# How do you combine #7/(x^2-x-2)+x/(x^2+4x+3)#?

##### 1 Answer

Here's what I got.

#### Explanation:

Your goal here is to find a **common denominator** for those two fractions. This will allow you to combine the two numerators and simplify the expression, if possible.

Take a look at the first denominator

#x^2 - x - 2#

If you make this equal to zero, you can use the **quadratic formula**

#color(purple)(|bar(ul(color(white)(a/a)color(black)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a))color(white)(a/a)|)))#

to find its roots. In this case, you have

#x_(1,2) = (-(-1) +- sqrt( (-1)^2 - 4 * 1 * (-2)))/(2 * 1)#

#x_(1,2) = (1 +- sqrt(9))/2 implies { (x_1 = (1 + 3)/2 = 2), (x_2 = (1 - 3)/2 = -1) :}#

You can thus rewrite the first denominator as

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

Do the same for the second denominator

#x^2 + 4x + 3#

to get

#x_(1,2) = (-4 +- sqrt( 4^2 - 4 * 1 * 3))/(2 * 1)#

#x_(1,2) = (-4 +- 2)/2 implies {(x_1 = (-4 - 2)/2 = -3), (x_2 = (-4 + 2)/2 = -1) :}#

The second denominator can be rewritten as

#x^2 + 4x + 3 = (x+3)(x+1)#

The expression becomes

#7/((x-2)(x+1)) + x/((x+3)(x+1))#

The **common denominator** will be

#overbrace((x+1))^(color(purple)("common to both denominators")) * (x-2) * (x+3)#

Multiply the first fraction by

#7/((x-2)(x+1)) * (x+3)/(x+3) + x/((x+3)(x+1)) * (x-2)/(x-2)#

#(7 * (x+3))/((x+1)(x-2)(x+3)) + (x * (x-2))/((x+1)(x-2)(x+3))#

Now focus on the numerator

#7(x+3) + x(x-2) = 7x+21 + x^2 - 2x#

#=x^2 + 5x + 21#

The expression can thus be rewritten as

#(x^2 + 5x + 21)/((x+1)(x-2)(x+3))#

The numerator cannot be factored in any *useful form*, i.e. the quadratic

#x^2 + 5x + 21 = 0#

has *complex roots*, so it's best to leave the answer like this.