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

1. anonymous

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

2. anonymous

3. anonymous

It's the second attach. sorry..

4. anonymous

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

5. anonymous

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

6. 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 $(x-r_1)(x-r_2)$

7. anonymous

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

8. anonymous

Okay

9. anonymous

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

10. anonymous

okay

11. 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"

12. anonymous

"first" is $9x^2$

13. anonymous

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

14. anonymous

yeah

15. anonymous

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

16. anonymous

because of + and - right?

17. anonymous

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

18. anonymous

yes because the zeros are $a+bi$ and $a-bi$

19. anonymous

ok

20. 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 $(3+\sqrt{5}i)(3-\sqrt{5}i)=3^2+5^2=34$

21. anonymous

yeah okay

22. 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 $-6x-6x=-12x$

23. anonymous

Oh okay!

24. anonymous

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

25. 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.

26. anonymous

Thanks!!

27. anonymous

welcome!

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