Expert: Richard J. Raridon - 8/5/2006

Question
Hi and thank you for your help. The problems are written correctly; however, I will write them again in parenthesis.

Reduce to lowest terms:

b) (3/4c+2) - (c-4/2c^2+c) - (5/6c)


Factor completely:

a)(x^2-5x-3) This is the correct problem. You say this cannot be factored, why?


b) (x^2+23+24) This is also correct. Why can't it be factored?

Thank you so much for all your help.

Is this problem correct?

Factor completely:

c) 6x^4-54x^2
= 6(x-3)x^2(x+3)
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Followup To

Question -
Reduce to lowest terms. Show all work performed.

a) x-1/x^2-x-12 = x+4/x^2+5x+6

b) 3/4c+2 - c-4/2c^2+c - 5/6c

Factor completely. Show all work.

a)x^2-5x-3

b)x^2+23x+24    


Answer -
a) for clarity, you need to write it
(x-1)/(x^2-x-12) = (x+4)/(x^2+5x+6)
x^2-x-12 =  (x-4)(x+3)
x^2+5x+6 = (x+3)(x+2)
so you have (x-1)(x+2) = (x+4)(x-4)
x^2+x-2 = x^2-16 or x=-14
b) Put in some parentheses so I can figure out what you mean.
a) and b) You can't factor either one of these.  Did you write them correctly?

Answer
b) you need to write it
3/(4c+2) -(c-4)/(2c^2+c) -5/(6c)
so the LCD is 6c(2c+1) and you have
[3(3c)-6(c-4) -5(2c+1)]/[6c(2c+1)]=(-7c+19)/[6c(2c+1)]
For any quadratic, you can always use
x = [-b +/-(b^2-4ac)^1/2]/(2a) to find values for x.  If the values for x are whole numbers, then the expression can be factored.  Both of these expressions fail and I could tell that by just looking at them.  If you had
x^2+23x-24 you could have factored it into (x+24)(x-1)