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When the polynomial $p(x)$ is divided by $x - 1,$ the remainder is 3. When the polynomial $p(x)$ is divided by $x - 3,$ the remainder is 5. What is the remainder when the polynomial $p(x)$ is divided by $(x - 1)(x - 3)$?

 Jan 7, 2021
 #1
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Remainder theorem says that:

β€œWhen a polynomial π‘(π‘₯)p(x) is divided by (π‘₯βˆ’π‘Ž)(xβˆ’a), the remainder is π‘(π‘Ž)p(a).”

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

Using the remainder theorem, we can write:

𝑝(1)=3;𝑝(3)=5p(1)=3;p(3)=5

𝑝(π‘₯)p(x) can be written as:

Dividend=(DivisorΓ—Quotient)+RemainderDividend=(DivisorΓ—Quotient)+Remainder

𝑝(π‘₯)=(π‘₯βˆ’1)(π‘₯βˆ’3)𝑄(π‘₯)+π‘Ÿ(π‘₯)p(x)=(xβˆ’1)(xβˆ’3)Q(x)+r(x)

π‘Ÿ(π‘₯)r(x) is the remainder polynomial and π‘„(π‘₯)Q(x) is the Quotient polynomial. Since π‘Ÿ(π‘₯)r(x) is linear, π‘Ÿ(π‘₯)=𝐴π‘₯+𝐡r(x)=Ax+B, where π΄A and π΅B are arbitrary constants.

βŸΉπ‘(π‘₯)=(π‘₯βˆ’1)(π‘₯βˆ’3)𝑄(π‘₯)+𝐴π‘₯+𝐡⟹p(x)=(xβˆ’1)(xβˆ’3)Q(x)+Ax+B

Now,

𝑝(1)=𝐴+𝐡=3⟹𝐴+𝐡=3p(1)=A+B=3⟹A+B=3

𝑝(3)=3𝐴+𝐡=5⟹3𝐴+𝐡=5p(3)=3A+B=5⟹3A+B=5

Solving the 22 equations, we get π΄=1A=1 and π΅=2B=2.

Therefore, π‘Ÿ(π‘₯)=𝐴π‘₯+𝐡=π‘₯+2r(x)=Ax+B=x+2

π‘Ÿ(βˆ’2)=βˆ’2+2=0r(βˆ’2)=βˆ’2+2=0

Hence, π‘Ÿ(βˆ’2)=0

 Jan 7, 2021
 #2
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Sorry are you indicating that you got -2 or 0?

 Jan 7, 2021

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