Find integers $a,b,c$ where $p(x) = x(x-a)(x-b)-13$ where $p(x)$ is divisible by $ (x-c)$

Question

Given integers $a,b,c$ such that $0<a<b$ if

$$ p(x)=x(x-a)(x-b)-13 $$

where $p(x)$ is divisible $(x-c)$

Find $a,b,c$

Hint: Consider what it means for the quantity $(x-c)$ to be a linear factor of $p(x)$

Hint: Consider whether your solution is unique or not if $13$ was not prime number.

I've been trying to solve this problem since a month. It is 'Concepts Connection Challenge' from a MOOC about pre-calculus. My last approach:

if $x=0$ then $p(x)=-13$ , as a result $c=13$

That would mean that $x(x-a)(x-b) = x$, hence $(x-a)(x-b)=1$ but no integer solutions apply for this equation.

Thanks for guiding this humble learner

Answer 1

Since $0<a< b$, we have $$c-b < c-a<c$$

Since $$c(c-a)(c-b)=13,$$

which is a prime number, it cannot be written as product of $3$ distinct positive numbers. Neither can it written as a product of $3$ distinct negative numbers or $2$ positive numbers and $1$ negative number.

Hence two of the numbers are negative and distinct.

Hence $c-b=-13$, $c-a=-1$, $c=1$.

That is $b=14, a=2, c=1$.

Remark about your approach:

By setting $x=0$, you are trying to find the $y$-intercept of the polynomial.

Divisibility by $x-c$ implies $c$ is an $x$-intercept of the polynomial.