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# Prove r divides b . . .?

Let r (an integer) be a nonzero solution of x^2 + ax + b =0 where a,b are integers

Prove r divides b

Not sure how to get to this solution...

### 3 Answers

- Anonymous8 years agoFavorite Answer
r solves x^2 + ax + b = 0

=> r*r + a*r = -b

=> (r+a)*r = -b

Since r,a are integers, -b is an integer multiple of r, and so is b, hence r divides b.

- 8 years ago
quadratic functions can be factorized, for example

x^2 + 3x +2 = 0, this can be factorized to

(x + 2)(x +1) = 0

and as we can see both 1 and 2 are solutions to this and both 1 and 2 divide 2,

so if we take x^2 + ax +b

you've already been given that r is a solution, so lets assume that some constant c (an integer) is also a solution, so we get

(x + r) ( x + c) = x^2 + ax +b = 0

as we can see from this r * c = b

then c = b/r

hence r divides b.

hope this helps.

- orometuaLv 48 years ago
y = (x-r)(x-s), so b=r*s and a=-(r+s). This shows that s is also an integer. Best of all, r | r*s so r|b.