Answer
$\frac{2(3k^2 - 2k - 7)}{(2k + 1)(k + 3)(4k - 5)}$
Work Step by Step
To add or subtract rational expressions, we need to make sure that the expressions have the same denominator.
We need to rewrite the two expressions with the same denominator, so we need to find the least common denominator (LCD) for both expressions.
The first thing we want to do is to make sure that each denominator is factored completely.
We see that the denominators are quadratic expressions, which are given by the formula:
$ax^2 + bx + c$, where $a$, $b$, and $c$ are all real numbers.
To factor the expression in the denominator of the first fraction, we want to find which factors when multiplied will give us the product of the $a$ and $c$ terms, which is $6$, but when added together will give us the $b$ term, which is $7$. This means that both factors are positive.
Let's look at possible factors:
$6$ and $1$
$3$ and $2$
It looks like the first combination will work. Let's split the middle term:
$2k^2 + 6k + k + 3$
Group the first two terms and the last two terms:
$(2k^2 + 6k) + (k + 3)$
Factor out what is common in both groups:
$2k(k + 3) + (k + 3)$
Group the factors:
$(2k + 1)(k + 3)$
To factor the expression in the denominator of the second fraction, we want to find which factors when multiplied will give us the product of the $a$ and $c$ terms, which is $-60$, but when added together will give us the $b$ term, which is $7$. This means that one factor is positive and the other negative, but the positive factor has the greater absolute value.
Let's look at possible factors:
$12$ and $-5$
$15$ and $-4$
$10$ and $-6$
It looks like the first combination will work. Let's split the middle term:
$4k^2 + 12k - 5x - 15$
Group the first two terms and the last two terms:
$(4k^2 + 12k) + (-5x - 15)$
Factor out what is common in both groups:
$4k(k + 3) - 5(k + 3)$
Group the factors:
$(4k - 5)(k + 3)$
The exercise can now be rewritten as:
$\frac{k + 2}{(2k + 1)(k + 3)} + \frac{k - 4}{(4k - 5)(k + 3)}$
Next, we want to find the least common denominator (LCD). We do this by taking the highest power of each factor in the denominators of the fractions. Before we can do that, we need to factor the denominators:
LCD = $(2k + 1)(k + 3)(4k - 5)$
Now that we have the least common denominator, we multiply the numerator of each fraction with the factor or factors it is missing in its denominator:
$\frac{(k + 2)(4k - 5)}{(2k + 1)(k + 3)(4k - 5)} + \frac{(k - 4)(2k + 1)}{(2k + 1)(k + 3)(4k - 5)}$
Rewrite the two fractions as one with the same denominator:
$\frac{(k + 2)(4k - 5) + (k - 4)(2k + 1)}{(2k + 1)(k + 3)(4k - 5)}$
Use the distributive property to rewrite the numerator:
$\frac{(4k^2 + 3k - 10) + (2k^2 - 7x - 4)}{(2k + 1)(k + 3)(4k - 5)}$
Combine like terms in the numerator:
$\frac{6k^2 - 4k - 14}{(2k + 1)(k + 3)(4k - 5)}$
Factor out what is common in the numerator:
$\frac{2(3k^2 - 2k - 7)}{(2k + 1)(k + 3)(4k - 5)}$
