Remainder theorem says that:
βWhen a polynomial π(π₯)p(x) is divided by (π₯βπ)(xβa), the remainder is π(π)p(a).β
Also, when a polynomial π(π₯)p(x) is divided by another polynomial π(π₯)q(x),the degree of the remainder is at most 11 less than the degree of π(π₯)q(x).
Using the remainder theorem, we can write:
π(1)=3;π(3)=5p(1)=3;p(3)=5
π(π₯)p(x) can be written as:
Dividend=(DivisorΓQuotient)+RemainderDividend=(DivisorΓQuotient)+Remainder
π(π₯)=(π₯β1)(π₯β3)π(π₯)+π(π₯)p(x)=(xβ1)(xβ3)Q(x)+r(x)
π(π₯)r(x) is the remainder polynomial and π(π₯)Q(x) is the Quotient polynomial. Since π(π₯)r(x) is linear, π(π₯)=π΄π₯+π΅r(x)=Ax+B, where π΄A and π΅B are arbitrary constants.
βΉπ(π₯)=(π₯β1)(π₯β3)π(π₯)+π΄π₯+π΅βΉp(x)=(xβ1)(xβ3)Q(x)+Ax+B
Now,
π(1)=π΄+π΅=3βΉπ΄+π΅=3p(1)=A+B=3βΉA+B=3
π(3)=3π΄+π΅=5βΉ3π΄+π΅=5p(3)=3A+B=5βΉ3A+B=5
Solving the 22 equations, we get π΄=1A=1 and π΅=2B=2.
Therefore, π(π₯)=π΄π₯+π΅=π₯+2r(x)=Ax+B=x+2
π(β2)=β2+2=0r(β2)=β2+2=0
Hence, π(β2)=0