Is this a new formula? If so, what is it called? – Khan Academy Centre d'aide

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Andrew

6 avril 2018 00:51

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You are given the zero, 5-5i.

Therefore you also have 5+5i.

You would then multiply the two complex zeros to find the polynomial function.

I spent a while attempting to more effectively multiply the two zeros and came up with the following:

(x - a - bi)(x - a +bi) = (x^2 - (2a)x + (a^2 + b^2)

So, for our numbers on lines 1 and 2, our answer would be:

f(x) = x^2 - 10x + 26

Is the formula I "discovered" "(x - a - bi)(x - a +bi) = (x^2 - (2a)x + (a^2 + b^2)" a known formula, or did I truly discover it?

Any feedback is appreciated!

-John

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3 commentaires

Doug Massey

6 avril 2018 01:12
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Both!

It is a known formula -- it's the only possible product when the complex conjugate root factors are multiplied together -- but you did discover it on your own, and you get credit for it!

I spend a lot of time tutoring students and do my best to help them discover things just like this -- because it has a bigger impact on them (and thus, is easier to remember or to be able to derive again at a later date). The educational journey is much stronger this way than if I were to just stand up in front of them and rattle of formulas and rules and laws for them to memorize.

Keep up the good work, John!
Andrew

7 avril 2018 06:10
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Thanks for the reply! It is definitely a formula that's stuck in my head! I was pretty excited to figure it out as it speeds the process up tremendously!

Thanks again for the reply Doug!
Doug Massey

7 avril 2018 09:32
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Careful with the arithmetic, though -- your function (by using your formula!) should actually be

f(x) = x^2 - 10x + 50

(and your formula has an extra parenthesis you don't need after the equal sign, but that's just a typo)

Check out more detail (and an alternate form of your formula) here:

https://www.wolframalpha.com/input/?i=y%3D(x-a-b*i)(x-a%2Bb*i)

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