Factoring x2+4 and Special Cubed Theorems: Homework Help

  • Thread starter Thread starter hockeyfghts5
  • Start date Start date
  • Tags Tags
    Factoring
AI Thread Summary
The expression x² + 4 can be factored using imaginary numbers, resulting in (x + 2i)(x - 2i). The discussion also covers the formulas for the sum and difference of cubes, which are x³ + a³ = (x + a)(x² - ax + a²) and x³ - a³ = (x - a)(x² + ax + a²). A user inquires about factoring a limit expression, specifically lim (x³ - 6x² + 11x - 6)/(x³ - 4x² - 19x + 14) as x approaches -1. It is suggested to evaluate the limit at x = -1 to determine if it results in an indeterminate form, which would indicate the need for factoring. The conversation emphasizes the importance of checking for factors by evaluating the expression at specific points.
hockeyfghts5
Messages
15
Reaction score
0

Homework Statement



can u factor x2+4
and can't seem to remember the the difference of cubes and isn't there another special cubed theorem.

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
hockeyfghts5 said:

Homework Statement



can u factor x2+4
and can't seem to remember the the difference of cubes and isn't there another special cubed theorem.

It can be factored but it won't be factored with real numbers. You'd need to introduce the imaginary constant,i, such that i2=-1 and use that to factor.
 
x2 + 4 = (x + 2i)(x - 2i), where i is the imaginary unit rock.freak667 mentioned.
Here are the sum and difference of cubes formulas:
x3 + a3 = (x + a)(x2 - ax + a2)
x3 - a3 = (x - a)(x2 + ax + a2)
 
thanks, so i have another question if someone can help me.

lim x3-62+11x-6
x->-1 x3-4x2-19x +14
i can't seem how to factor the top out if i even need to do that or the bottom
 
hockeyfghts5 said:
thanks, so i have another question if someone can help me.

lim x3-62+11x-6
x->-1 x3-4x2-19x +14
i can't seem how to factor the top out if i even need to do that or the bottom

Why would you need to factor it? Did you evaluate it at x=-1? Was it in the indeterminate form 0/0?
 
If the numerator, or denominator, goes to zero when x = 1, then (x-1) is a factor.

You can factor it out by dividing (x-1) into the numerator, say, in the style of long-division.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

Similar threads

Back
Top