# How do you simplify this equation?: 2x^2-9x-5/ 4x^2-1

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`(2x^2-9x-5)/(4x^2-1)`

We note that the part below ratio can be write as `4(x+1/2)(x-1/2)` The part above ratio as: `2(x+1/2)(x-5)`

So: `(2x^2-9x-5)/(4x^2-1)= (2(x+1/2)(x-5))/(4(x+1/2)(x-1/2))=` `(x-5)/(2(x-1/2))` `=(x-5)/(2x-1)`

We can use Factoring here.

For the top, we can apply the ac method.

Identify first what is a, b, and c.

a = 2, b = -9 and c = -5. Take note that the top is in the form ax^2 + bx + c.

Multiply a and c. 2 *-5 = -10.

We will find two numbers which when we multiply we get - 10 (value of ac), and when we add we get -9 (value of b).

Let us lists the pair factors of -10 and their sum:

-10 and 1 : - 10 + 1 = -9

10 and - 1 ; 10 + (-1) = 9

-5 and 2 ; -5 + 2 = -3

5 and - 2 ; 5 + (-2) = 3

Therefore the two numbers we are looking for are -10 and 1.

We will rewrite the middle terms using the numbers we got.

`2x^2 -10x + 1x - 5`

Split the terms into two groups, where each group has a gcf.

`(2x^2 - 10x) + (1x - 5)`

Factor out the gcf from each group.

`2x(x - 5) + 1(x - 5)`

Factor out the common factor.

`(x - 5)(2x + 1)`

For the bottom, we will use the Difference of Two Squares formula.

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

`4x^2 - 1 = (2x)^2 - (1)^2 = (2x - 1)(2x + 1)`

So, we will have:

`((x - 5)(2x + 1))/((2x - 1)(2x+1))`

Cancel the common factor on top and bottom.

`(x - 5)/(2x - 1)`