(9x^2 -1)/(12x^2 +7x+1)

So firstly, I want to factor these polynomials. Starting with the numerator, I will be applying the difference of squares rule, which is [tex]x^2-y^2=(x+y)(x-y)[/tex] . In this case:

[tex]9x^2-1=(3x+1)(3x-1)\\\\\frac{(3x+1)(3x-1)}{12x^2+7x+1}[/tex]

Next, with the denominator I will be factoring by grouping. Firstly, what two terms have a product of 12x² and a sum of 7x? That would be 3x and 4x. Replace 7x with 3x + 4x:

[tex]\frac{(3x+1)(3x-1)}{12x^2+3x+4x+1}[/tex]

Next, factor 12x² + 3x and 4x + 1 separately. Make sure that they have the same quantity on the inside of the parentheses:

[tex]\frac{(3x+1)(3x-1)}{3x(4x+1)+1(4x+1)}[/tex]

Now you can rewrite it as:

[tex]\frac{(3x+1)(3x-1)}{(3x+1)(4x+1)}[/tex]

Now that we factored both expressions, we can start to divide. Any common factors between the numerator and the denominator get canceled out. In this case:

[tex]\frac{(3x+1)(3x-1)}{(3x+1)(4x+1)}=\frac{3x-1}{4x+1}[/tex]

Answer:

Your final answer is: [tex]\frac{3x-1}{4x+1}[/tex]

OmegaAzzy OmegaAzzy · Answer 2 · 2018-11-03T19:21:35+01:00

First, we can factory both sides of this division problem by using important rules.

First, (9x^2 - 1) can be simplified using difference of squares, which states:

(a^2 - b^2) = (a + b)(a - b)

In this case, the numerator can be simplified to (3x + 1)(3x - 1)

Now, we can also simplify the denominator using the AC method.

By splitting the middle term to factors of 12, we can factor this trinomial has a pair of binomials.

12x^2 + 4x + 3x + 1

Factor binomials.

4x(3x + 1) + 1(3x + 1)

Rearrange.

(4x + 1)(3x + 1)

Let's bring our binomial groups together.

(3x + 1)(3x - 1)/(4x + 1)(3x + 1)

We have the same factor, (3x + 1), in both the numerator and the denominator.

Because of this, we can cancel both factors out, which leaves:

(3x - 1) / (4x + 1), which is the solution.