Question

Write a polynomial function of least degree with integral coefficients that has the given zeros. –2, –3,3 – 6i

Answer 1

Answer:

f(x) = (x+2)(x+3)(x-(3-6i))(x-(3+6i))

f(x) = 270 + 189 x + 21 x^2 - x^3 + x^4

Step-by-step explanation:

First of all, we must know that complex roots come in conjugate pairs.

So the zeros of your equation would be

x = -2

x = -3

x = 3 - 6i

x = 3 + 6i

Your polynomial is of fourth degree.

f(x) = (x-(-2))(x-(-3))(x-(3-6i))(x-(3+6i))

f(x) = (x+2)(x+3)(x-(3-6i))(x-(3+6i))

Please , see attached image below for full expression

f(x) = 270 + 189 x + 21 x^2 - x^3 + x^4

Answer 2

Answer:

The required polynomial is $P(x)=a\left(x^4-x^3+21x^2+189x+270\right)$.

Step-by-step explanation:

The general form of a polynomial is

$P(x)=a(x-c_1)^{m_1}(x-c_2)^{m_2}...(x-c_n)^{m_n}$

where, a is a constant, $c_1,c_2,..c_n$ are zeroes with multiplicity $m_1,m_2,..m_n$ respectively.

It is given that  –2, –3,3 – 6i are three zeroes of a polynomial.

According to complex conjugate root theorem, if a+ib is a zero of a polynomial, then a-ib is also the zero of that polynomial.

3 – 6i is a zero. By using complex conjugate root theorem 3+6i is also a zero.

The required polynomial is

$P(x)=a(x-(-2))(x-(-3))(x-(3-6i))(x-(3+6i))$

$P(x)=a(x+2)(x+3)(x-3+6i)(x-3-6i)$

$P(x)=a\left(x^2+5x+6\right)\left(x-3+6i\right)\left(x-3-6i\right)$

On further simplification, we get

$P(x)=a\left(x^3+6ix^2+2x^2+30ix-9x+36i-18\right)\left(x-3-6i\right)$

$P(x)=a\left(x^4-x^3+21x^2+189x+270\right)$

Therefore the required polynomial is $P(x)=a\left(x^4-x^3+21x^2+189x+270\right)$.