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# Polynomial

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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 8. What is the remainder when the polynomial p(x) is divided by (x - 1)(x - 3)?

Jan 7, 2022

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+118117
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$$p(x)=(x-1)(x-3)Q(x)+ax+b\\ \text{So the remainder when p(x) is divided by }(x-1)(x-3) \;\;is\;\; ax+b\\~\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{ax+b}{x-1}\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{ax-a+b+a}{x-1}\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+a+\frac{b+a}{x-1}\\ \qquad \text {But we already know that the remainder is 3 so}\\ \qquad a+b=3\qquad (1)\\~\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{ax+b}{x-3}\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+a+\frac{b+3a}{x-3}\\ \qquad \text {But we already know that the remainder is 8 so}\\ \qquad b+3a=8\qquad \\ \qquad b+a+2a=8\qquad \\ \qquad 3+2a=8\qquad \\ \qquad a=2.5 \qquad \\~\\ \qquad b=3-a\\ \qquad b=3-2.5\\\qquad b=0.5 \qquad \\~\\$$

So when  p(x) is divided by (x-1)(x-3)  the remainder is   2.5x+0.5

I have not checked that this is correct,   lets see

$$p(x)=(x-1)(x-3)Q(x)+2.5x+0.5\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{2.5x+0.5}{x-1}\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{2.5x-2.5+3}{x-1}\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+2.5+\frac{3}{x-1}\qquad Cool!\\~\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{2.5x+0.5}{x-3}\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{2.5x-7.5+0.5+7.5}{x-3}\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{2.5(x-3)+0.5+7.5}{x-3}\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+2.5+\frac{8}{x-3}\qquad Double\;\;Cool!\\$$

LaTex:

p(x)=(x-1)(x-3)Q(x)+ax+b\\
\text{So the remainder when p(x) is divided by }(x-1)(x-3) \;\;is\;\; ax+b\\~\\
\frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{ax+b}{x-1}\\
\frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{ax-a+b+a}{x-1}\\
\frac{p(x)}{(x-1)}=(x-3)Q(x)+a+\frac{b+a}{x-1}\\
\qquad \text {But we already know that the remainder is 3 so}\\

\frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{ax+b}{x-3}\\
\frac{p(x)}{(x-3)}=(x-1)Q(x)+a+\frac{b+3a}{x-3}\\
\qquad \text {But we already know that the remainder is 8 so}\\

p(x)=(x-1)(x-3)Q(x)+2.5x+0.5\\
\frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{2.5x+0.5}{x-1}\\
\frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{2.5x-2.5+3}{x-1}\\