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maybe factoring out the x in the numerator and denominator can help..yes?
it might see these are the answer choices..... a. 1/(x-3)(x+1) b. 1/x2(x-1) c. x-2/x2(x-1) d. -1/x2
\[( x^3-x^2-2x)/(x^5-2x^4-3x^3)\] First, take out the common factors \[x( x^2-x-2)/x^3(x^2-2x-3)\] Then, do the factorization \[x( x-2)(x+1)/x^3(x-3)(x+1)\] Can you work out the answer your own? (Note that i don't think any of the above choices are correct.. :S )
well my teacher taught to like reduce the powers from top to bottom is that right????
What do you mean by ' reduce the powers from top to bottom'?
i mean like well.....im not sure she......its kinda confusing to explain... like take away an "x" 4rm the top like how the problem is one of the xs is x3 and the is x5 by taking an x away u would have no x on the top???? i dont know im confused...
Actually, i think it is like this Just take a look at the numerator, x^3-x^2-2x You can see every term contain an 'x', that is the common factor, so take is out and group it like x(x^2 - x -2x) Do you understand this part?
Sorry, i made a mistake. It should be x(x^2 - x -2) . Well can you do the same for the denominator? Take out the common factor of x^5-2x^4-3x^3
its ok yeah i think i can
so would the denominator be x^3(x^2-2x-3)?
So next, you need to do factorization. Take a look at the denominator again, you can further factorise it x(x^2 - x -2), consider (x^2 - x -2) |dw:1333510991772:dw| So, (x^2 - x -2) = (x-2)(x+1) x(x^2 - x -2) = x(x+1)(x-2) Do you understand this part?
yeah i've gotten to the part where i cancel x+1 and x+1 but since none of the answer choices are there do i just keep the answer i go????? even though its a pratice book for the CST's?????
Perhaps you can write down the answer you get and ask your teacher?
ok :) thnx