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# Help is appreciated

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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When a polynomial π(π₯)p(x) is divided by (π₯βπ)(xβa), 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=(DivisorΓQuotient)+RemainderDividend=(DivisorΓQuotient)+Remainder

π(π₯)=(π₯β1)(π₯β3)π(π₯)+π(π₯)p(x)=(xβ1)(xβ3)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.

βΉπ(π₯)=(π₯β1)(π₯β3)π(π₯)+π΄π₯+π΅βΉp(x)=(xβ1)(xβ3)Q(x)+Ax+B

Now,

π(1)=π΄+π΅=3βΉπ΄+π΅=3p(1)=A+B=3βΉA+B=3

π(3)=3π΄+π΅=5βΉ3π΄+π΅=5p(3)=3A+B=5βΉ3A+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(β2)=β2+2=0r(β2)=β2+2=0

Hence, π(β2)=0

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

Jan 7, 2021