Posted by **Anon** on Friday, May 24, 2013 at 5:49am.

Create a 3rd degree polynomial with real coefficients that has roots -1 and 4i. Write your answer in form ax^3 + bx^2 + cx+ d .

- Math -
**Bosnian**, Friday, May 24, 2013 at 10:01am
The general equation for the function is:

f( x ) = a ( x - x1 ) ( x - x2 )( x - x3 )

where x1, x2 and x3 are the roots and a is the coefficient of x ^ 3

There is a theorem in algebra called the Imaginary Zeros Theorem.

It says that for a polynomial that is restricted to real coefficients then for any imaginary zero (root) that exists then its conjugate also exists.

This means that the roots you gave :

-1 and ( 4 + i ), also includes the congugate root:

( 4 - i )

This is the conjugate to ( 4 + i )

Since there are no other restrictions to the polynomial you can construct the roots in factored form like this:

f ( x ) = a [ x - ( - 1 ) ] ( x - 4 i ) ( x + 4 i ) =

a ( x + 1 ) ( x - 4 i ) ( x + 4 i ) =

a ( x + 1 ) ( x * x - x * 4 i + 4 i * x + 4 i * ( - 4 i ) =

a ( x + 1 ) ( x ^ 2 - 4 i x + 4 i x - 16 i ^ 2 ) =

a ( x + 1 ) [ x ^ 2 - 4 i x + 4 i x - 16 * ( - 1 ) ] =

a ( x + 1 ) ( x ^ 2 + 16 ) =

a ( x ^ 2 * x + x ^ 2 * 1 + 16 * x + 16 * 1 ) =

a ( x ^ 3 + x ^ 2 + 16 x + 16 )

a can be any real number.

If a = 1 then

f ( x ) = x ^ 3 + x ^ 2 + 16 x + 16

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