Advanced Algebra

Question

Form a? third-degree polynomial function with real coefficients such that 5 plus i and 2 are zeros.

Answer 1

You zeros are

 

x₁ = 2

x₂ = 5 + i

x₃ = 5 - i

 

Then your polynomial will be

 

f(x) = (x - 2)(x - [5 - i])(x - [5 + i])

 

f(x) = (x - 2)(x² - x(5 + i) - x(5 - i) + (25 + 1))

 

f(x) = (x - 2)(x² - 10x + 26)

 

f(x) = x(x² - 10x + 26) - 2(x² - 10x + 26)

 

f(x) = x³ - 10x² + 26x - 2x² + 20x - 52

 

f(x) = x³ - 12x² + 46x - 52

 

 

Answer 2

Hi Devin.

 

The trick to these questions is that you know complex roots come in conjugate pairs. So if (5+i) is a root, then (5-i) must also be a root. You also know 2 is a root. So there you have your three roots for your 3rd degree polynomial.

 

f(x) = (x - 2)(x - (5+i))(x - (5-i))

 

f(x) = (x - 2)(x² - x(5+i) - x(5-i) + (5+i)(5-i))

 

f(x) = (x - 2)(x² - 5x - xi - 5x + xi + 25 - i²)

 

f(x) = (x - 2)(x² - 10x + 25 - (-1))

 

f(x) = (x - 2)(x² - 10x + 26)

 

f(x) = x³ - 10x² + 26x - 2x² + 20x - 52

 

f(x) = x³ - 12x² + 46x - 52

 

There's your polynomial functions, all real coefficients, whose roots are 2, 5 - i, 5 + i.

 

Hope this helps.