Question

ಅಧಿಕ ವೆಚ್ಚವಾಗುವ ಸಂಘರ್ಷ ವಿಧಾನ............​

Answer 1

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

GIVEN :

$\sf{p(x)= 6 {x}^{4} + 8 {x}^{3} + 17 {x}^{2} + 21x + 7}$

$\sf{g(x) = 3 {x}^{2} + 4x + 1}$

TO FIND :

Value of 'a' and 'b'

SOLUTION :

Division refer to the attachment.

After division we get x + 2 as remainder.

But According to question, remainder is given as ax + b

Therefore,

=> x + 2 = ax + b

By comparing LHS and RHS we get

=> x = ax and 2 = b

=> a = 1 and b = 2.

$\boxed{ \sf{Hence, a = 1 \: and \: b = 2}}$