Question

Find a polynomial function that has zeros -10, 2i, -2i.

This is what I have
(X+10)(x^2+2i)
X^3+20x^2+2x+20

Am I close?

Answer 1

Uh, I'm not convinced by your imaginary root combination...

Answer 2

\[(x+2i)(x-2i) = x(x-2i) + 2i(x-2i) = x^2 -2ix + 2ix - 4i^2\]

Answer 3

In a polynomial with only real coefficients, the complex roots always come as complex conjugate pairs, which always annihilate the imaginary bit because they are the difference of squares when multiplied...

Your multiplication of \((x+10)(x^2+2i) = x^3 + 20x^2+2x+20\) somehow manages to get \(i\) to vanish, but not by allowable means :-)

Answer 4

Thanks, I'm going to rework.

Answer 5

Remember that the solution to \[x^2-4 = 0\] is \(x = \pm2\) which implies that \[x^2+4 =0\] has solutions \[x=\pm 2i\]right?

Answer 6

Is X^3+8x-4 closer'

Answer 7

Close only counts with hydrogen bombs and horseshoes :-)

What do you get when you simplify that multiplication of \((x+2i)(x-2i)\) that I did earlier? Multiply that by \((x+10)\) and you should have your answer.

Answer 8

I don't know why I can't get this, here is what I have now and it doesn't seem right.

(X+10)(x^2 -4i)
X^3 -4ix + 10x^2 -40i
X^3 +10x^2 -4ix -40i

I have myself all confused

Answer 9

ok so do you want me to start from the beginning?

Answer 10

I need to find a polynomial function that has zeros -10, 2i and -2i

Answer 11

ok, so you were on the right track there were just some simple errors.

Answer 12

I figured but I have gotten myself all messed up

Answer 13

that's fine math can sometimes be confusing especially if you are trying to solve a problem that doesn't want to be solved. trust me i've been there ;P.

so with the zeros you want to plug them in to a standard form: (x-r) r=root or zeros

so we can start with(x-2i)(x+2i)

Answer 14

So I have (x+10) (x-2i) (x+2i)

X^3-2ix+2ix+10x+10x-4i

X^3+20x-4i

Answer 15

ok so (x+10)(x-2i)(x+2i) is right… to make your life easier multiply only 2 at a time so try multiplying only (x-2i)(x+2i)… and as a head's up \[i^2\] is -1.

Answer 16

So would the answer be (x+10)(x^2 -4)
X^3 - 4x + 10x^2 -40
X^3+10x^2 - 4x -40.

Answer 17

so close!!!! one small mistake it would be (x+10)\[(x^2+4)\] because \[-4i^2\] is in essence -4 X -1 (-1 because i^2 is -1)

Answer 18

Let's go through this part again, because it is where your troubles seem to arise:

\[(x+2i)(x-2i) = x(x-2i)+2i(x-2i) = x*x - 2i*x +2i*x - 2i(2i)\]\[\qquad = x^2\cancel{-2ix} \cancel{+ 2ix} - 4i^2\]\[\qquad=x^2-4i^2\]but \[i=\sqrt{-1}\]\[i^2=-1\]so we further simplify to \[x^2-4(-1) = x^2+4\]

Now we have \[(x+10)(x^2+4) = x(x^2+4) + 10(x^2+4) = x^3+4x + 10x^2 + 40 \]\[\qquad = x^3+10x^2+4x+40\]

Answer 19

Plug in the roots as a check:

\[(-10)^3 + 10(-10)^2 + 4(-10) + 40 = -1000 + 1000 -40 + 40 = 0\checkmark\]
\[(2i)^3+10(2i)^2 + 4(2i) + 40 = 8i^3 + 10*4i^2 + 8i + 40 \]\[\qquad = 8(-1)i+40(-1)+8i+40 = -8i+8i -40+40 = 0\checkmark\]\[(-2i)^3+10(-2i)^2+4(-2i)+40 = -8i^3 + 40i^2 -8i + 40\]\[\qquad = 8i-40-8i+40 = 0\checkmark\]

Answer 20

Thank you ever so much! I see where I was getting confused !

Answer 21

Always multiply the complex roots first. If when you're done, you still see any \(i\)'s running around, that's the tip-off that you've either forgotten to fully simplify, or made a mistake.

Answer 22

You can take a shortcut if you remember that they are always the difference of squares:

\[(x - a-bi)(x - a+bi) = ((x-a) - bi))(x-a) + bi) = (x-a)^2 -b^2i^2 \]\[\qquad = (x-a)^2 +b^2\]\[\qquad = x^2 + a^2 + b^2 - 2ax\]

Not needed so much for this problem, but if we'd had complex roots instead of just imaginary roots, I think you can see how this might make life easier.

Say we had \(x = 1\pm2i\) as our roots instead:

\[x^2+1^2 + 2^2 -2(1)x = x^2 -2x+5\]beats\[(x-1-2i)(x-1+2i) = \text{<cloud of dust>} = x^2-2x+5\]:-)