How can I factorize x^2-x-2/x^2-3x+2?

Question

Is for this exercise to give you an idea.

http://archives.math.utk.edu/visual.calculus/1/continuous.9/index.html

I am stuck at factoriig the botton of the equation. which is x^2-3x+2

Thanks.

factoring the bottom*

You can't see it in the page but is exercise #1 from the basic quizz.

(x^2-x-2)/(x^2-3x+2)?

Answer 1

Factor: (x^2-x-2)/(x^2-3x+2)

First: factor the numerator---multiply the 1st & 3rd coefficient to get "-2." Find two numbers that give you "-2" when multiplied &
"-1" (2nd/middle coefficient) when added/subtracted. The numbers are (-2 & 1).

*Rewrite the numerator with the new middle coefficients...

x^2 - 2x + x - 2

*With 4 terms-group "like" terms & factor both sets...

(x^2 + x) - (2x - 2)
x(x+1) - 2(x+1)
(x+1)(x-2)

*Now, you have: [(x+1)(x-2)]/(x^2-3x+2)

Sec: factor the denominator---multiply the 1st & 3rd coefficient to get "2." Find two numbers that give you "2" when multiplied &
"-3" (2nd/middle coefficient) when added/subtracted. The numbers are (-1 & -2).

*Rewrite the denominator with the new middle coefficients....

x^2 - x - 2x + 2

*With 4 terms-group "like" terms & factor both sets...

(x^2 - x) - (2x + 2)
x(x-1) - 2(x-1)
(x-1)(x-2)

*Now, you have: [(x+1)(x-2)]/[(x-1)(x-2)]

Third: cross cancel "like" terms---cancel the sets of "x-2"...

= (x+1)/(x-1)

Answer 2

this is an easy sum.
first we factorize the numerator x^2-x-2
=x^2-x-4+2=x^2-4-x+2=(x+2)(x-2)-(x-2)
=(x-2)(x+2-1)=(x-2)(x+1)
next we factorize the denominator x^2-3x+2=x^2-2x-x+2
=x(x-2)-1(x-2)=(x-1)(x-2)
while dividing the factor(x-2) being common cancels out and thus the answer is(x+1)/(x-1)

Answer 3

The bottom factors into (x-2)(x-1)

Answer 4

factoring the expression is basic:
x^2-3x+2
obviously x^2 factors as x(x) now you need 2 numbers when multiplied times x and added togther results in minus -3x and when multiplied times each other is 2.
this is satisfied by -2 and -1
therefore the factors are (x-1)(x-2)

Answer 5

x^2-3x+2 can be factorised as (x-1)(x-2)
the top is simplified as (x-2)(x+1)
thus the answer after cancelling the common factors[(x-2)]
is (x-1)/(x+1)

Answer 6

hirs the solution

first factor both the numerator and the denominator using sqt (simple quadratic trinomial)

=(x-2)(x+1)/(x-2)(x-1)

then cancell

=x+1/x-1

that's the final answer
hope it may help you
^_^

Answer 7

I'll factor both:

(x-2)(x+1)/((x-2)(x-1))

You should be able to solve it from here.

Answer 8

x² - x - 2 / x² - 3x + 2 =

(x - 2)(x + 1) / (x - 2)(x - 1) =

(x + 1) / (x - 1)

- - - - - - - - - -s-

Answer 9

You can't factor the bottom, the answer you got is the final answer.