Let f(x) be a polynomial with integer coefficients such that f(5) = f(7) = 20. What is the smallest possible value of |f(0)|?

Guest Jan 5, 2020

#1**+1 **

A bit of a trick question! If f(x) is the constant polynomial 20, then you get |f(0)| = 20, which is the minimum.

Guest Jan 6, 2020

#3**0 **

If it did not have integer coefficients the answer would be 0.

BUT

with integer coefficients I do not know ???

If I had time and inclination I would go watch some videos on Rational Root Theorum and maybe I could figure it out.

BUT

I'd really like someone to show my how to do it here.

Any takers?

Melody Jan 8, 2020

#4**+1 **

The minimum is, I think, 15.

Suppose that \(f(x)=a+bx+cx^{2}+...\)

then \(|f(0)|=a.\)

Substituting x=5 and x=7 gets you

\(20-a=5b+25c+...\)

and

\(20-a=7b+49c+...\)

For integer a, b and c, it's necessary that 20 - a should be divisible by both 5 and by 7.

The smallest, in magnitude, a, is -15.

The quadratic \(f(x)= -15+12x-x^{2}\)

works, I haven't looked for a cubic.

Guest Jan 8, 2020

#6**+2 **

**Let f(x) be a polynomial with integer coefficients such that \(f(5) = f(7) = 20\). What is the smallest possible value of \(|f(0)|\)?**

**a) **

\(f(x) = 20 + (5-x)(7-x) \qquad \Big(~ f(x)=x^2-12x+55 ~\Big)\)

\(\begin{array}{|rcll|} \hline f(0) &=& 20 + (5-0)(7-0) \\ f(0) &=& 20 + 35 \\ f(0) &=& 55 \\ \mathbf{|f(0)|} &=& \mathbf{55} \\ \hline \end{array}\)

**b)**

\(f(x) = 20 - (5-x)(7-x) \qquad \Big(~ f(x)=-x^2+12x-15 ~\Big)\)

\(\begin{array}{|rcll|} \hline f(0) &=& 20 - (5-0)(7-0) \\ f(0) &=& 20 - 35 \\ f(0) &=& -15 \\ \mathbf{|f(0)|} &=& \mathbf{15} \\ \hline \end{array}\)

The smallest possible value of \(|f(0)|\) is **15**

heureka Jan 9, 2020