Question

plz solve this question hint also given in attachment

Answer 1

$\huge{\pink{\underline{\ulcorner{\star\:Solution\: \star}\urcorner}}}$

$\huge{\underline{\green{PART \:-\:1}}}$

$Polynomial \: given \: \colon \\ f(x) = x^{4} -2x^{3}+3x^{2}-ax+b$

Now,

We have to given that
remainder is given for two different cases :

$\red{\underline{Case \:1}}$

Remainder = 5
Division by = (x-1)

Zero,
x-1 = 0
x = +1

$f(+1) = x^{4} -2x^{3}+3x^{2}-ax+b\\ 5= (1)^{4} -2(1)^{3}+3(1)^{2}-a(1)+b\\5= 1-2+3-a+b\\ 5 = 2-a+b\\3=-a+b$

$\red{\underline{Case \:2}}$

Zero,
x+1 = 0
x = -1

$f(-1) = x^{4} -2x^{3}+3x^{2}-ax+b\\ 19= (-1)^{4} -2(-1)^{3}+3(-1)^{2}-a(-1)+b\\19= 1-2(-1)+3+a+b\\ 19= 1+2+3+a+b\\19=6+a+b\\19-6 =a+b \\a+b = 13$

-a+b = 3..(1) and a+b =13..(2)

now using substitution,

b=3+a...(3)

put value in 2 equation,

a+(3+a) = 13
2a = 13-3
2a = 16
$a= \frac{10}{2} \\a \implies 5\\ b= 3+a\\\\ b \implies 3+5 \\a \implies 8$

$\huge{\underline{\green{PART \:-\:2}}}$

Attempt second :

Remainder when equation is divided by (x-2):

Now , we put the value in equation :

a = 5 and b = 8

By remainder theorem

x-2 = 0
x = 2

$f(x) = x^{4} -2x^{3}+3x^{2}-5x+8//f(2) =2^{4} -2(2)^{3}+3(2)^{2}-5(2)+8\\f(2) = 16-2(8)+3(4)-5(2)+8\\f(2) = \cancel{16}-\cancel{16}+12-10+8\\f(2) = 12-10+8\\f(2)= 2+8\\f(2) = 10\\ Now, \:we \:get\: a\: remainder\: of\: 10\:when\: divided\: by \:(x-2)$