Polynomial with rational roots

$x^3+13=ax-a$

$x^3+13=a(x-1)$

$a=\frac{x^3+13}{x-1}=\frac{(x-1)(x^2+x+1)+14}{x-1} = x^2+x+1 +\frac{14}{x-1}$

since $a$ and $x$ are both positive integers,

$x-1$ must be a positive divisor of $14$

$x-1=1, 2, 7, 14$

$x=2, 3, 8, 15$

Totaling up,

$a=\boxed {358}$

Rearrange the equation to solve for $a$ as follows: $x^3+13 = a(x-1) \leadsto a = \dfrac{x^3+13}{x-1} = x^2+x+1+\dfrac{14}{x-1}$ .

If $x$ is an integer, then $a$ is an integer if and only if $x-1$ divides $14$ .

Since $x$ is positive, $x-1 = 1$ , $2$ , $7$ , $14$ . Thus, $x = 2$ , $3$ , $8$ , $15$ .

These values of $x$ yield $a = 21$ , $20$ , $75$ , $242$ respectively.

The sum of these values of $a$ is $21+20+75+242 = \boxed{358}$ .

$x^3+13 = a(x-1) \implies (x^3-1)+14 = a(x-1)$ Since $a$ and $x-1$ are both integers and since $x^3-1$ is divisible by $x-1$ means that $x-1|14$ . Since we want positive integer solutions only , thus ,possible values of $x$ are $2,3,8,15$ . Since $a= \frac{x^3+13}{x-1}$ gives possible values of , which are $a = 21 , 20 , 75 , 242$ . adding them gives sum of possible values of $a$ which is $\boxed{358}$

$x^{3}$ -ax+a+13=0

a= $\frac{x^3+13}{x-1}$

a= $\frac{x^3-1+14}{x-1}$

a= $\frac{x^3-1}{x-1}$ + $\frac{14}{x-1}$

a= $\frac{(x-1)(x^2+x+1)}{x-1}$ + $\frac{14}{x-1}$

a= $(x^2+x+1)$ + $\frac{14}{x-1}$

By looking at the expression $(x^2+x+1)$ , we notice that this expression is always an integer for x integer. Therefore, (x-1) must be a positive divisor of 14 because "a" has to be an integer and x is positive. So, we have:

x-1=1 =>x=2 and a=21

x-1=2 =>x=3 and a=20

x-1=7 =>x=8 and a=75

x-1=14 =>x=15 and a=242

The sum of all integer values of a for which the equation has at least one positive solution for x is:

21+20+75+242=358

Solving for $a$ , we get that $a=\dfrac{x^3+13}{x-1}$ We further reduce this by noticing that $x^2(x-1)=x^3-x^2$ , $a=x^2+\dfrac{x^2+13}{x-1}$ Similarly, $a=x^2+x+\dfrac{x+13}{x-1}$ $a=x^2+x+1+\dfrac{14}{x-1}$ Now, since $x$ is a positive integer, if $a$ is to be integer, then $x-1|14$ . We can work this out for the different possibilities. $x=2:\ a=21$ $x=3:\ a=20$ $x=8:\ a=75$ $x=15:\ a=242$ The answer is $21+20+75+242=\boxed{358}$ .

Since $x$ needs to be an integer, we have by the Rational root Theorem that $x | (a + 13)$ So we can write $a + 13 = kx$ , where $k$ is an integer. $\\$ Substituting $a + 13 = kx$ and $a = kx - 13$ in the original equation, we get $x^3 - (kx-13)\cdot x + kx = 0$ or, since $x \neq 0$ , $x^2 - kx + k + 13 = 0$ Solving for $k$ , we get $k = \frac{x^2+13}{x-1} = x + 1 + \frac{14}{x-1}$ Since $k$ and $x$ are both integers, and $x > 0$ , x can assume only 4 different values: $15, 8, 3$ and $2$ . $\\$ The corresponding values of $k$ are $17, 11, 11$ and $17$ , and the corresponding values for $a$ are $242, 75, 20$ and $21$ . $\\$ Finally we have that the sum over all possible values of $a$ is equal to $sum = 242 + 75 + 20 + 21 = 358$

The given equation can be rearranged as

$a = \frac{x^3 +13}{x -1} ......... (1)$

$\Rightarrow a = \frac{x^3 -1 +14}{x - 1} = x^2 + x+1 + \frac{14}{x- 1}$

Now for $a$ to be an integer $x -1$ must divide 14.

Since $x$ must be a positive integer , We can have

$x-1 = 1 \Rightarrow x =2$

$x-1 = 2 \Rightarrow x =3$

$x-1 =7 \Rightarrow x =8$

$x-1 =14 \Rightarrow x =15$

Substituting these values of $x$ in $(1)$ the values of $a$ are

$21 , 20 , 75$ and $242$ which sum up to $358.$

Obviously, x=1 cannot be the root of the given equation.

From the given equation, we have: $a=\frac{x^3+13}{x-1}=x^2+x+1+\frac{14}{x-1}$ .

Because a is integer and x is positive integer, there are 4 possible values of $x-1$ : 14,7,2,1.

The values of a are: 242,75,20,21 respectively.

The answer is: $242+75+20+21=358$ .

Using the given polynomial x^3 - ax + a + 13 = 0 find a with respect to x. a = (x^3 +13)/(x-1)= x^2 + x + 1 + 14/(x-1) Since x is a positive integer therefore x^2 + x + 1 will be an integer. Also a is an integer. For a to be integer 14/(x-1) shoul be an integer. Therefore possible values of x are 2,3,8,15. By using these values of x we can find the values of a : 20,21,75,242. Adding these values we get the answer. 20 + 21 + 75 + 242 = 358

Solving for $a$ gives $a = \frac{x^3+13}{x-1}=\frac{x^3-1}{x-1} + \frac{14}{x-1}$

Since $x-1$ is a factor of $x^3-1$ , and $a, x$ are positive integers, $x-1$ is a factor of $14$ .

This gives 4 solutions: $x \in \{2, 3, 8, 15\}$ .

Evaluating the expression with each of these 4 solutions gives $a\in \{21, 20, 75, 242\}$ , and the sum of these values of $a$ is $\boxed{358}$ .