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