Solving Polynomials

Algebra Level 4

Suppose that $P(x)$ is a polynomial of degree 4 with integer coefficients, and there exists distinct integers $a, b, c, d$ such that

$P(a) = P(b) = P(c) = P(d) = 5$

How many integers $n$ are there so that $P(n) = 10$ ?

0 1 2 3 Depends on the leading coefficient Depends on the constant term

1 solution

Saswata Naha
Nov 20, 2016

Assume a new polynomial F(x) as---- F(x)=P(x)-5 clearly, F(a)=F(b)=F(c)=F(d)=0

Therefore.,F(x)=u.(x-a)(x-b)(x-c)(x-d)

Now, when P(x)=10, then, F(x)=5 let us assume, a solution x'

F(x)=u(x'-a)(x'-b)(x'-c)(x'-d)=5

we know , 5=(-1)(-5)(1)

Therefore, we can write 5 as at most 3 factors. so ,equation will not be satisfied by any solution. Therefore, no. of solution =0

$5=(1)(-1)(5)(1)(-1)$

What a about this ????

Kushal Bose - 4 years, 6 months ago

a,b,c,d is distinct. sorry! I miss the fact

Saswata Naha - 4 years, 6 months ago

The solution is essentailly correct. Can you clean it up further to make it easier to read / understand, esp for those who cannot solve it? Thanks!

Calvin Lin Staff - 4 years, 6 months ago