anonymous
  • anonymous
Which polynomial has roots at... (see attachment)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
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anonymous
  • anonymous
A. 11x^2-8x+16 B. 11x^2-12x+12 C. 9x^2-12x+16 D. 9X^2-8x+12
anonymous
  • anonymous
anonymous
  • anonymous
It's the second attach. sorry..

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anonymous
  • anonymous
it is \[9x^2-12x+16\] if you want i will show you how to get it without solving each one.
anonymous
  • anonymous
sure. i never got up to learning that in alg 2. so it should be good to learn
anonymous
  • 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
  • anonymous
so we can just write \[(x-\frac{2+\sqrt{3}i}{3})(x-\frac{2-\sqrt{3}i}{3})\]
anonymous
  • anonymous
Okay
anonymous
  • 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
  • anonymous
okay
anonymous
  • 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
  • anonymous
"first" is \[9x^2\]
anonymous
  • anonymous
"last" is \[(2+2\sqrt{3}i)(2-\sqrt{3}i)\]
anonymous
  • anonymous
yeah
anonymous
  • anonymous
oops i meant \[(2+2\sqrt{3}i)(2-2\sqrt{3}i)\]
anonymous
  • anonymous
because of + and - right?
anonymous
  • 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
  • anonymous
yes because the zeros are \[a+bi\] and \[a-bi\]
anonymous
  • anonymous
ok
anonymous
  • 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
  • anonymous
yeah okay
anonymous
  • 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
  • anonymous
Oh okay!
anonymous
  • anonymous
Ooooh okay! Jeez, that was a lot. LOL. You figured it out quick. Are you a math teacher?
anonymous
  • 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
  • anonymous
Thanks!!
anonymous
  • anonymous
welcome!

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