Question 683260: X squared minus 5x plus six divided by x minus two
Answer by RedemptiveMath(80)   (Show Source):
You can put this solution on YOUR website! Simplifying these expressions are not as hard as they may appear. The key thing to look at is the part of the fraction with no squared, cubed, or higher operations. Looking at the entire fraction, we can see that the numerator is more complex than the denominator. What we want to simply do is factor the part that has the squared or cubed "x" if possible. Looking at the numerator alone:
x^2-5x+6
(x-2)(x-3).
Reviewing the fraction:
(x^2-5x+6)/(x-2) = [(x-2)(x-3)]/(x-2) = x-3.
The new numerator is the quadratic factored. Now we can see similarities between the numerator and the denominator. We have two (x-2). Now, we can simply cancel those parts out, and we are left with the answer (x-3). It would be wise to note that if we want to cancel something inside parentheses, we must cancel everything inside of those parentheses. We cannot pick what we can cancel out of a set of parentheses. In other words, if we have the fraction 3/(x-3), we cannot reduced the 3s and say that the simplified version is 1/(x-1). We would have to have the entire piece (x-3) in the numerator to reduce: [3(x-3)]/(x-3) = 3. Terms that are not being added, subtracted, or being placed in parentheses can reduce if applicable. For example, 3/[3(x-3)] can reduce to 1/(x-3). The 3 in the denominator is not being added or subtracted (it's technically being multiplied), and 3 over 3 can be reduced to 1 over 1. The (x-3) cannot be reduced, so it remains in the denominator. Finally, coefficients (numbers in front of variables) can be reduced given the rules above are still applied. 12x/[4(x-4)] becomes 3x/(x-4). The "x" remains unaffected in the numerator, but the coefficient is reduced because 12 and 4 can be reduced to 3 and 1.
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