An algebra problem by K. J. W.

Algebra Level 5

$\large x^5+ax^4+bx^3+cx^2+dx-210=0$

Let all the five distinct roots of the equation above be positive integers. Find the value of $b+d-a-c$ .

The answer is 941.

1 solution

K. J. W.
Jul 18, 2014

Let us call the five distinct roots e, f, g, h and i.

By Vieta's Formula,

-efghi=-210

i.e. efghi=210

Note that e, f, g, h and i must be distinct.

Also note that 210=1x2x3x5x7 and there is no way to further factorise it without repeating 1 or using negative roots.

Therefore, e, f, g, h, i=1, 2, 3, 5, 7 (in no particular order).

Expanding (x-1)(x-2)(x-3)(x-5)(x-7)=0, we get x^5-18(x^4)+118(x^3)-348(x^2)+457x-210=0.

Therefore we have b+d-a-c=118+457+18+348 =941.

Why must the roots be distinct? Why can't we have roots of 1, 1, 1, 1, 210?

Calvin Lin Staff - 6 years, 11 months ago

Because there are 5 roots

K. J. W. - 6 years, 10 months ago

Note that repeated roots are counted with multiplicity, unless explicitly stated. This way, we can say that polynomials of degree $n$ always have $n$ roots (over the complex numbers), without having to care about repeated factors.

Calvin Lin Staff - 6 years, 10 months ago

More detailed explanation: If one or more of the roots are the same, then similar roots count as one. For example, if the solutions are 1,1,1,1,210, then there are only two roots, 1 and 210. However, I will still change the words to "five distinct roots".

K. J. W. - 6 years, 10 months ago

It says that all roots are distinct

Figel Ilham - 6 years, 10 months ago

I changed it afterwards

K. J. W. - 6 years, 10 months ago

An algebra problem by K. J. W.

The answer is 941.

1 solution

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