Which polynomial has roots at... (see attachment)

- anonymous

Which polynomial has roots at... (see attachment)

- schrodinger

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- anonymous

A. 11x^2-8x+16
B. 11x^2-12x+12
C. 9x^2-12x+16
D. 9X^2-8x+12

##### 1 Attachment

- anonymous

##### 1 Attachment

- anonymous

It's the second attach. sorry..

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## More answers

- anonymous

it is \[9x^2-12x+16\] if you want i will show you how to get it without solving each one.

- anonymous

sure. i never got up to learning that in alg 2. so it should be good to learn

- anonymous

ok we start slow, but if it is too slow let me know.
we know the zeros are \[\frac{2+2\sqrt{3}i}{3}\] and
\[\frac{2+2\sqrt{3}i}{3}\] and we know that if a quadratic has zeros \[r_1\] and \[r_2\] it factors as \[(x-r_1)(x-r_2)\]

- anonymous

so we can just write \[(x-\frac{2+\sqrt{3}i}{3})(x-\frac{2-\sqrt{3}i}{3})\]

- anonymous

Okay

- anonymous

get rid of the 3 in the denominator by multiplying each factor by 3 to get
\[(3x-(2+\sqrt{3}i)(3x-(2-\sqrt{3}i)\]

- anonymous

okay

- anonymous

now this might look like a pain to multiply out but it is not. do not distribute to get 3 terms , just use the old 'first outer inner last"

- anonymous

"first" is \[9x^2\]

- anonymous

"last" is \[(2+2\sqrt{3}i)(2-\sqrt{3}i)\]

- anonymous

yeah

- anonymous

oops i meant \[(2+2\sqrt{3}i)(2-2\sqrt{3}i)\]

- anonymous

because of + and - right?

- anonymous

and when you multiply a complex number by its conjugate you get \[a^2+b^2\] so in this case you get \[2^2+(2\sqrt{3})^2=4+12=16\]

- anonymous

yes because the zeros are \[a+bi\] and \[a-bi\]

- anonymous

ok

- anonymous

so we know first term is \[9x^2\]
and last term is \[16\]
hope it is clear how i got 16
another example would be \[(3+\sqrt{5}i)(3-\sqrt{5}i)=3^2+5^2=34\]

- anonymous

yeah okay

- anonymous

ok now for the middle terms, the 'outer - inner'
outer is \[3x(-2+\sqrt{3}i)\] inner is
\[3x(-2-\sqrt{3}i)\] and when you add the imaginary part will cancel (add to zero, it always does!) and you just get \[-6x-6x=-12x\]

- anonymous

Oh okay!

- anonymous

Ooooh okay! Jeez, that was a lot. LOL. You figured it out quick. Are you a math teacher?

- anonymous

may seem like a lot of work, but multiplying a complex number by its conjugate is easy, and seeing that in the middle terms the imaginary part drops out is not too hard. easier than solving 4 quadratics using the formula and risking an arithmetic error. good luck.

- anonymous

Thanks!!

- anonymous

welcome!

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