Question

May someone help me with this factor equation?

Answer 1

Answer:

$\frac{x^2+8x+9}{(x+3)(x-1)}$

Step-by-step explanation:

$\frac{2x}{x^2+2x-3}$ + $\frac{x+3}{x-1}$ ← factor the denominator of first fraction

= $\frac{2x}{(x+3)(x-1)}$  + $\frac{x+3}{x-1}$

Before adding we require the denominators to be the same

Multiply numerator/denominator of second fraction by (x + 3)

= $\frac{2x}{(x+3)(x-1)}$ + $\frac{(x+3)(x+3)}{(x+3)(x-1)}$

Add the numerators leaving the common denominator ( LCD = (x + 3)(x - 1) )

= $\frac{2x+(x+3)(x+3)}{(x+3)(x-1)}$ ← expand factors on numerator using FOIL

= $\frac{2x+x^2+6x+9}{(x+3)(x-1)}$

= $\frac{x^2+8x+9}{(x+3)(x-1)}$