Identity in x x

Algebra Level 3

If the following is an identity in x x for some polynomial f f , what is the value of b − a b-a : ( x − 1 ) ( x 2 − 6 ) f ( x ) = x 4 + a x 2 + b ? (x-1)(x^2-6)f(x)=x^4+ax^2+b?

7 13 9 11

Ajay Dutta by Ajay Dutta , 24, India

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3 solutions

Nikhil Kashyap
Mar 16, 2014

Since, it is an identity in x.

So, the given equation satisfies every value of x.

So, putting x=1 and x= 6 \sqrt { 6 }

We get 1 + a + b = 0 36 + 6 a + b = 0 1+a+b=0\\ 36+6a+b=0\\

So, a = − 7 , b = 6 b − a = 13 a=-7,b=6\\ b-a=\boxed { 13 }

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Moshiur Mission
Mar 30, 2014

x^4+ax^2+b if two root of x^2 is α,β then α+β=-a and αβ=b again if x is identical for (x-1)(x^2-6)f(x)= x^4+ax^2+b or (x-1)(x^2-6)(cx+d)= x^4+ax^2+b solving we get c=1, d=1 and the equation is x^4-7x^2+6 thus a =-7, b=6 hence b-a=13

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Ankush Gogoi
Mar 15, 2014

let f be px+q as it must be a polynomial of degree 1.. now (x-1)(x^2-6)(px+q)=px^4+(q-p)x^3-(6p+q)x^2+(6p-6q)x+6q=x^4+ax^2+b....comparing the co-efficients,we get p=1,q=1....giving b=6,a=-7....hence b-a=13

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It is no where given that f is a polynomial of degree 1.

Nikhil Kashyap - 7 years, 2 months ago

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but it shoud be as what you give is an identity so if you bring everything in the one place and equal them with 0 this should be valid

Kon Tim - 7 years, 2 months ago

Yes ! But, if u analyze, that's (degree 1) is the only possibility for poly.f

Suman Kumar - 7 years, 2 months ago

Note that since it is an identity, both the sides must be equivalent and for that, the degree of the expanded polynomials on both sides must be the same. The degree of the monic polynomial on R.H.S is 4 4 while the one on L.H.S is a product of two polynomials: one monic of degree 3 3 and the other ( f ( x ) ) (f(x)) of unknown degree.

If d e g ( f ( x ) ) ≠ 1 deg(f(x))\neq 1 , we can never have the the degree of R.H.S polynomial = 4 =4 since we'll never get a x 4 x^4 term for any other degree of f ( x ) f(x) . So, d e g ( f ( x ) ) = 1 deg(f(x))=1 . Also, if g ( x ) g(x) isn't a monic polynomial, the sides cannot be equivalent. So, f ( x ) f(x) has to be monic. From these conclusions, we can imply that f ( x ) f(x) has the following form:

f ( x ) = x + q , for some constant q f(x)=x+q~,\quad \textrm{for some constant }q

Prasun Biswas - 6 years, 3 months ago

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