`k/(k^2-7k+12)+3/(k^2-2k+3)` I have no idea how to do this.    

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lemjay eNotes educator| Certified Educator

`k/(k^2-7k+12)+3/(k^2-2k+3)`

Since the given fractions have unlike denominators, let's determine the LCD. To do so, factor the quadratic expression.

`k^2-7k+12=(k-4)(k-3)`

`k^2-2k+3` - not factorable

Since the second quadratic expression is not factorable, this indicates that the denominators have no common factor. So, the LCD is the product of the two denominators. And, let's re-write the fractions as:

`= [k(k^2-2k+3)]/[(k^2-7k+12)(k^2-2k+3)] + [3(k^2-7k+12)]/[(k^2-7k+12)(k^2-2k+3)]`

`=(k^3-2k^2+3k)/[(k^2-7k+12)(k^2-2k+3)] + (3k^2-21k+36)/[(k^2-7k+12)(k^2-2k+3)]`

Now that the denominators are the same, add the fractions.

`= (k^3+k^2-18k+36)/[(k^2-7k+12)(k^2-2k+3)]`

Hence, `k/(k^2-7k+12)+3/(k^2-2k+3)= (k^3+k^2-18k+36)/[(k^2-7k+12)(k^2-2k+3)]` .

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