Writing Polynomial Functions With Given Zeros

Introduction

Hopefully something I can just do. I will try with the first question he had

Example 1

Write a polynomial with these characteristics

n =2 (degree), x = 1, 2 (values with zero) and where f(3) = 6

So guessing we start with this is what it will look like

f(3) = (x-1)(x-2) = 6
     = (x-1)(x-2) - 6 = 0
     = x² -3x -6

And wrong. Like being wrong because I hopefully will remember this and not make the same mistake again. So where did I go wrong. We need another step to find the leading coefficient a so we write

f(x) = a(x-1)(x-2)

So we know when x = 3 the function = 6

   6 = a(3-1)(3-2)
   6 = 2a
   3 = a     

So

f(x) = 3(x-1)(x-2)
     = 3[x² -3x +2]
     = 3x² -9x + 6

Example 2

n = 3   x=3,2i  f(4) = 40

So the factors are

 (x - 3) and hmmmm
   2i = +/- $\sqrt{2}$ * $\sqrt{-1}$
     = +/- $\sqrt{-2}$
     = (x² - 4)
     = (x - 2)(x + 2)
f(x) = (x - 2)(x + 2)(x - 3)
  40 = a(4 - 2)(4 + 2)(4 - 3)
     = a(2)(6)(-1)
  40 = -12a

So given how this would be a silly number, I watched the video and the mistake was the handling of the imaginary numbers. We leave them as this 2i notation and proceed

f(x) = a(x - 3)(x - 2i)(x + 2i)

So

(x - 2i)(x + 2i) = x² +2ix - 2ix - 4i² = (x² + 4)

So

f(x) = a(x - 3)(x² + 4)
  40 = a(4 - 3)(4² + 4)
  40 = a(1)(20) = 20a
   2 = a

So

f(x) = 2(x - 3)(x - 2i)(x + 2i)   
     = 2(x³ +4x -3x² -12)
     = 2x³ - 6x² + 8x -24