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Minimizing the number of terms of the square of a polynomial

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Suppose a, b, c, and d belong to the set of nonzero integers.

Let P(x) = (x4 + ax3 + bx2 + cx + d)2.

Determine one of the sets of values of a, b, c, and d, such that when P(x) is

expanded into individual terms of an 8th degree polynomial, that polynomial

will have the fewest number of nonzero terms possible.

Bonus:

Write P(x) in its expanded form.

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Posted · Report post

Pickett,

 

can you come up with a different set with smaller magnitudes (as in absolute values) for a, b, c, and d?

But my first answer still solved the OP...

a=-2

b=-2

c=-4

d=4

or

 

a=2

b=-2

c=4

d=4

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Posted (edited) · Report post

First off, expanded out:

P(x) = x8 + 2ax7 + (a2 + 2b)x6 + (2ab + 2c)x5 + (2ac + b2 + 2d)x4 + (2ad + 2bc)x3 + (2bd + c2)x2 + 2cdx + d2

 

So that means all 9 coefficients are:

  1. (1)
  2. (2a)
  3. (a2 + 2b)
  4. (2ab + 2c)
  5. (2ac + b2 + 2d)
  6. (2ad + 2bc)
  7. (2bd + c2)
  8. (2cd)
  9. (d2)

Since we know a, b, c, and d are non-zero...that means coefficients 1, 2, 8, and 9 can never be 0. Therefore, the theoretical max would be 5 coefficients equaling zero:

  1. (a2 + 2b) = 0
  2. (2ab + 2c) = 0
  3. (2ac + b2 + 2d) = 0
  4. (2ad + 2bc) = 0
  5. (2bd + c2) = 0

By equation 1, we can see that b = -(a2/2)
Then by substitution in equation 2, we can find that c = a3/2
Then using substitution in equations 3, 4, and 5 we find the values of d...which end up being:
By equation 3: d = -(5a4/8)
By equations 4 and 5: d = a4/4

Since there are no situations where -(5a4/8) = a4/4 are true (except when a = 0, which we know it doesn't), we have to choose the one that gives us more valid solutions, which is that d = a4/4...which means that we can have a maximum of 4 coefficients that equal 0 in this expanded equation (which means the minimum non-zero would be 5):

  • a = x
  • b = -(x2/2)
  • c = x3/2
  • d = x4/4

We now need to find values of x that give us integer values...so an obvious answer would simply be x=4 and/or x=-4, which gives us these two solutions (resulting in 4 zero coefficients, aka 5 non-zero terms):

SOLUTION 1:
a = -4
b = -8
c = -32
d = 64

SOLUTION 2:

a = 4
b = -8
c = 32
d = 64

Edited by Pickett
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Posted · Report post

Pickett,

 

can you come up with a different set with smaller magnitudes (as in absolute values) for a, b, c, and d?

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