Can someone please explain how to do these problems.

1)write a polynomial function of least degree with intregal coefficients whose zeros include 4 and 2i.

2)list all of the possible rational zeros of f(x)= 3x^3-2x^2+7x+6.

3)Find all of the rational zeros of f(x)= 4x^3-3x^2-22x-15.

Because all the coefficients of the polynomial are real, if 2i is a zero, then the complex conjugate -2i must also be a zero. This means that the least order polynomial is proportional to:

(x-4)(x-4i)(x+4i)= (x-4)(x^2 + 16)=
x^3 - 4x^2 + 16 x - 64

To find the rational roots, use the Rational Roots Theorem: if x = p/q is a rational root (and p and q don't have any divisors in common), then p must divide the constant term and q the coefficient of the highest power of x.

In case of the polynomial

3x^3-2x^2+7x+6.

Ths means that p is a divisor of 6 and q a divisor of 3.

The possible rational roots are therefore:

p/q = ±1, ±2, ±3, ±6

For the polynomial

4x^3-3x^2-22x-15

The possible rational roots are:

p/q = ±1, ±3, ±5, ±15

1) To find a polynomial function with the given zeros, start by using the zero 4. This means that (x - 4) is a factor of the polynomial.

Next, since the zeros also include 2i, this means that the conjugate -2i is also a zero. Thus, the factors (x - 2i) and (x + 2i) should also be included in the polynomial.

Now, multiply all the factors together to obtain the polynomial function:

(x - 4)(x - 2i)(x + 2i) = (x - 4)(x^2 + 4) = x^3 - 4x^2 + 4x - 16

Therefore, the polynomial function of least degree with integral coefficients and the given zeros is f(x) = x^3 - 4x^2 + 4x - 16.

2) To find the possible rational zeros of the polynomial function f(x) = 3x^3 - 2x^2 + 7x + 6, we need to consider the factors of the constant term (6) divided by the factors of the leading coefficient (3).

The factors of 6 are ±1, ±2, ±3, and ±6.
The factors of 3 are ±1 and ±3.

Therefore, the possible rational zeros are:
±1/1, ±1/3, ±2/1, ±2/3, ±3/1, and ±3/3, which simplifies to:
±1, ±1/3, ±2, ±2/3, ±3, and ±1.

These are all the possible rational zeros of f(x) = 3x^3 - 2x^2 + 7x + 6.

3) To find all the rational zeros of the polynomial function f(x) = 4x^3 - 3x^2 - 22x - 15, we can use the same process as in question 2.

The factors of 15 are ±1, ±3, ±5, and ±15.
The factors of 4 are ±1 and ±2.

Therefore, the possible rational zeros are:
±1/1, ±1/2, ±3/1, ±3/2, ±5/1, and ±5/2, which simplifies to:
±1, ±1/2, ±3, ±3/2, ±5, and ±5/2.

These are all the possible rational zeros of f(x) = 4x^3 - 3x^2 - 22x - 15.

To find all possible rational zeros, we need to consider all the factors of 6 which are possible numerators (p) and all the factors of 3 which are possible denominators (q). The factors of 6 are:

1, 2, 3, 6

And the factors of 3 are:

1, 3

Now, we can create all the possible combinations of p/q.

Possible rational zeros:

±1/1, ±2/1, ±3/1, ±6/1, ±1/3, ±2/3

In this case, it is important to note that if the polynomial has a rational zero, it will be in this list. However, not all numbers in this list are guaranteed to be zeros of the polynomial. To determine the actual rational zeros of the polynomial, they need to be tested by substituting them into the polynomial and checking if the result is zero.