Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\text{Problem: }{x^3-x^2-x-2 \over x^2 + x - 6}$$ My textbook was able to come up with $(x-2)(x^2+x+1)$ $$\text{Textbook: }{(x-2)(x^2+x+1) \over (x-2)(x+3)}$$

I've tried grouping and using the first and last 3 terms as trinomial but it doesn't work. How do I get that solution? Thank you in advance.

share|improve this question
    
So what's the question? Put the fraction in least terms? –  Ian Coley Apr 3 at 18:44
    
@IanColey well, if I got (x-2). I could eliminate them. I'm trying to solve a limit problem –  Mouse Hello Apr 3 at 18:45
    
Which limit?$\,$ –  Ian Coley Apr 3 at 18:46
    
Question unclear –  Jason Zimba Apr 3 at 18:46
1  
Have you had polynomial or synthetic division? (That should have come up somewhere in pre-calculus.) Dividing $ \ x-2 \ $ into $ \ x^3 - x^2 - x + 2 \ $ does give a quotient of $ \ x^2 + x + 1 \ $ ( or using "2" in synthetic division on $ \ 1 \ -1 \ -1 \ -2 \ $ produces $ \ 1 \ 1 \ 1 \ 0 \ $ ) . –  RecklessReckoner Apr 3 at 19:28

1 Answer 1

up vote 2 down vote accepted

For the cubic polynomial, by checking the value of it at some integer values of $x$ like $x = 2$ then $2^3 - 2^2 - 2 - 2 = 0$. This means $x - 2$ is a factor, and using long division you get the other factor. For the quadratic expression, you also have $2$ is a root and also $-3$ is also a root. Then it can be factored as $(x - 2)(x + 3)$.

share|improve this answer
    
I don't get it. How can you know x-2 is a factor? And if you know it's a factor, how can you come up with x^2+x+1 –  Mouse Hello Apr 3 at 18:52
    
@MouseHello: you carry out the long division. –  Dark-Chocolate Apr 3 at 18:58
    
What do you mean by long division? Do you mean $x^3-x^2-x-2 \over (x-2)$ –  Mouse Hello Apr 3 at 19:04
    
    
@MouseHello: yes. –  Dark-Chocolate Apr 3 at 22:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.