Could someone please help me with this question? Thanks!

Reduce (x^2-5x+6)/(x-2) to lowest terms

You have to first factorise the numerator, then a term in the numerator will cancel out the (x-2) in d denominator.

let me give you an example with a similar problem...

(x^2+5x+6)/(x+3)
To do this, you have to first factorize the numerator:
To factor the numerator, you first find the products of the constant (+6) that when added together will give you the coefficient of x (+5). I found the factors of +6 that when added together will give me +5 to be +2 and +3...Now replace that plus 5x with +2x and +3x...so this is what you will have:
x^2+3x+2x+6.
Now, you factor the terms out and what you'll have is:
x(x+3) + 2(x+3)...then, your answer will be: x^2+5x+6 = (x+2)(x+3); Now replace this with the original numerator and simplify...
(x+2)(x+3)/(x+3) = (x+2).

Ok, now try the one you were given as an assignment. I will be glad to check your work.

Sure! To reduce the expression (x^2-5x+6)/(x-2) to its lowest terms, we need to simplify it as much as possible.

First, let's check if x-2 is a factor of the numerator, x^2-5x+6, using synthetic division:

2 | 1 -5 6
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2 -6 0

Since the remainder is 0, we can conclude that x-2 is indeed a factor of the numerator. Therefore, we can rewrite the expression as:

(x^2-5x+6)/(x-2) = (x-2)(x-3)/(x-2)

Next, we can cancel out the common factor of (x-2) from the numerator and denominator:

(x-2)(x-3)/(x-2) = (x-3)

Therefore, the expression (x^2-5x+6)/(x-2) reduced to its lowest terms is simply (x-3).