# Thread: imaginary numbers

1. ## imaginary numbers

Hello! I don't know how to foil and add/subtract an imaginary numbers.

$\displaystyle x^2 (x+2i) (x-2i) (x - (3-i)) (x-(3+i))$

thanks...

2. Originally Posted by Anemori
Hello! I don't know how to foil and add/subtract an imaginary numbers.

$\displaystyle x^2 (x+2i) (x-2i) (x - (3-i)) (x-(3+i))$

thanks...
$\displaystyle x^2(x+ 2i)(x-2i)[(x-3)+i][(x-3)-i]$

$\displaystyle x^2(x^2 - 4i^2)[(x-3)^2 - i^2]$

note that $\displaystyle i^2 = -1$ ... finish.

3. Originally Posted by skeeter
$\displaystyle x^2(x+ 2i)(x-2i)[(x-3)+i][(x-3)-i]$

$\displaystyle x^2(x^2 - 4i^2)[(x-3)^2 - i^2]$

note that $\displaystyle i^2 = -1$ ... finish.

$\displaystyle x^2(x^2 - 4i^2)[(x-3)^2 - i^2]$

$\displaystyle x^2(x^2+4)(x^2-6x+10)$

$\displaystyle x^2(x^4 - 6x^3+14x^2-24x+40)$

$\displaystyle x^6 - 6x^5+14x^4-24x^3+40x^2$

This is what i get... i hope this is right...

4. Originally Posted by Anemori
$\displaystyle x^2(x^2 - 4i^2)[(x-3)^2 - i^2]$

$\displaystyle x^2(x^2+4)(x^2-6x+10)$

$\displaystyle x^2(x^4 - 6x^3+14x^2-24x+40)$

$\displaystyle x^6 - 6x^5+14x^4-24x^3+40x^2$

This is what i get... i hope this is right...
I dont know if this is right...

when i plug in back the factor from this polynomial, it doesn't go to zero.

5. Originally Posted by Anemori
I dont know if this is right...

when i plug in back the factor from this polynomial, it doesn't go to zero.
the zeros are $\displaystyle x = \pm 2i$ , $\displaystyle x = 3 \pm i$ , and $\displaystyle x = 0$

... are these the values what you "plugged" in for x ?

6. Originally Posted by skeeter
the zeros are $\displaystyle x = \pm 2i$ , $\displaystyle x = 3 \pm i$ , and $\displaystyle x = 0$

... are these the values what you "plugged" in for x ?

yes. but its not real zero. maybe im doing it wrong. can you show me one of them

7. Originally Posted by Anemori
yes. but its not real zero. maybe im doing it wrong. can you show me one of them
$\displaystyle f(x) = x^6 - 6x^5 + 14x^4 - 24x^3 + 40x^2$

$\displaystyle f(-2i) = (-2i)^6 - 6(-2i)^5 + 14(-2i)^4 - 24(-2i)^3 + 40(-2i)^2$

$\displaystyle f(-2i) = 64i^6 + 192i^5 + 224i^4 + 192i^3 + 160i^2$

$\displaystyle f(-2i) = -64 + 192i + 224 - 192i - 160 =0$

8. Originally Posted by skeeter
$\displaystyle f(x) = x^6 - 6x^5 + 14x^4 - 24x^3 + 40x^2$

$\displaystyle f(-2i) = (-2i)^6 - 6(-2i)^5 + 14(-2i)^4 - 24(-2i)^3 + 40(-2i)^2$

$\displaystyle f(-2i) = 64i^6 + 192i^5 + 224i^4 + 192i^3 + 160i^2$

$\displaystyle f(-2i) = -64 + 192i + 224 - 192i - 160 =0$
oh ok... maybe my calculator if messed up. im getting huge numbers heheheh. but it wouldnt make any difference if I use the exponents first before multiplying the outside paranthesis?

9. Originally Posted by skeeter
$\displaystyle f(x) = x^6 - 6x^5 + 14x^4 - 24x^3 + 40x^2$

$\displaystyle f(-2i) = (-2i)^6 - 6(-2i)^5 + 14(-2i)^4 - 24(-2i)^3 + 40(-2i)^2$

$\displaystyle f(-2i) = 64i^6 + 192i^5 + 224i^4 + 192i^3 + 160i^2$

$\displaystyle f(-2i) = -64 + 192i + 224 - 192i - 160 =0$

only -2i and 3-i will result of zero in the function. the rest are not zero.

10. Originally Posted by Anemori
only -2i and 3-i will result of zero in the function. the rest are not zero.
is that so?

$\displaystyle f(2i) = (2i)^6 - 6(2i)^5 + 14(2i)^4 - 24(2i)^3 + 40(2i)^2$

$\displaystyle f(2i) = 64i^6 - 192i^5 + 224i^4 - 192i^3 + 160i^2$

$\displaystyle f(2i) = -64 - 192i + 224 + 192i - 160 = 0$

... and I guarantee you that $\displaystyle f(3+i) = 0$ ... complex zeros always come in conjugate pairs.

11. There is enough information for the user to solve the problem. There is no need to carry on this discussion anymore.