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Question Number 135624 by liberty last updated on 14/Mar/21
  A polynomial p(x) leaves a remainder of -1 when divided by x - 1, a remainder of 3 when divided by x - 2, and a remainder of 4 when divided by x + 3. Let r(x) be the remainder when p(x) is divided by (x - 1) (x - 2) (x + 3). What is r(6)?
$$ \\ $$A polynomial p(x) leaves a remainder of -1 when divided by x – 1, a remainder of 3 when divided by x – 2, and a remainder of 4 when divided by x + 3. Let r(x) be the remainder when p(x) is divided by (x – 1) (x – 2) (x + 3). What is r(6)?
Answered by EDWIN88 last updated on 14/Mar/21
Let p(x)=(x−1)(x−2)(x+3)+r(x)  with degree of r(x)is 2  and r(1)=−1 ; r(2)=3 and r(−3)=4  use Remainder Theorem   we get r(x)=(((x−2)(x+3))/((1−2)(1+3))).(−1)+(((x−1)(x+3))/((2−1)(2+3))).(3)  + (((x−1)(x−2))/((−3−1)(−3−2))).(4)  therefore r(6)=(((4)(9))/(−4)).(−1)+(((5)(9))/5).(3)+(((5)(4))/((−4)(−5))).(4)  = 9 + 27 + 4 = 40
$$\mathrm{Let}\:\mathrm{p}\left(\mathrm{x}\right)=\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}+\mathrm{3}\right)+\mathrm{r}\left(\mathrm{x}\right) \\ $$$$\mathrm{with}\:\mathrm{degree}\:\mathrm{of}\:\mathrm{r}\left(\mathrm{x}\right)\mathrm{is}\:\mathrm{2} \\ $$$$\mathrm{and}\:\mathrm{r}\left(\mathrm{1}\right)=−\mathrm{1}\:;\:\mathrm{r}\left(\mathrm{2}\right)=\mathrm{3}\:\mathrm{and}\:\mathrm{r}\left(−\mathrm{3}\right)=\mathrm{4} \\ $$$$\mathrm{use}\:\mathrm{Remainder}\:\mathrm{Theorem}\: \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{r}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}+\mathrm{3}\right)}{\left(\mathrm{1}−\mathrm{2}\right)\left(\mathrm{1}+\mathrm{3}\right)}.\left(−\mathrm{1}\right)+\frac{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)}{\left(\mathrm{2}−\mathrm{1}\right)\left(\mathrm{2}+\mathrm{3}\right)}.\left(\mathrm{3}\right) \\ $$$$+\:\frac{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)}{\left(−\mathrm{3}−\mathrm{1}\right)\left(−\mathrm{3}−\mathrm{2}\right)}.\left(\mathrm{4}\right) \\ $$$$\mathrm{therefore}\:\mathrm{r}\left(\mathrm{6}\right)=\frac{\left(\mathrm{4}\right)\left(\mathrm{9}\right)}{−\mathrm{4}}.\left(−\mathrm{1}\right)+\frac{\left(\mathrm{5}\right)\left(\mathrm{9}\right)}{\mathrm{5}}.\left(\mathrm{3}\right)+\frac{\left(\mathrm{5}\right)\left(\mathrm{4}\right)}{\left(−\mathrm{4}\right)\left(−\mathrm{5}\right)}.\left(\mathrm{4}\right) \\ $$$$=\:\mathrm{9}\:+\:\mathrm{27}\:+\:\mathrm{4}\:=\:\mathrm{40} \\ $$

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