IIT JEE , AIEEE Forum - Mathematics , Algebra

himanshu2006

Q14:- If g(x) & h(x) are polynomials such that g(x^2) & x^2h(x^2) is divisible by x^2+x+1, then ( here

is the imaginary
cube root of unity).
a) g(

) =h(

)=0 b) g(

)

h(

) c) g(

) = -h(

) d) g(

)+ h(

)

0
(ans:-a)

krish

Let g(x^2) = f(x) (x^2+x+1) and x^2 h(x^2) = k(x) (x^2+x+1)
Putting x=w^2
g(w^4) = f(w^2) (w^4+w^2+1) = 0 = g(w) (w^4 = w)
w^2 h(w^4) = k(w^2) (w^4+w^2+1) = 0 = h(w) since , w is not equal to 0
Therefore , g(w) = h(w) = 0
plz reply if u find anything wrong

gorakavipraveen

Its easy, infact
look g(x^2)=p[x^2+x+1] and g(w^4)=g(w)=p*0=0
and for h(x^2)*x^2=q[x^2+x+1] and substitue w^2 for x to get the result.

I mea omega for w friend. I think you understood. SO ANSWERE IS A .

vinu

Hi himanshu2006,
x^2+x+1=(x-w)(x-w^2)
As g(x^2) , h(x^2) 've w,w^2 as their factors;
{ (w^2)^2 =w} ,
g(w) =h(w) =0.

puneet

hiiiiiii

There have beeen correct answers to this query already .. however lemme make things a bit more clear and neat ...

well ... now we know x² + x + 1 divides g(x²) and x²h(x²)

Also x²+ x + 1 is (x-

)(x-

²)

So this means that

and

² must make g(x²) and x²h(x²) zero ...

Now, g(

²) = g(

⁴) = 0 ..... (1)

and,

²h(

²) =

^4.h(

⁴) = 0 ...... (2)

Now

⁴=

And thus from (1) we get g(

) = 0

And from (2) we get

.h(

) = 0 or since

is not zero

we have h(

) = 0

So, the correct answer is (a)

cheers

And so we get

kuldeep

puneet sir can u please give ur e-mail id.

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