Name: Quinn
Who is asking: Student

Level: Secondary Question:

1. Without fully factoring the following show that they all have the same zeros:

     I:  x⁴-x²+2x+6
    II: -x⁴+x²-2x-6
   III: -4x⁴+4x²-8x-24
   IV:  10x⁴-10x²+20x+60

2. When P(x)=x³-3x²+5x+1 and G(x)=x³-2x²-x+10 are each divided by(x-a) the remainders are equal. At what coordinate point does the graph of P(x) intersect G(x)?

Hi Quinn,

1. I organize my thoughts here with a picture. I 'see' the graphs and the question says they have cross the x-axis 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!

2. 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) = (x-a)Q(x) + R1, G(x) = (x-a) H(x) + R2 and R1=R2.

   Check out what happens when you substitute x=a !

Walter