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Test of Mathematics Solution Subjective 175 - Integer Roots

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 175 (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. & C.M.I. Entrance Course of Cheenta

Problem

 Let \text{P(x)}=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_{1}x+a_{0} be a polynomial with integer coefficients,such that,\text{P(0) and P(1)}  are odd integers.Show that

(a)\text{P(x)} does not have any even integer roots. 

(b)\text{P(x)} does not have any odd integer roots.


Solution

Given the two statements (a) and (b)  above it is clear that if we can prove that \text{P(x)} has no integer roots,then we are done.

 Let us assume \text{P(x)} has an integer root \text{ a}.

Then we can write,

    \[ \text{P(x)=(x-a)Q(x)}\dots(1)\]

where,\text{Q}is any function of \text{ x}.

Now ,putting \text{x=0,1} in \text{ 1},we get

    \[\text{P(0)=(-a)Q(0)}\dots(2)\]

and,

    \[\text{P(1)=(1-a)Q(1)}\dots(3)\]

now as \text{(1-a)} and \ \text{(-a)},are consecutive integers \text{(2)} and \text{(3)}

cannot be both odd,

which means that \text{P(0),P(1)} cannot be both odd,given whatever \text{Q(0) and Q(1)} are.

So,it is a contradiction!!.

So,we can conclude that there exists no integer solution of \text{P(x)}.

Hence,we are done.

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 175 (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. & C.M.I. Entrance Course of Cheenta

Problem

 Let \text{P(x)}=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_{1}x+a_{0} be a polynomial with integer coefficients,such that,\text{P(0) and P(1)}  are odd integers.Show that

(a)\text{P(x)} does not have any even integer roots. 

(b)\text{P(x)} does not have any odd integer roots.


Solution

Given the two statements (a) and (b)  above it is clear that if we can prove that \text{P(x)} has no integer roots,then we are done.

 Let us assume \text{P(x)} has an integer root \text{ a}.

Then we can write,

    \[ \text{P(x)=(x-a)Q(x)}\dots(1)\]

where,\text{Q}is any function of \text{ x}.

Now ,putting \text{x=0,1} in \text{ 1},we get

    \[\text{P(0)=(-a)Q(0)}\dots(2)\]

and,

    \[\text{P(1)=(1-a)Q(1)}\dots(3)\]

now as \text{(1-a)} and \ \text{(-a)},are consecutive integers \text{(2)} and \text{(3)}

cannot be both odd,

which means that \text{P(0),P(1)} cannot be both odd,given whatever \text{Q(0) and Q(1)} are.

So,it is a contradiction!!.

So,we can conclude that there exists no integer solution of \text{P(x)}.

Hence,we are done.

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