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ankoles [38]
3 years ago
7

Help would be truly appreciated. Write the polynomial in standard form from the given zeroes of lest degree that has rational co

efficients, 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!!!

Mathematics
1 answer:
pishuonlain [190]3 years ago
5 0

Answer:

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.

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