Asked by Kristen
                I need help factoring a problem and cant seem to get it right. The one I'm stumped on is 2x^2+7x+2. I have to compute as indicated and write final in the lowest terms.
The whole problem is
x^2-7x-18/x^2-6x-27*2x^2+7x+2/2x^2+5x+2
            
        The whole problem is
x^2-7x-18/x^2-6x-27*2x^2+7x+2/2x^2+5x+2
Answers
                    Answered by
            Reiny
            
    2x^2 + 7x + 2 does not factor using rational numbers. 
Here is how I know
Multiply the first times the last ---> 2(2) = 4
Are there two numbers which when multiplied will give you 4 and which added will give you 7 ?
Nope!
I think your problem is meant to say:
(x^2-7x-18)/(x^2-6x-27) * (2x^2+7x+2)/(2x^2+5x+2)
HUGE difference from what you wrote
= (x-9)(x+2)/((x-9)(x+3) * (2x^2 + 7x + 2)/((2x+1)(x+2))
= (2x^2 + 7x + 2)/((x+3)(2x+1))
notice that if we expand the bottom we get
2x^2 + 7x + 3
are you sure you don't have a typo in the top and that 2 is not a 3 ????
If not, the book probably has that typo.
    
Here is how I know
Multiply the first times the last ---> 2(2) = 4
Are there two numbers which when multiplied will give you 4 and which added will give you 7 ?
Nope!
I think your problem is meant to say:
(x^2-7x-18)/(x^2-6x-27) * (2x^2+7x+2)/(2x^2+5x+2)
HUGE difference from what you wrote
= (x-9)(x+2)/((x-9)(x+3) * (2x^2 + 7x + 2)/((2x+1)(x+2))
= (2x^2 + 7x + 2)/((x+3)(2x+1))
notice that if we expand the bottom we get
2x^2 + 7x + 3
are you sure you don't have a typo in the top and that 2 is not a 3 ????
If not, the book probably has that typo.
                    Answered by
            Kristen
            
    Yes it was a typo it was a 3! Thanks for the help this does make scense. 
    
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