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Mathematics 5 Online
OpenStudy (anonymous):

Using the given zero, find all other zeros of f(x). -2i is a zero of f(x) = x^4 - 45x^2 - 196

OpenStudy (whpalmer4):

Complex zeros come in conjugate pairs. Do you know what the conjugate zero of -2i is?

OpenStudy (anonymous):

Is it 2i?? I'm not really sure.... :/

OpenStudy (whpalmer4):

That's correct! If you have a complex number \(a+bi\), the conjugate will be \(a-bi\). Here we've got \(b = \pm2\) and \(a = 0\). A polynomial with \(n\) zeros \(z_1, z_2, ... z_n\) can be written as a product of binomials like this: \[(x-z_1)(x-z_2)(x-z_3)...(x-z_n) =0\] We know that two of the terms are \[(x-2i)(x+2i)\]If we multiply that expression out, we get \[x^2-2ix+2ix-4i^2= x^2 + 4\]. Divide our original polynomial by that and you'll have a new polynomial which if factored will give us the other zeros. Do you know how to divide polynomials?

OpenStudy (anonymous):

No... I'm sort of confused! :P

OpenStudy (whpalmer4):

Do you agree that if we can figure out how to divide\( x^4 - 45x^2 - 196\) by \(x^2 + 4\) we'll have a polynomial with just the remaining 2 zeros?

OpenStudy (anonymous):

Yes.... So, how would you do that?

OpenStudy (whpalmer4):

Just like you would plain old long division, it turns out. What is the biggest factor you can multiply with \(x^2\) without going over \(x^4\)?

OpenStudy (anonymous):

x^3??????

OpenStudy (whpalmer4):

\[x^2*x^3 = x^5\] doesn't it?

OpenStudy (anonymous):

oh, so, x^2?

OpenStudy (whpalmer4):

Right! So what is x^2 * (x^2 + 4)?

OpenStudy (anonymous):

x^4 + 4x^2? (:

OpenStudy (whpalmer4):

Exactly. So what is\[( x^4 - 45x^2 - 196 ) - (x^4 + 4x^2)\]

OpenStudy (anonymous):

-41x^2 -196 ?? I feel like I did something wrong! :P

OpenStudy (anonymous):

I'm sorry, but, I have to go! D: Here are my answer choices: 2i, 14i, -14i 2i, 7, -7 2i, 14, -14 2i, 7i, -7i Could you please help me find the answer? (:

OpenStudy (whpalmer4):

Close, but not quite. \[x^4 - 45x^2 - 196 - x^4 - 4x^2\]Collect like terms \[-49x^2 - 196\]Now, are there any common factors there? Hint: there are :-)

OpenStudy (whpalmer4):

\[-49(x^2 + 4) = -49x^2 - 196\] So our division yielded\[ (x^2 -49)\] as the quotient. If you factor that, you'll get the other two zeros.

OpenStudy (whpalmer4):

If we had written it like a long division problem, it might look like this: x^2 - 49 -------------------- x^2 + 4 | x^4 - 45x^2 - 196 -( x^4 + 4x^2 ) --------------- 0x^4 -49x^2 - 196 - ( -49x^2 - 196 ) -------------------- 0

OpenStudy (whpalmer4):

Hint for factoring the rest: \[(a-b)(a+b) = a^2 -ab + ab - b^2 = a^2 - b^2\]

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