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# Help on hard division problem

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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)?

Guest Mar 29, 2018

#1
+7339
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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) .

hectictar  Mar 29, 2018
#1
+7339
+2

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) .

hectictar  Mar 29, 2018