Question

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

Answer 1

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