SOLUTION: Use the given zero to find all other zeros. Zero:2i Equation: F(x)=2x^3+3x^2+8x+12 Thanks in advance!

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Use the given zero to find all other zeros. Zero:2i Equation: F(x)=2x^3+3x^2+8x+12 Thanks in advance!      Log On


   

Question 1178287: Use the given zero to find all other zeros.
Zero:2i
Equation: F(x)=2x^3+3x^2+8x+12
Thanks in advance!
Found 3 solutions by josgarithmetic, MathLover1, MathTherapy:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Quadratic factor of F is . Use this to help find your other two zeros. Try polynomial division or whatever other method you find helpful.

. 
              2x    3
          ______________________________
x^2+0x+4  |   2x^3 +3x^2  + 8x + 12
          |   2x^3 + 0x^2 + 8x
          |   -----------------
               0    3x^2   0x   12
                    3x^2   0x   12
                   ----------------
                     0     0    0

The zero found here, 2x%2B3 for x=-3%2F2.


Or this could work too:
2x%5E3%2B8x%2B3x%5E2%2B12
2%28x%5E3%2B4x%29%2B3%28x%5E2%2B4%29
2x%28x%5E2%2B4%29%2B3%28x%5E2%2B4%29
%28x%5E2%2B4%29%282x%2B3%29----------and the other zero is -3%2F2.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Use the given zero to find all other zeros.
Zero: x%5B1%5D=2i=>so, you also have x%5B2%5D=-2i
using root product rule we have
%28x-x%5B1%5D%29%28x-x%5B2%5D%29=%28x-2i%29%28x-%28-2i%29%29
%28x-x%5B1%5D%29%28x-x%5B2%5D%29=%28x-2i%29%28x%2B2i%29
%28x-x%5B1%5D%29%28x-x%5B2%5D%29=x%5E2-%282i%29%5E2
%28x-x%5B1%5D%29%28x-x%5B2%5D%29=x%5E2-4%28-1%29
%28x-x%5B1%5D%29%28x-x%5B2%5D%29=x%5E2%2B4

equal given quation to zero and factor it: note one factor should be x%5E2%2B4
0=2x%5E3%2B3x%5E2%2B8x%2B12
0=%282x%5E3%2B8x%29%2B%283x%5E2%2B12%29
0=2x%28x%5E2%2B4%29%2B3%28x%5E2%2B4%29
0+=+%282x+%2B+3%29+%28x%5E2+%2B+4%29
=>third root is:
0+%282x+%2B+3%29=0=>2x+=-3=>x+=-3%2F2

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Use the given zero to find all other zeros.
Zero:2i
Equation: F(x)=2x^3+3x^2+8x+12
Thanks in advance!
matrix%281%2C3%2C+f%28x%29%2C+%22=%22%2C+2x%5E3+%2B+3x%5E2+%2B+8x+%2B+12%29.

It's NOT as complex as the other 2 people state. 

You don't even need to use the 2 roots that're given (2i and its conjugate, - 2i) because when EACH binomial, from left to right, is factored, we get: 

You MIGHT be able to recognize x%5E2+%2B+4 as the EXPANDED form of the given zero, 2i and its conjugate.

With that being said, the other factor, as seen above, is 2x + 3, which gives us: 2x + 3 = 0 ====> 2x = - 3, and ultimately, the other zero, or: highlight_green%28matrix%281%2C3%2C+x%2C+%22=%22%2C+highlight%28-+3%2F2%29%29%29