 1}{2} )}^{3} + 4 \times {( \frac{ - 1}{2}) }^{2} - ( \frac{ - 1}{2} ) - 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4 \times (\frac{ - 1}{8} ) + 4 \times \frac{1}{4} + \frac{1}{2} - 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 1}{2} + 1 + \frac{1}{2} - 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 1 + 2 + 1 - 2}{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 0$
Therefore, Remainder = 0

Hope it'll help.