HELP with my calculus summer project.I need a fourth degree polynomial that has four real roots...ASAP thnx?

Question

this is an extremely long project so i will almost definitely be asking more questions in the future...your help with this one will be most appreciated!!

OH!! Also, if you can give me these other polynomials i would appreciate it!

fifth degree polynomials that have 2,3, and 5 real roots (one of each)

Answer 1

Oh, come on. You know that a polynomial can be expressed as a serie of monomes multiplied together, don't you?

(x-a) * (x-b) * (x-c) * (x-d)
will have the real roots a, b, c and d
Now, substitude a, b, c and d for the values you want, multiply averything as you will have a 4th degree polynomial with 4 real roots.

For the 5th order one, do the same thing as for the 4th order, and to reduce the number of roots, use a parabolic where the discriminant is negative, i.e. a parabolic with no real roots. That takes care of dropping the number of roots by 2. To have a5th order polynomial with 2 roots, you need one root to actually be a double, that means you have

(x-a) * (x-a)

with only has a double root "a".

Answer 2

Easy way to generate a polynomial is to remember that if one of the roots is substituted for the independent variable the value of the polynomial goes to 0.

In general

P(x) = (x-r1)*(x-r2)*(x-r3).. (x-rn)

will give you an n'th degree polynomial with roots r1, r2, r3.....rn.

Example:

(x-3)*(x-5)*(x+2) = x^3 - 6x^2 - x + 30

is a 3'rd degree polynomial with roots of 3, 5, and -2 (as can be easily checked by substituting 3, 5, and -2 for x in the polynomial and doing the arithmetic)

If you need repeated roots, just repeat the term(s)

(x-4)*(x+5)*(x+5) = x^3 + 6x^2 - 15x -100

is a 3'rd order polynomial with a root at 4 and a repeated root (of order 2) at -5.

This also works for functions with a complex variable and complex coefficients and roots. (If you want complex functions with real coefficients and real or complex roots, it gets *much* deeper. But you won't have to worry about that for a couple of years )

And when you're doing the algebraic multiplications, keep close track of the terms (ask me how I know )


Doug

Answer 3

(x - a)(x - b)(x - c)(x - d) will give you a 4th degree polynomial for all distinct real numbers a, b, c, d.

If you want to have a 5th degree polynomial with less than five roots, you can use the same technique only using complex numbers. For instance, here is an example of a 2nd degree polynomial that has no real roots:

(x - i)(x + i) = x^2 - i^2 = x^2 + 1. This polynomial obviously has no real roots.

Answer 4

This is real easy to do. Just decide what you want the roots to be then multiply as follows. If the roots are r1, r2, r3, and r4. Then multiply (x-r1)(x-r2)(x-r3)(x-r4) this will give you a 4th order polynomial with those 4 roots. For example if all four roots are 1 (just to make my life easier) then the polynomial is

(x -1)(x-1)(x-1)(x-1) or x^4 - 4x^3 + 6x^2 - 4x + 1

Answer 5

5 real roots
say a b c d e
(x-a)(x-b)(x-c)(x-d)(x-e) = 0
put any value in a b c d e and get the result

3 real roots
say a b c
(x-a)(x-b)(x-c)(x^2+1) =0

expand

it cannot have 2 real roots then it shall have 3 complex roots. but complex roots appear as conjugates. I presume that you expect real coefficient.

for 4th oder with 4 real roots
(x-a)(x-b)(x-c)(x-d) = 0 and expand
for example 1 2 3 4
you should get(x-1)(x-2)(x-3)(x-4) = 0
or expanding

(x-1)(x-2)(x^2-7x+12) =0

and keep multiplying

Answer 6

You have to have a function that has two roots that are equivalent but different in sign (+) or (-). So, as specified you start with x to the fourth, and since you simply need two parabolas ( imagine two x^2 functions side by side, you have four roots), add x^2.

Answer 7

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Answer 8

(x^2 - 5)(x^2 - 9)(x - 2) = x^5 - 2x^4 - 14x^3 + 28x^2 + 45x - 90