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anonymous
 5 years ago
Which polynomial has roots at... (see attachment)
anonymous
 5 years ago
Which polynomial has roots at... (see attachment)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A. 11x^28x+16 B. 11x^212x+12 C. 9x^212x+16 D. 9X^28x+12

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's the second attach. sorry..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is \[9x^212x+16\] if you want i will show you how to get it without solving each one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure. i never got up to learning that in alg 2. so it should be good to learn

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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 \[(xr_1)(xr_2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we can just write \[(x\frac{2+\sqrt{3}i}{3})(x\frac{2\sqrt{3}i}{3})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0get 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
 5 years ago
Best ResponseYou've already chosen the best response.0now 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
 5 years ago
Best ResponseYou've already chosen the best response.0"last" is \[(2+2\sqrt{3}i)(2\sqrt{3}i)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops i meant \[(2+2\sqrt{3}i)(22\sqrt{3}i)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because of + and  right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 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
 5 years ago
Best ResponseYou've already chosen the best response.0yes because the zeros are \[a+bi\] and \[abi\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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 \[6x6x=12x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ooooh okay! Jeez, that was a lot. LOL. You figured it out quick. Are you a math teacher?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0may 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.
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