Add 3 / (x2 - x - 30) + x / (2x2 + 7x - 4)

Question

the x is squared in (x2-x-30) and the 2x is squared in (2x2+ 7x -4)

Answer 1

it seems strange, but the first thing you should do is factor the denominators. It seems easier to just to multiply each fraction by the denominator of the other over itself, but unless you want to search for a common factor between the new cubic numerator and the resulting quartic denominator, you are much better off factoring first.

 

3/[(x-6)(x+5)] + x/[(2x-1)(x+4)]  

 

Now that you know there is no way to simplify the problem up front, you proceed to convert ot a common denominator and add.  It is best to leave the denominator in factored form, for now.

 

3(2x²+7x-4) + x(x²-x-30)/[(x-6)(x+5)(2x-1)(x+4)] =

 

(x³ + 5x² - 9x - 12)/[(x-6)(x+5)(2x-1)(x+4)]

 

which was always an acceptable form to me as a teacher: once I have asked you to factor something I have little interest in asking you to multiply it back together.  Your teacher may want you to simplify the denominator.

 

(x³ + 5x² - 9x -12)/(2x⁴ + 5x³ - 71x² - 206x + 120)

 

Answer 2

Samantha, this is kind of a long problem. I am going to start with a simple example and let you figure out the rest.

 

1/14 + 2/15 = 1/(2 x 7) + 2/(3 x 5)

 

Since there are no common factors between our denominators, our common denominator is 14 x 15, or 210.

 

((1 x 15) + (2 × 14))/210 = (15 + 28)/210 = 43/210

 

If I factor our denominators, which you should know how to do...

 

x² - x - 30 = (x - 6)(x + 5)

2x² + 7x - 4 = (2x - 1)(x + 4)

 

Again, we have no common factors, as none of our binomials match up.

 

So, the common denominator is (x² - x - 30)(2x² + 7x - 4)

 

(3(2x² + 7x - 4)) + x(x² - x - 30))/((x² - x - 30)(2x² + 7x - 4))

 

Multiply out your numerator and denominator and combine like terms for your answer.