133k views
5 votes
Form a third-degree polynomial function with real coefficients, with leading coefficient 1, such that - 2 + i and 7 are zeros.

1 Answer

5 votes

In standard form, this third-order polynomial - call it p(x) - has real coefficients a, b, c, d so that it is written as

p(x) = a x ³ + b x ² + c x + d

We're told the leading coefficient is a = 1, so

p(x) = x ³ + b x ² + c x + d

Since the coefficients are all real, any complex roots with non-zero imaginary part occurs alongside its complex conjugate, so the fact that -2 + i is a root means that its conjugate, -2 - i is also a root.

Then by the factor theorem, we can write p(x) as the product of linear terms involving its roots:

p(x) = (x - (-2 + i )) (x - (-2 - i )) (x - 7)

Now just expand this to get

p(x) = x ³ - 3 x ² - 23 x - 35

User CustomCalc
by
8.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories