Question

When polynomial x⁴+ax³+bx²+cx+d is divided by each x-4,x-3,x+3,x+4 the remainder in each case is 102.Find the sum of all possible values of x for whichf(x)=246?​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

let f(x)= x^4 +ax^3 + bx^2 +cx + d

If the above polynomial is a perfect square then it can be represented as

f(x)= (x-p)^2*(x-q)^2;

where p and q are real numbers

So f(x)=0 at x=p and x=q.

Now let us take the derivative of f(x). On simplification we get

f’(x)= 2(x-p)(x-q)(2x-p-q)

for condition of extrema of f(x),

f’(x)=0 at x=p; x=q; x=(p+q)/2.

so we observe that at x=p and at x=q f(x)=0 and f’(x)=0.

therefore at x=p and at x=q f(x) has repeated roots.

so roots of f(x) are p,p,q,q.

now from f(x)= x^4 +ax^3 + bx^2 +cx + d,

we get sum of roots=-a. so 2(p+q)=-a …(1)

similarly p^2 +q^2+4pq =b …(2)

2(p^2)q + 2p(q^2) =-c

and (p^2)(q^2) =d

Now solving these equations you can get the relationship between a,b,c,d.

Now choosing suitable values of any two co-efficient , the values of other two coefficients can be found out and the polynomial can be constructed.

(Tried my best to keep my handwriting good…Sorry if it still sucks…)

Hope it helps…

Let it be =(x2+px+q)2=x4+2px3+(p+2q)x2+2pqx+q2

This requires

a=2p

b=p2+2q

c=2pq and

d=q2 .Algebraic simplification leads to

c=a(4b−a2)8d=(4b−a2)264