Question

this is a multi - part item.
complete the following steps to find the lcd and write the sum of the numerators for the given problem:
$\frac{3}{x^{2}-3x - 4}+\frac{3}{x^{2}-6x + 8}$
factor each denominator.
$x^{2}-3x - 4=quaddownarrow$
$x^{2}-6x + 8=quaddownarrow$

Explanation:

For factoring \(x^2 - 3x - 4\):

Step1: Find two numbers

We need two numbers that multiply to \(-4\) and add up to \(-3\). The numbers are \(-4\) and \(1\) since \(-4\times1 = -4\) and \(-4 + 1=-3\).

Step2: Factor the quadratic

Using the numbers from Step 1, we can factor \(x^2 - 3x - 4\) as \((x - 4)(x+ 1)\) because \(x^2 - 3x - 4=x^2+(-4 + 1)x+(-4\times1)=(x - 4)(x + 1)\).

For factoring \(x^2 - 6x + 8\):

Step1: Find two numbers

We need two numbers that multiply to \(8\) and add up to \(-6\). The numbers are \(-2\) and \(-4\) since \((-2)\times(-4)=8\) and \(-2+(-4)=-6\).

Step2: Factor the quadratic

Using the numbers from Step 1, we can factor \(x^2 - 6x + 8\) as \((x - 2)(x - 4)\) because \(x^2 - 6x + 8=x^2+(-2-4)x+(-2\times(-4))=(x - 2)(x - 4)\).

Answer:

For \(x^2 - 3x - 4\), the factored form is \((x - 4)(x + 1)\).
For \(x^2 - 6x + 8\), the factored form is \((x - 2)(x - 4)\).