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 .
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
To create a 3rd degree polynomial with real coefficients that has roots -1 and 4i, we know that since the coefficients are real, the complex conjugate of 4i must also be a root.
The complex conjugate of 4i is -4i. Therefore, the roots of the polynomial are -1, 4i, and -4i.
To find the polynomial, we can use the fact that if r is a root of a polynomial, then (x-r) is a factor of the polynomial.
So, since the roots are -1, 4i, and -4i, we have the following factors:
(x - (-1)) = (x + 1)
(x - 4i)
(x - (-4i)) = (x + 4i)
To obtain the polynomial, we multiply these factors together:
(x + 1)(x + 4i)(x + 4i)
Expanding this expression, we get:
(x + 1)(x^2 + 4ix + 4ix + 16i^2)
(x + 1)(x^2 + 8ix - 16)
Next, we multiply out the terms:
x(x^2 + 8ix - 16) + 1(x^2 + 8ix - 16)
x^3 + 8ix^2 - 16x + x^2 + 8ix - 16
Combining like terms, the polynomial is:
x^3 + (8i + 1)x^2 + (8i + 8i)x - 16
x^3 + (8i + 1)x^2 + 16ix - 16
x^3 + (1 + 8i)x^2 + 16ix - 16
So, the polynomial with real coefficients that has roots -1 and 4i is:
x^3 + (1 + 8i)x^2 + 16ix - 16
To create a 3rd degree polynomial with real coefficients that has the roots -1 and 4i, we must also have the root -4i since complex roots always occur in conjugate pairs.
The polynomial can be written in factored form as follows:
(x - r)(x - s)(x - t),
where r, s, and t represent the roots of the polynomial.
Using the given roots, we have:
(x - (-1))(x - 4i)(x - (-4i))
Simplifying this expression, we get:
(x + 1)(x - 4i)(x + 4i)
Now, let's multiply the factors to expand the expression:
(x + 1)(x^2 - (4i)^2)
Applying the difference of squares formula ((a^2 - b^2) = (a + b)(a - b)), we get:
(x + 1)(x^2 - 16i^2)
Simplifying further, we know that i^2 is equal to -1:
(x + 1)(x^2 - 16(-1))
(x + 1)(x^2 + 16)
Expanding this further by distributing, we have:
x^3 + 16x + x^2 + 16
Now, let's combine like terms to get the final form of the polynomial:
x^3 + x^2 + 16x + 16
Therefore, the 3rd degree polynomial with real coefficients and roots -1 and 4i is:
x^3 + x^2 + 16x + 16.