I've been using the quadratic equation to factor 2 z 2 + 5 i z − 2. Obviously the roots are z = − 2 i and z = − i / 2.Why can't I then write it out as ( z + 2 i ) ( z + i / 2 ) which is not equivalent to the initial function?

Question

I&#039;ve been using the quadratic equation to factor   2      z    2    +  5  i  z  −  2. Obviously the roots are   z  =  −  2  i and   z  =  −  i      /    2.Why can&#039;t I then write it out as   (  z  +  2  i  )  (  z  +  i      /    2  ) which is not equivalent to the initial function?

Kristen Garza · Accepted Answer

Explanation:You can write it out as   2  (  z  +  2  i  )  (  z  +      i    2    ).

Howard Nelson · Accepted Answer

Step 1If a quadratic equation has roots       x    1   and       x    2   then it can be expressed as:  c  (  x  −      x    1    )  (  x  −      x    2    )Step 2For instance,       x    2    +  5  x  +  6 has the same solution as   5      x    2    +  25  x  +  30. So multiplying by a constant term does not affect the roots of the equation. Here, if you multiply   (  z  +  2  i  )  (  z  +  i      /    2  ) by 2, then you get your original function.

I've been using the quadratic equation to factor 2z^2+5iz-2. Obviously the roots are z=-2i and z=-i/2. Why can't I then write it out as (z+2i)(z+i/2) which is not equivalent to the initial function?

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