Help would be truly appreciated. Write the polynomial in standard form from the given zeroes of lest degree that has rational coefficients, a leading coefficient of 1. Don't have to answer all just what you can. I would appreciate if work is shown that way I can do the same for the rest. I really just need a tutorial and I'm good. Thank you so much!!!

1) The polynomial in standard form is f(x) = x³ - 9·x² + 23·x - 15

2) The polynomial in standard form is f(x) = x³ - (2 - 2·i)·x² + 4·i·x

3) The polynomial in standard form is f(x) = x² + (3·i - 3)·x + 2 - 6·i

4) The polynomial in standard form is f(x) = x² - (5 + √5)·x + 6 + 3·√5

Step-by-step explanation:

Given that that the polynomial is of least degree, we have;

The standard form is the form, f(x) = a·xⁿ + b·xⁿ⁻¹ +...+ c

1) The zeros of the polynomial are x = 5, 3, 1

Which gives the polynomial in factored form as_f(x) = (x - 5)·(x - 3)·(x - 1)

From which we have;

f(x) = (x - 5)·(x - 3)·(x - 1) = (x - 5)·(x² - 4·x + 3) = x³ - 4·x² + 3·x - 5·x² + 20·x - 15

f(x) = x³ - 9·x² + 23·x - 15

The polynomial in standard form is therefore f(x) = x³ - 9·x² + 23·x - 15

2) The zeros of the polynomial are x = 2, 0, 2·i

Which gives the polynomial in factored form as_f(x) = (x - 2)·(x)·(x - 2·i)

From which we have;

f(x) = (x - 2)·x·(x - 2·i) = (x² - 2·x)·(x - 2·i) = x³ - 2·i·x² + 2·x² + 4·ix

The polynomial in standard form is therefore f(x) = x³ - (2 - 2·i)·x² + 4·i·x

3) The zeros of the polynomial are x = 2, 1 - 3·i

Which gives the polynomial in factored form as_f(x) = (x - 2)·(x - 1 - 3·i)

From which we have;

f(x) = (x - 2)·(x - (1 - 3·i)) = x² - x + 3·i·x - 2·x + 2 -6·i = x² + 3·i·x - 3·x + 2 - 6·i

The polynomial in standard form is therefore f(x) = x² + (3·i - 3)·x + 2 - 6·i

4) The zeros of the polynomial are x = 3, 2 + √5

Which gives the polynomial in factored form as_f(x) = (x - 3)·(x - (2 + √5))

From which we have;

f(x) = (x - 3)·(x - (2 + √5)) = x² - 2·x - x·√5 - 3·x + 6 + 3·√5

f(x) = x² - 5·x - x·√5 + 6 + 3·√5 = x² - (5 + √5)·x + 6 + 3·√5

The polynomial in standard form is therefore f(x) = x² - (5 + √5)·x + 6 + 3·√5.

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.