Part 1:
Let f(x) and g(x) be polynomials.
Suppose f(x)=0 for exactly three values of x: namely, x=-3,4, and 8.
Suppose g(x)=0 for exactly five values of x: namely, x=-5,-3,2,4, and 8.
Is it necessarily true that g(x) is divisible by f(x)? If so, carefully explain why. If not, give an example where g(x) is not divisible by f(x).
Part 2:
Generalize: for arbitrary polynomials f(x) and g(x), what do we need to know about the zeroes (including complex zeroes) of f(x) and g(x) to infer that g(x) is divisible by f(x)?
+9468 Part 1:
If f(x) = 0 when x = -3, 4, or 8 , then f(x) in its factored form is...
f(x) = (x + 3)a(x - 4)b(x - 8)c where a, b, and c are positive integers.
If g(x) = 0 when x = -5, -3, 2, 4, or 8 , then g(x) in its factored form is...
g(x) = (x + 5)d(x + 3)e(x - 2)f(x - 4)g(x - 8)h where d, e, f, g, and h are positive integers.
So for example, g(x) / f(x) could be...
\(\frac{g(x)}{f(x)}=\frac{(x+5)(x+3)(x-2)(x-4)(x-8)}{(x+3)(x-4)(x-8)}\) which reduces to....
\(\frac{g(x)}{f(x)}=(x+5)(x-2)\)
In this case, g(x) is divisible by f(x) with no remainder.
However, g(x) / f(x) could also be...
\(\frac{g(x)}{f(x)}=\frac{(x+5)(x+3)(x-2)(x-4)(x-8)}{(x+3)^2(x-4)(x-8)}\) which reduces to...
\(\frac{g(x)}{f(x)}=\frac{(x+5)(x-2)}{(x+3)}\)
In this case, g(x) is not divisble by f(x) with no remainder.
So it is not necessarily true that g(x) is divisible by f(x) with no remainder.
An example is f(x) = (x + 3)2(x - 4)(x - 8) and g(x) = (x + 5)(x + 3)(x - 2)( x - 4)(x - 8)
Part 2:
For g(x) to be divisible by f(x) , f(x) has to have the same zeros as g(x) and each zero of f(x) has to have a multiplicity less than or equal to that of g(x) .
+9468 Part 1:
If f(x) = 0 when x = -3, 4, or 8 , then f(x) in its factored form is...
f(x) = (x + 3)a(x - 4)b(x - 8)c where a, b, and c are positive integers.
If g(x) = 0 when x = -5, -3, 2, 4, or 8 , then g(x) in its factored form is...
g(x) = (x + 5)d(x + 3)e(x - 2)f(x - 4)g(x - 8)h where d, e, f, g, and h are positive integers.
So for example, g(x) / f(x) could be...
\(\frac{g(x)}{f(x)}=\frac{(x+5)(x+3)(x-2)(x-4)(x-8)}{(x+3)(x-4)(x-8)}\) which reduces to....
\(\frac{g(x)}{f(x)}=(x+5)(x-2)\)
In this case, g(x) is divisible by f(x) with no remainder.
However, g(x) / f(x) could also be...
\(\frac{g(x)}{f(x)}=\frac{(x+5)(x+3)(x-2)(x-4)(x-8)}{(x+3)^2(x-4)(x-8)}\) which reduces to...
\(\frac{g(x)}{f(x)}=\frac{(x+5)(x-2)}{(x+3)}\)
In this case, g(x) is not divisble by f(x) with no remainder.
So it is not necessarily true that g(x) is divisible by f(x) with no remainder.
An example is f(x) = (x + 3)2(x - 4)(x - 8) and g(x) = (x + 5)(x + 3)(x - 2)( x - 4)(x - 8)
Part 2:
For g(x) to be divisible by f(x) , f(x) has to have the same zeros as g(x) and each zero of f(x) has to have a multiplicity less than or equal to that of g(x) .
