I need help with a few questions, please explain.

1. Write a polynomial function in standard form with zeros -1, -1, 6.

2. Find the roots of the equation
x^3 – 3x^2 + x + 5

3. Describe the number and type of roots for the polynomial (how many real and complex?). x^3 + 5x^2 – 4x – 2 = 0

4. One zero of x^3 – 3x^2 – 6x + 8 = 0 is -2. What are the other zeros of the function?

#1. a zero at x=h means that (x-h) is a factor of f(x). So, a simple function which satisfies the conditions is

f(x) = (x+1)^2 (x-6)

#2. Since x^3-3x^2+x+5 = (x+1)(x^2-4x+5)
it has one real root at x = -1
Since the quadratic has no real roots (the discriminant is negative), -1 is the only root.

#3. f(x) = (x-1)(x^2+6x+2)
work it the same way as #2.

#4. f(x) = (x+2)(x^2-5x+4)
so see what you can do with that.

1. To write a polynomial function in standard form with zeros -1, -1, and 6, we can start by setting up the factors of the polynomial. Since -1 is a zero, the factor (x + 1) is repeated twice. Additionally, the zero 6 can be written as (x - 6). We can create the polynomial function by multiplying these factors:

f(x) = (x + 1)(x + 1)(x - 6)

To simplify, we multiply the factors:

f(x) = (x + 1)^2(x - 6)

Expanding this expression further, we get:

f(x) = (x + 1)(x + 1)(x - 6)
= (x^2 + 2x + 1)(x - 6)
= x^3 - 4x^2 - 11x - 6

Therefore, a polynomial function in standard form with zeros -1, -1, 6 is f(x) = x^3 - 4x^2 - 11x - 6.

2. To find the roots of the equation x^3 – 3x^2 + x + 5, we can use the Rational Root Theorem to test the possible rational roots. The rational roots are all the possible fractions that can be formed with the factors of the constant term (5) divided by the factors of the leading coefficient (1).

The possible rational roots for this equation are: ±1, ±5

We can test these values by plugging them into the equation and checking if they make the equation equal to zero. By testing each value, we find that x = -1 is a root of the equation.

To continue, we can perform synthetic division (or long division) to divide the equation x^3 - 3x^2 + x + 5 by (x + 1):

-1 | 1 -3 1 5
| -1 4 -5
_________________
1 -4 5 0

The result of the division is: x^2 - 4x + 5

Now we have a quadratic equation x^2 - 4x + 5, which can be factored easily using the quadratic formula or by factoring it as (x - 1)(x - 5).

Therefore, the roots of the equation x^3 – 3x^2 + x + 5 are x = -1, x = 1, and x = 5.

3. To determine the number and type of roots for the polynomial x^3 + 5x^2 – 4x – 2 = 0, we need to look at the discriminant of the equation.

The discriminant is calculated as b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation in standard form (ax^2 + bx + c). For our equation, the coefficients are a = 1, b = 5, and c = -2.

Plugging these values into the discriminant formula, we have:

Discriminant = (5^2) - 4(1)(-2)
= 25 + 8
= 33

Since the discriminant (33) is positive, the equation has two distinct, real roots and one complex root.

4. Given that one zero of x^3 – 3x^2 – 6x + 8 = 0 is -2, we can find the other zeros of the function by using polynomial division. Dividing the polynomial by (x + 2), we get:

-2 | 1 -3 -6 8
| -2 10 -8
___________________
1 -5 4 0

The result of the division is: x^2 - 5x + 4.

We can now factor the quadratic equation x^2 - 5x + 4 as (x - 1)(x - 4).

Therefore, the other zeros of the function x^3 – 3x^2 – 6x + 8 = 0 are x = 1 and x = 4.