When a polynomial $f(x)$ is divided by $x-5$ or $x-3$ or $x-2$ it leaves a remainder of $1$. Which of the following would be the polynomial?
a. $ x^3 - 10x^2 + 31x + 31$
b. $x^3 - 10x^2 + 31x - 31 $
c. $ x^3 - 10x^2 + 31x - 29$
d. $x^3 - 10x^2 + 31x + 29$
On
Set $f(x)=x^3+bx^2+cx+d$. we know $f(2)=f(3)=f(5)=1$. Hence $$\left\{ \begin{align} & 4b+2c+d=-7 \\ & 9b+3c+d=-26 \\ & 25b+5c+d=-124 \\ \end{align} \right.$$ in other words $$\left( \begin{matrix} 4 & 2 & 1 \\ 9 & 3 & 1 \\ 25 & 5 & 1 \\ \end{matrix} \right)\left( \begin{matrix} b \\ c \\ d \\ \end{matrix} \right)=\left( \begin{matrix} -7 \\ -26 \\ -124 \\ \end{matrix} \right) \\$$ By application of Cramer's rule, we have \begin{align} & b=\frac{\left| \begin{matrix} -7 & 2 & 1 \\ -26 & 3 & 1 \\ -124 & 5 & 1 \\ \end{matrix} \right|}{\left| \begin{matrix} 4 & 2 & 1 \\ 9 & 3 & 1 \\ 25 & 5 & 1 \\ \end{matrix} \right|}=\frac{60}{-6}=-10 \\ \\ & c=\frac{\left| \begin{matrix} 4 & -7 & 1 \\ 9 & -26 & 1 \\ 25 & -124 & 1 \\ \end{matrix} \right|}{\left| \begin{matrix} 4 & 2 & 1 \\ 9 & 3 & 1 \\ 25 & 5 & 1 \\ \end{matrix} \right|}=\frac{-183}{-6}=31 \\ \\ & d=\frac{\left| \begin{matrix} 4 & 2 & -7 \\ 9 & 3 & -26 \\ 25 & 5 & -124 \\ \end{matrix} \right|}{\left| \begin{matrix} 4 & 2 & 1 \\ 9 & 3 & 1 \\ 25 & 5 & 1 \\ \end{matrix} \right|}=\frac{174}{-6}=-29 \end{align}
On
Set $\;g(x)=x^3-10x^2+31x$. As the remainder upon division by $x-\alpha$ is $g(\alpha)$, it is enough to determine whether $$g(5)=g(3)=g(2)= -30,\;32,\; 30\text{ or }-28, \;\text{ respectively}.$$ We'll check this with Horner's scheme: $$\begin{array}{*{5}{r}} &1&-10&3&0\\ \hline &&5&-25&30\\ \hline \times5&1&-5&6&\color{red}{30}\\ \hline\\\hline &&3&-21&30\\ \hline \times3&1&-7&10&\color{red}{30}\\ \hline\\\hline &&2&-16&30\\ \hline \times2&1&-8&15&\color{red}{30}\\ \hline \end{array}$$
Thus $f(x)$ would be polynomial c). Naturally, if we're confident such a polynomial exists in the list, one check will do.
It would be $$ (x-5)(x-3)(x-2)+1=x^3 - 10x^2 + 31x - 29. $$