Harder Polynomial (1 Viewer)
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spice girl
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anyone want the solution to this one yet?
i thought of one using mod-arithmetic...i dun think they teach that for 4u...
i thought of one using mod-arithmetic...i dun think they teach that for 4u...
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Spice Girl Quote:
i thought of one using mod-arithmetic...i dun think they teach that for 4u...
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I could hear Spice Girl chuckling already.
I didn't really want to post a non-4U question, so the following should make it a fair q7 type problem.
a) If a_1+a_2+...+a_n is an odd number, prove that
a_1*b_1+a_2*b_2+...+a_n*b_n is an odd number if each b_i is an odd number.
b)P(x) is a polynomial with integral coefficients. The leading coefficient, the constant term, and P(1) are all odd. Show that P(x) has no rational roots. You can use part a).
Spice Girl Quote:
i thought of one using mod-arithmetic...i dun think they teach that for 4u...
___________________________________________
I could hear Spice Girl chuckling already.
I didn't really want to post a non-4U question, so the following should make it a fair q7 type problem.
a) If a_1+a_2+...+a_n is an odd number, prove that
a_1*b_1+a_2*b_2+...+a_n*b_n is an odd number if each b_i is an odd number.
b)P(x) is a polynomial with integral coefficients. The leading coefficient, the constant term, and P(1) are all odd. Show that P(x) has no rational roots. You can use part a).
underthesun
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excuse my ignorance, but what's integral co-efficients?
last year I wagged most complex number and polynomial classes, and this year I joined the class late..
last year I wagged most complex number and polynomial classes, and this year I joined the class late..
spice girl
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hell yeh it comes out nicely...
suppose x=p/q is a solution
then by rational root thm, since both leading coefficient and constant term is odd, p and q must both be odd (p | a_0, q | a_n)
So P(p/q) = 0
=> a_n*p^n/q^n + a_n-1*p^n-1/q^n-1 + ... + a_1*p/q + a_0 = 0
=> a_n*p^n + a_n-1*p^n-1*q + ... + a_1*p*q^n-1 + a_0*q^n = 0
(notation ~= is "congruent to")
So P(p/q) ~= a_n*(1) + a_n-1*(1) + ... + a_1*(1) + a_0*(1) (mod 2)
(since p, q are both odd)
~= (1)*(a_n + a_n-1 + ... + a_1 + a_0)
~= (1)*P(1)
~= 1
so P(p/q) is an odd integer which is certainly not = 0
thus contradiction.
suppose x=p/q is a solution
then by rational root thm, since both leading coefficient and constant term is odd, p and q must both be odd (p | a_0, q | a_n)
So P(p/q) = 0
=> a_n*p^n/q^n + a_n-1*p^n-1/q^n-1 + ... + a_1*p/q + a_0 = 0
=> a_n*p^n + a_n-1*p^n-1*q + ... + a_1*p*q^n-1 + a_0*q^n = 0
(notation ~= is "congruent to")
So P(p/q) ~= a_n*(1) + a_n-1*(1) + ... + a_1*(1) + a_0*(1) (mod 2)
(since p, q are both odd)
~= (1)*(a_n + a_n-1 + ... + a_1 + a_0)
~= (1)*P(1)
~= 1
so P(p/q) is an odd integer which is certainly not = 0
thus contradiction.
Never thought it will be done this quickly -two solutions at that on an unfair question. If this was a real fair q7, it would have an additional lead up: i) Let x=p/q, a root, where p and q are integers having no common divisor...explain why p divides a_n, and q divides a_0.
Now Affinity :take a leisurely stroll through 2001 Q7 (b).
Now Affinity :take a leisurely stroll through 2001 Q7 (b).
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spice girl: did you learn that stuff before uni? If so was it at all helpful for the hsc?
spice girl
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no usually we're not allowed to introduce too much weird new stuff into our HSC 4u, but i learnt this in yr7 maths enrichment...surprising NSW doesn't teach it at all...