Question

Anyone can solve it. ​

Answer 1

Answer:

❤️ Hey mate ❤️

Check attachment....

Answer 2

$\huge{\pink{\green {\sf\boxed{ANSWER, R1=7 and\:R2\: is\: also \:7}}}}$

$\huge{\pink {\sf\boxed{Step\: by\: step \:explanations}}}$

Let polynomial $P(x)={ax} ^{3}+{3x}^{2}-{13} \:\:\:\:\:\:\:\:\: {and}$

${q} ={2x}^{3}-{5x}+{a}\:\:\:\:\:\:is\: divided\: by \:(x-2)$

So, ${x-2}=0$

${x=2}$

Putting (x=2) in above polynomial.

$P(x) ={ax} ^{3}+{3x}^{2}-{13}$

$P(2)=a{(2)}^{3}+3{(2)}^{2}-{13}$

$\:\:"=8a + 12 - 13$

$\:\:" =8a - 1$

$\bold{\pink {\sf\underline {Hence ,R1 = 8a - 1}}}\:\:\:\:\:\:\:\:[R = Remainder]$

To find R2,

$q(x) = {2x}^{3}-{5x}+{a}$

$q(2)=2{(2)}^{3} - 5 (2) + a$

$\:\:"=16 - 10 + a$

$\:\:" =6 + a$

$\bold {\pink {\sf\underline{And, R2 = 6 + a}}}$

$\small\boxed{A.T.Q ==) }$

$\:\:\:\:\:\:\:{R1\:\:\:\:=\:\:\:\:R2}$

$\:\:\:\:\:\:\:{8a - 1\:\:\:\:=\:\:\:\:6 + a}$

$\:\:\:\:\:\:\:{7a\:\:\:\:=\:\:\:\:7}$

$\:\:\:\:\:\:\:{a\:\:\:\:=\:\:\:\:\frac{7}{7}}$

$\:\:\:\:\:\:\:{a\:\:\:\:=\:\:\:\:1}$

$\Large\boxed{Hence, }$

$\bold {\pink{\green {\sf{\underline{R1= 8a - 1 = 8*1 - 1} ={\boxed {7}}}}}}$

$\bold {\pink{\green {\sf{\underline{R2= 6 + a = 6 + 1} ={\boxed {7}}}}}}$

$\huge{\pink {\sf\boxed{Solved}}}$