Pale Maths (Polynomials and divisibility)
a)P(x)=a_nx^n+a_n-1x^(n-1)+...+a1x+a0 with integral coeff and r,s are 2 relatively prime integers. Suppose a0 and an are odd
If r is not 0 and s/r is a root of P(x)=0, show both r,s are both odd.----------------b)let a,b,c be integers and f(x) =15x^4+ax^3+bx^2+cx^+35-----------------------
suppose f(x)=0 has rational root h/k , h,k are 2 relativly prime integers
(i)Use (a), show f(h) is even
(ii)let u,v be 2 distinct integers, show f(u)-f(v) is divisible by u-v (I can do it)
(iii) use a, b(i), b(ii) show f(1) is even
(iv)use b(ii), show if f(1) is odd, then k is not 1
why????????? ( although h is odd)
ah3 + bh2 + ch is even , i dont no what are a, ,b ,c
- ?Lv 51 decade agoFavorite Answer
an(s/r)n + an-1(s/r)n-1 + ... + a0 = 0
ansn + an-1sn-1r + ... + a0rn = 0
ansn = -(an-1sn-1r + ... + a0rn)
Since r divides the R.H.S, it also divides L.H.S. By since r, s are relatively prime, r divides an. Since an is odd, r must be odd.
a0rn = -(ansn + an-1sn-1r + ... + a1s )
so s divides a0, and hence must be odd.
(b) (i) Since the leading coeff. and the constant term of f(x) are both odd, by (a), both h and k must be odd.
Since 15(h/k)4 + a(h/k)3 + b(h/k)2 + c(h/k) + 35 = 0
15h4 + ah3k + bh2k2 + chk3 + 35k4 = 0
ah3k + bh2k2 + chk3 = -15h4 -35k4 is even.
Write k=2t+1, then
ah3k + bh2k2 + chk3
= ah3(2t+1) + bh2(2t+1)2 + ch(2t+1)3
= ah3 + bh2 + ch + 2(ah3t + bh2(2t2+2t) + ch(4t3+6t2+3t))
Since ah3k + bh2k2 + chk3 and 2(ah3t + bh2(2t2+2t) + ch(4t3+6t2+3t)) are even,
ah3 + bh2 + ch is even. Therefore
f(h) = 15h4 + ah3 + bh2 + ch + 35 is even.
(ii) as you can do it, I will skip this part.
(iii) f(h)-f(1) is divisible by h-1, which is even. Hence f(h)-f(1) is even.
by (i) f(h) is even. Hence f(1) is also even.
(iv) f(k)-f(h) is divisible by k-h. Since both k and h are odd by part (a), k-h is even. Since f(h) is even by (i), f(k) is even. Hence if f(1) is odd, k is not 1.