Help is appreciated

Remainder theorem says that:

“When a polynomial 𝑝(𝑥)p(x) is divided by (𝑥−𝑎)(x−a), the remainder is 𝑝(𝑎)p(a).”

Also, when a polynomial 𝑝(𝑥)p(x) is divided by another polynomial 𝑞(𝑥)q(x),the degree of the remainder is at most 11 less than the degree of 𝑞(𝑥)q(x).

Using the remainder theorem, we can write:

𝑝(1)=3;𝑝(3)=5p(1)=3;p(3)=5

𝑝(𝑥)p(x) can be written as:

Dividend=(Divisor×Quotient)+RemainderDividend=(Divisor×Quotient)+Remainder

𝑝(𝑥)=(𝑥−1)(𝑥−3)𝑄(𝑥)+𝑟(𝑥)p(x)=(x−1)(x−3)Q(x)+r(x)

𝑟(𝑥)r(x) is the remainder polynomial and 𝑄(𝑥)Q(x) is the Quotient polynomial. Since 𝑟(𝑥)r(x) is linear, 𝑟(𝑥)=𝐴𝑥+𝐵r(x)=Ax+B, where 𝐴A and 𝐵B are arbitrary constants.

⟹𝑝(𝑥)=(𝑥−1)(𝑥−3)𝑄(𝑥)+𝐴𝑥+𝐵⟹p(x)=(x−1)(x−3)Q(x)+Ax+B

Now,

𝑝(1)=𝐴+𝐵=3⟹𝐴+𝐵=3p(1)=A+B=3⟹A+B=3

𝑝(3)=3𝐴+𝐵=5⟹3𝐴+𝐵=5p(3)=3A+B=5⟹3A+B=5

Solving the 22 equations, we get 𝐴=1A=1 and 𝐵=2B=2.

Therefore, 𝑟(𝑥)=𝐴𝑥+𝐵=𝑥+2r(x)=Ax+B=x+2

𝑟(−2)=−2+2=0r(−2)=−2+2=0

Hence, 𝑟(−2)=0

Sorry are you indicating that you got -2 or 0?