Question

Write the polynomial P(x) of degree 4, in standard form, with integer coefficients, and zeros 5i, 4, and -1/3 as some of its zeros. Show all work.

Answer 1

`P(x)=3x^4-11x^3+71x^2-275x-100`

Explanation:

Given information:

• Polynomial function with real integer coefficients.
• Zeros: 5i, 4, and -1/3
• Degree: 4

For any complex number
`z=a+bi`, the complex conjugate of the number is defined as
`z*=a-bi`.

If f(z) is a polynomial with real coefficients, and z₁ is a root of f(z)=0, then its complex conjugate z₁* is also a root of f(z)=0.

Therefore, if P(x) is a polynomial with real coefficients, and 5i is a root of P(x)=0, then its complex conjugate -5i is also a root of P(x)=0.

So the zeros of the function are: 5i, -5i, 4, and -1/3.

The zeros of a function P(x) are the x-values of the function such that the P(x)=0. According to the Factor Theorem, if P(x) is a polynomial, and P(a)=0, then (x - a) is a factor of P(x).

Therefore, the factors of the function are:

`x=5i \implies (x-5i)`

`x=-5i \implies (x+5i)`

`x=4 \implies (x-4)`

`x=-(1)/(3) \implies (3x+1)`

So the polynomial in factored form is:

`P(x)=a(x-5i)(x+5i)(x-4)(3x+1)`

As we have not been given a specific leading coefficient, let us assume it is one.

`P(x)=(x-5i)(x+5i)(x-4)(3x+1)`

To write the polynomial in standard form, expand and simplify.

`\begin{aligned}P(x)&=(x-5i)(x+5i)(x-4)(3x+1)\\&=(x^2+5ix-5ix-25i^2)(3x^2+x-12x-4)\\&=(x^2-25(-1))(3x^2-11x-4)\\&=(x^2+25)(3x^2-11x-4)\\&=3x^4-11x^3-4x^2+75x^2-275x-100\\&=3x^4-11x^3+71x^2-275x-100\end{aligned}`