Choose 5 coefficients, a can't = 0, so that, ax^4+bx^3+cx^2+dx+e has 4 distinct, real roots?

Really? Wow.

You don't get it do you?

A polynomial can be decomposed to factors (A1+B1x)(A2+B2x)(A3.....

Where at some value of x, Ai = -Bix and that term equals ZERO and so the whole polynomial equals zero.

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Putting it another way: a polynomial of degree n can be reduced to n terms (Ai+Bix) where i ranges from 1 to n.

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so how hard can it be to choose values for Ai and Bi that satisfy the two equations?

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here is an example: choose 3 coefficients a, b and c so that ax²+bx+c has two distinct real roots, then choose them so that the same form has one root of multiplicity two

OK first pick whatever numbers you want, lets keep them positive

Lets just pick the primes, but as I said I could pick ANY integers (any numbers actually)

(2+3x)(5+7x) ← see the first 4 primes?

do the math: 10 +15x + 14x +21x² combine 10 + 29x + 21x² then rearrance to get the form right

21x² + 29x + 10 and the coefficients are a =21, b = 29 and c = 10. HOw is that hard?

for an quadratic equation with one root you can pick ANY two numbers

say the next primes which are 11 and 13. so (11+13x)(11 + 13x) and I'll leave you to do the math to figure out what a, b, and c are. (hint c = 121)

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Of course the same thing needs to be done with a fourth order polynomial except, of course, you will have four terms (8 numbers) instead of 4. By the way: if you choose 1 for all the terms Bi, the math gets simpler. (eg (1+x)(2+x)(3+x)(4+x) )

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