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December 20, 2014

December 20, 2014

Posted by **Tabitha** on Monday, September 5, 2011 at 8:52am.

1.What are theroots of the following polynomial equation?

(x+2i)(x-5)(x-2i)(x+8) = 0

I think the roots are -2i,5,2i,-8-you take out the roots as opposites of what are in the equaion correct?

2. Write a polynomial of the smallest degree with roots 3,3i and -3i

Final answer can't have parentheses

(x-3)(x-3i)(x+3i)=0 and then I'm stuck I don't know what to do when you multiply the x times the -3i

- Pre-Calculus/Trig Check please and help -
**Mgraph**, Monday, September 5, 2011 at 8:59amFirst find (x-3i)(x+3i)=x^2+9 and then

(x-3)(x^2+9)=x^3-3x^2+9x-27

- Pre-Calculus/Trig Check please and help -
**bobpursley**, Monday, September 5, 2011 at 9:01amDo the last two first: (x-3i)(x+3i) You can do FOIL (the anwer to x*(-3i) is -3xi), however, notice x+9 is the answer. (x^2+9) is the difference of two squares, x and 3i, so (x^2+9)=(x+3i)*(x-3i)

so you now have (x-3)(x^2+9)

= x^3+9x -3x^3-27 Using the FOIL method.

- Pre-Calculus/Trigbobpursley please recheck -
**Tabitha**, Monday, September 5, 2011 at 9:10amWhenn I get to (x-3)(x^2+9) and FOIL wouldn't it be x^3 +9x-3x^2 -27? I get (-3x^2) not (3x^3) when I multiply -3 times x^2 or am I missing something

Thank you

- Pre-Calculus/Trig Check please and help -
**bobpursley**, Monday, September 5, 2011 at 9:41amcorrec, typo. -3x^2

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