An algebra problem by João Vitor

Algebra Level 3

Let $p(x)$ be a monic polynomial of ${ 4 }^\text{th}$ degree with non-negative integer coefficients, such that $p(0)=1007$ and $p(1)=2016$ . How many $p(x)$ do exist if $x=-1$ is a root of that polynomial?

Note: A polynomial is said to be monic if its leading coefficient (the coefficient of the term of greatest degree) is equal to 1.

The answer is 1009.

1 solution

Rishabh Jain
Dec 23, 2016

Knowing $(1).$ $p(x)$ has $-1$ as one of its zeroes, $(2).$ leading coefficient as $1$ and $(3).f(0)$ (Constant term) as $1007$ ,

$p(x):(x+1)(x^3+ax^2+bx+1007)$

Using $f(1)=2016$ $\implies a+b=0\implies a=-b$ . Using this relation(substituting for a) and expanding $p(x)$ ,

$p(x):x^4+(1-b)x^3+(b+1007)x+1007$

Given this equation has non negative integral coefficients , we get $1-b\ge 0$ and $b+1007\ge 0$ $\implies -1007\le b\le 1$ i.e $\boxed{1009}$ values of $b$ and corresponding $p(x)$ for each of those values of $b$ .

An algebra problem by João Vitor

The answer is 1009.

1 solution

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