Question

add.
\\(\dfrac{6}{3x^2 + 14x + 8} + \dfrac{3}{3x^2 - x - 2}\\)
simplify your answer as much as possible.

Explanation:

Step1: Factor the denominators

Factor \(3x^2 + 14x + 8\): We need two numbers that multiply to \(3\times8 = 24\) and add to \(14\). The numbers are \(12\) and \(2\). So, \(3x^2+12x + 2x + 8=3x(x + 4)+2(x + 4)=(3x + 2)(x + 4)\).

Factor \(3x^2 - x - 2\): We need two numbers that multiply to \(3\times(-2)=-6\) and add to \(-1\). The numbers are \(-3\) and \(2\). So, \(3x^2-3x + 2x - 2 = 3x(x - 1)+2(x - 1)=(3x + 2)(x - 1)\).

So the expression becomes \(\frac{6}{(3x + 2)(x + 4)}+\frac{3}{(3x + 2)(x - 1)}\).

Step2: Find the common denominator

The common denominator is \((3x + 2)(x + 4)(x - 1)\).

Rewrite each fraction with the common denominator:

\(\frac{6(x - 1)}{(3x + 2)(x + 4)(x - 1)}+\frac{3(x + 4)}{(3x + 2)(x + 4)(x - 1)}\).

Step3: Add the numerators

\(6(x - 1)+3(x + 4)=6x-6 + 3x + 12=9x + 6\).

So the expression is \(\frac{9x + 6}{(3x + 2)(x + 4)(x - 1)}\).

Step4: Factor the numerator and simplify

Factor the numerator: \(9x + 6 = 3(3x + 2)\).

Now we have \(\frac{3(3x + 2)}{(3x + 2)(x + 4)(x - 1)}\).

Cancel out the common factor \((3x + 2)\) (assuming \(3x+2
eq0\)): \(\frac{3}{(x + 4)(x - 1)}\).

Expand the denominator: \((x + 4)(x - 1)=x^2+3x - 4\). So the simplified form is \(\frac{3}{x^2 + 3x - 4}\).

Answer:

\(\frac{3}{x^2 + 3x - 4}\)