
Subject: polynomial functions
Name: Quinn
Who is asking: Student
Level: Secondary
Question:
 Without fully factoring the following show that they all have the same zeros:
I: x^{4}x^{2}+2x+6
II: x^{4}+x^{2}2x6
III: 4x^{4}+4x^{2}8x24
IV: 10x^{4}10x^{2}+20x+60
 When P(x)=x^{3}3x^{2}+5x+1 and G(x)=x^{3}2x^{2}x+10 are each divided by(xa) the remainders are equal. At what coordinate point does the graph of P(x) intersect G(x)?
Hi Quinn,
 I organize my thoughts here with a picture.
I 'see' the graphs and the question says they have cross the xaxis at
the same points. One way this could happen would be if they were
multiples of one another:
f(a) = 0 if and only if c f(x) = 0 (c not 0).
Now look again at these expressions, I, II, III, IV.
II = (1) I.
III = (4) I
etc. No factoring of anything. In fact, this principle does apply
even if the expression does not factor. There is a big fancy theory
of higher algebra, about polynomials, that says something of this
sort  provided you check real and complex solutions to f(a) = 0.
If you want to impress you teacher, refer to Hilbert's Nullstellensatz!
 Again, I see a picture. They are asking for (at least) one point where
the graphs cross. Let that point be c .
At that point, P(c) = G(c).
Is it possible that they have told you, in disguise, that
P(a) = G(a)? [After all, there is not much else to work with.]
Well yes they have. The division stuff says:
P(x) = (xa)Q(x) + R1, G(x) = (xa) H(x) + R2 and R1=R2.
Check out what happens when you substitute x=a !
Walter
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