3 years ago

Help me, I will give the brainlyest to the smartest and the fastest!

Mathematics

1 answer:

lora16 [44]3 years ago

4 0

Answer:

{(1,4) (2,8) (3,12) (4,16)}

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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?

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Hello,

If $X=x^2$ it becomes

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

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So,

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

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d = 17

e = 0

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