Question

Let r1 and r2 be the remainders when the polynomial x^3-6x^2+2x-kand kx^3+12x^2+14x-3 are divided by 1-2x and 2x+1respectively if r1 -r2 =25/8 ,find the value of k.

Answer 1

for calculating r1 put the value of x=1/2
${ \frac{1}{2} }^{3} - 6 \times { \frac{1}{2} }^{2} + 2 \times \frac{1}{2} - k \\ \frac{1}{8} - \frac{3}{2} + 2 - k$

$( \frac{1 - 12 + 16}{8} ) - k \\ \frac{5}{8} - k = r1$

for second equation put the value x=-1/2
$\frac{ - 1}{8}k + 12 \times \frac{1}{4} - \frac{14}{2} - 3 \\ \frac{ - k}{8} + 3 - 7 - 3$
$\frac{ - k}{8} - 7 = r2$
given that
$r1 - r2 = \frac{25}{8}$
$\frac{5}{8} - k + \frac{k}{8} + 7 = \frac{25}{8} \\ ( \frac{5 - 8k + k + 56}{8} ) = \frac{25}{8}$
$- 7k = 25 - 61 \\ - 7k = - 36 \\ k = \frac{36}{7} \: \: \: \: ans$

Answer 2

Step-by-step explanation:

for calculating r1 put the value of x=1/2

\begin{gathered} { \frac{1}{2} }^{3} - 6 \times { \frac{1}{2} }^{2} + 2 \times \frac{1}{2} - k \\ \frac{1}{8} - \frac{3}{2} + 2 - k \\ \end{gathered}213−6×212+2×21−k81−23+2−k

\begin{gathered}( \frac{1 - 12 + 16}{8} ) - k \\ \frac{5}{8} - k = r1\end{gathered}(81−12+16)−k85−k=r1

for second equation put the value x=-1/2

\begin{gathered} \frac{ - 1}{8}k + 12 \times \frac{1}{4} - \frac{14}{2} - 3 \\ \frac{ - k}{8} + 3 - 7 - 3 \\ \end{gathered}8−1k+12×41−214−38−k+3−7−3

\frac{ - k}{8} - 7 = r28−k−7=r2

given that

r1 - r2 = \frac{25}{8}r1−r2=825

\begin{gathered} \frac{5}{8} - k + \frac{k}{8} + 7 = \frac{25}{8} \\ ( \frac{5 - 8k + k + 56}{8} ) = \frac{25}{8} \end{gathered}85−k+8k+7=825(85−8k+k+56)=825

\begin{gathered} - 7k = 25 - 61 \\ - 7k = - 36 \\ k = \frac{36}{7} \: \: \: \: ans\end{gathered}−7k=25−61−7k=−36k=736ans