Question

the polynomial p(x) =x^4-2x^3x-ax+b when divided by (x-1) and (x+1) leaves the remainder 5and 9 respectively . find the values of a and b . hence, find the remaider when p(x) is divided by (x-3)​

Answer 1

$given \\ \\ p(x) = {x}^{4} - 2 {x}^{3} - ax + b \\ \\ also \: p(1) = 5 \\ \\ {1}^{4} - 2 ({1}^{3}) - a(1) + b= 5 \\ \\ b - a = = 6 \\ \\ and \: \: p( - 1) = 9 \\ \\ {( - 1)}^{4} - 2 {( - 1)}^{3} - a( - 1) + b = 9 \\ \\ a + b = 6 \\ \\ = > a = 0 \\ \\ b = 6 \\ \\ now \\ \\ p(x) = {x}^{4} - 2 {x}^{3} + 6 \\ \\ p(3) = {3}^{4} - 2 {(3)}^{3} + 6 \\ \\ p(3) = 33$

hope it helps.

Answer 2

ANSWER

We have,

p(x)=x^4–2x^3+3x^2-ax+b

By remainder theorem, when p(x) is divided by (x-1) and (x+1) , the remainders are equal to p(1) and p(-1) respectively.

By the given condition, we have

p(1)=5 and p(-1)=19

=> (1)^4–2(1)^3+3(1)^2-a(1)+b=5 and (-1)^4–2(-1)^3+3(-1)^2-a(-1)+b=19

=> 1–2+3-a+b=5 and 1-(-2)+3+a+b=19

=> -a+b=5–1+2–3 and 1+2+3+a+b=19

=> -a+b=3 and a+b=19–1–2–3

=> -a+b=3 and a+b=13

Adding these two equations,we get

-a+b+a+b=3+13

=> 2b=16

=> 2b/2=16/2

=> b=8

Putting b=8 in a+b=13 , we get

a+8=13

=> a=13–8

=> a=5

Therefore, a=5 and b=8 .