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# Test of Mathematics Solution Subjective 42- Polynomial with Integer Coefficients

This is a Test of Mathematics Solution of Subjective 42 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also Visit: I.S.I & CMI Entrance Course of Cheenta

## Problem

Let f(x) be a polynomial with integer coefficients. Suppose that there exist distinct integers $a_1 , a_2 , a_3 , a_4$ , such that $f( a_1 ) = f( a_2 ) = f( a_3 ) = f( a_4 ) = 3$ . Show that there does not exist any integer b with f(b) = 14.

## Solution:

Consider the auxiliary polynomial g(x) = f(x) - 3. Clearly, according to the problem,  g(x) has four distinct integer roots $a_1 , a_2 , a_3 , a_4$. Hence we may write $g(x) = (x - a_1)(x- a_2)(x- a_3)(x- a_4) Q(x)$ where Q(x) is an integer coefficient polynomial. (Since by factor theorem if 't' is a root of a polynomial f(x) then (x-t) is it's factor).

Suppose there exists an integer b such that f(b) = 14; then g(b) = 14-3 =11. Hence $g(b) = 11 = (b- a_1 )(b- a_2 )(b- a_3 )(b- a_4 ) Q(b)$. $a_1 , a_2 , a_3 , a_4$ are distinct so are $(b- a_1 ) , (b- a_2 ) , (b- a_3 ) , (b- a_4 )$. Therefore the equation $g(b) = 11 = (b- a_1 ) (b- a_2 )(b- a_3 )(b- a_4 ) Q(b)$ indicates that 11 can be written as the product of at least 4 different integers which is impossible (as 11 is a prime). Indeed even if we consider -1, 1 and 11 , we have only three integers whose product is 11.

Thus we have a contradiction and proving that there does not exist an integer b such that f(b) 14.

Key Idea: Factor Theorem

This is a Test of Mathematics Solution of Subjective 42 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also Visit: I.S.I & CMI Entrance Course of Cheenta

## Problem

Let f(x) be a polynomial with integer coefficients. Suppose that there exist distinct integers $a_1 , a_2 , a_3 , a_4$ , such that $f( a_1 ) = f( a_2 ) = f( a_3 ) = f( a_4 ) = 3$ . Show that there does not exist any integer b with f(b) = 14.

## Solution:

Consider the auxiliary polynomial g(x) = f(x) - 3. Clearly, according to the problem,  g(x) has four distinct integer roots $a_1 , a_2 , a_3 , a_4$. Hence we may write $g(x) = (x - a_1)(x- a_2)(x- a_3)(x- a_4) Q(x)$ where Q(x) is an integer coefficient polynomial. (Since by factor theorem if 't' is a root of a polynomial f(x) then (x-t) is it's factor).

Suppose there exists an integer b such that f(b) = 14; then g(b) = 14-3 =11. Hence $g(b) = 11 = (b- a_1 )(b- a_2 )(b- a_3 )(b- a_4 ) Q(b)$. $a_1 , a_2 , a_3 , a_4$ are distinct so are $(b- a_1 ) , (b- a_2 ) , (b- a_3 ) , (b- a_4 )$. Therefore the equation $g(b) = 11 = (b- a_1 ) (b- a_2 )(b- a_3 )(b- a_4 ) Q(b)$ indicates that 11 can be written as the product of at least 4 different integers which is impossible (as 11 is a prime). Indeed even if we consider -1, 1 and 11 , we have only three integers whose product is 11.

Thus we have a contradiction and proving that there does not exist an integer b such that f(b) 14.

Key Idea: Factor Theorem