Question

Find the polynomial of lowest degree with only real coefficients and having the given zeros. -7i and square root of 2

Answer 1

Answer:

x³ - (√2)x² + 49x - 49√2

Step-by-step explanation:

If one root is -7i, another root must be 7i.  You can't just have one root with i.  The other roos is √2, so there are 3 roots.  

x = -7i       is one root,

   (x + 7i) = 0    is the factor

x = 7i       is one root

  (x - 7i) = 0     is the factor

x = √2      is one root

    (x - √2) = 0   is the factor

So the factors are...

(x + 7i)(x - 7i)(x - √2) = 0

Multiply these out to find the polynomial...

(x + 7i)(x - 7i) =  x² + 7i - 7i - 49i²

Which simplifies to

  x² - 49i²        since i² = -1 , we have

    x² - 49(-1)  

 

      x² + 49

Now we have...

(x² + 49)(x - √2) = 0

Now foil this out...

 x²(x) - x²(-√2) + 49(x) + 49(-√2) = 0

     x³ + (√2)x² + 49x - 49√2