Question

$\textsf{\huge{\underline{QUESTION}}}}$

$\textsf{Find an integer k such that :}\\\\\mathsf{x^2-x+k \:\:divides\:\: x^{13}+x+90}$


DON'T DO BY TRIAL AND ERROR METHOD PLZZZ !

Answer 1

Find all integers n for which x^2 - x + n divides x^13 + x + 90.

 

Solution

Answer: n = 2.

If n is negative or zero, then the quadratic has two real roots. But we can easily check that the other polynomial has derivative everywhere positive and hence only one real root. So n must be positive.

If x^2 - x + n divides x^13 + x + 90, then x^13 + x + 90 = p(x) (x^2 - x + n), where p(x) is a polynomial with integer coefficients. Putting x = 0, we see that n must divide 90. Putting x = 1, we see that it must divide 92. Hence it must divide (92 - 90) - 2. So the only possibilities are 1 and 2. Suppose n = 1. Then putting x = 2, we have that 3 divides 2^13 + 92. But 2^odd is congruent to 2 mod 2, so 2^13 + 92 is congruent to 1 mod 3. So n cannot be 1.

To see that n = 2 is possible, we write explicitly: (x^2 - x + 2) (x^11 + x^10 - x^9 - 3 x^8 - x^7 + 5 x^6 + 7 x^5 - 3 x^4 - 17 x^3 - 11 x^2 + 23 x + 45) = x^13 + x + 90.

(I had used n in place of K)

Answer 2

Answer

here=>

(x^13 + x + 90)/(x2-x+k) = p

Where p is an integer.

Now,

When we put x =0

We get =>

90/k = p1

-where p1 is an integer.

When we put x =1

We get=>

1+1+90/k = p2

92/k = p2

-Where p2 is an integer.

Hence k is an common factor of 90 and 92.

Hence we get k =2