factor 8x^3 - 1, simplify 2+6i/3-4i, solve w/ remainder thm

hollerback1

Junior Member
Joined
Dec 21, 2005
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Hola, I have a final exam tommorow and I just have a few questions.

1) factor: 8x<sup>3</sup>-1 i forgot how to do these cubes lol

2) simplify: 2+6i/3-4i

3) solve by using remainder therom: P(x)=3x<sup>3</sup>-2x<sup>2</sup>+x+4, find P(-2)

Please help asap, thanks.
 
1) The Difference of Cubes formula (which you should memorize now) is as follows:

. . . . .a<sup>3</sup> - b<sup>3</sup> = (a - b)(a<sup>2</sup> + ab + b<sup>2</sup>)

You should also know the Difference of Squares formula and the Sum of Cubes formula:

. . . . .a<sup>2</sup> - b<sup>2</sup> = (a - b)(a + b)

. . . . .a<sup>3</sup> + b<sup>3</sup> = (a + b)(a<sup>2</sup> - ab + b<sup>2</sup>)

2) What you have posted, "2 + 6i/3 - 4i", means the following:

. . . . .\(\displaystyle \L 2\,+ \,\frac{6i}{3}\, - \,4i\)

Is this what you meant? If so, then there isn't much simplifying to do: divide the 6 by the 3, and combine "like" terms. If not, then please reply with clarification. (You probably need to use grouping symbols.) Please also show what you have tried so far and clarify where you are stuck.

3) "Solve"? Or "evaluate"? I would suspect you are supposed to do the latter.

Using the Remainder Theorem to evaluate a polynomial at a given value of x is a fairly straight-forward matter of plugging the x-value into the synthetic-division algorithm. Where are you stuck in this process?

Thank you.

Eliz.
 
No, that you have been mistaken I wrote the problem wrong i think...

This is it:
2+6i
-------
3-4i

it looks like that... hard to draw on the post.

it's 2 plus 6i divided by 3 minus 4i
 
correct way to type a rational expression ... (2 + 6i)/(3 - 4i) ... clear?

multiply numerator and denominator by the conjugate of the denominator ...

(2 + 6i)/(3 - 4i)*(3 + 4i)/(3 + 4i) = (-18 + 26i)/25
 
wait...I dont get what to do when it's: (2+6i)/(3-4i)*(3+4i)/(3+4i)

How did you get (-18 + 26i)/25?

did you cancel anything? or cross multiply?

Im confused :oops:
 
multiply the fractions straight across

first the numerators ...
(2 + 6i)(3 + 4i) = 6 + 8i + 18i + 24i<sup>2</sup> = 6 + 26i - 24 = -18 + 26i

then the denominators ...
(3 - 4i)(3 + 4i) = 9 + 12i - 12i - 16i<sup>2</sup> = 9 + 16 = 25

the product is ... (-18 + 26i)/25

kapish?
 
6 + 8i + 18i + 24i2 = 6 + 26i - 24 = -18 + 26i

lol, no kapish yet.

How did u cancel out the 24i<sup>2</sup> to -24?

:lol: this is my final question....hopefully.
 
first of all, what is i? second, what does i<sup>2</sup> = ?

here's a BIG hint ... \(\displaystyle \L \sqrt{-1}\)
 
oHHHHH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I get it.

Gracias, adios amigo.
 
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