Question

If a,b,c,d, and e are non negative such that x^6-289x^4-x^2+289=(x-a)(x+b)(x_c)(x+d)(x^2+ex+1), what is the value of a+b+c+d+e?

Answer 1

Answer:

36

Step-by-step explanation:

Hello,

If $X=x^2$ it becomes

$X^3-289X^2-X+289=(X^2-1)(X-289)$

and $289=17^2$

$X^2-1=(X-1)(X+1)=(x^2-1)(x^2+1)=(x+1)(x-1)(x^2+1) \\\\x^2-289=x^2-17^2=(x-17)(x+17)$

So,

$x^6-289x^4-x^2+289=(x^2-1)(x^2+1)(x^2-289)=(x-1)(x+1)(x+17)(x-17)(x^2+1)$

a = 1

b = 1

c = 17

d = 17

e = 0

Then a + b + c + d + e = 36

Thanks