Question

Simplify completely quantity 12 x plus 36 over quantity x squared minus 4 x minus 21 and find the restrictions on the variable.

12 over quantity x minus 7, x ≠ 7
12 over quantity x minus 7, x ≠ 7, x ≠ −3
quantity x plus 3 over quantity x minus 7, x ≠ 7
quantity x plus 3 over quantity x minus 7, x ≠ 7, x ≠ −3

Answer 1

ANSWER

$\frac{12x + 36}{ {x}^{2} - 4x - 21 } = \frac{12}{x - 7} \: , x\ne -3,x\ne7$

EXPLANATION

We want to simplify

$\frac{12x + 36}{ {x}^{2} - 4x - 21 }$,

This Rational function is defined, for

${x}^{2} - 4x - 21\ne0$,

$(x+3)(x-7)\ne0$

$x\ne -3,x\ne7$

In order to simplify the above rational function, we need to factor both the numerator and the denominator.

Let us factor the numerator first to get,

$\frac{12x + 36}{ {x}^{2} - 4x - 21 } = \frac{12(x + 3)}{ {x}^{2} - 4x - 21}$

As for the denominator, we need to split the middle term of the quadratic trinomial to get,

$\frac{12x + 36}{ {x}^{2} - 4x - 21 } = \frac{12(x + 3)}{ {x}^{2} - 7x + 3x- 21}$


We factor to obtain,

$\frac{12x + 36}{ {x}^{2} - 4x - 21 } = \frac{12(x + 3)}{ x(x - 7) + 3(x- 7)}$

We factor further to obtain,

$\frac{12x + 36}{ {x}^{2} - 4x - 21 } = \frac{12(x + 3)}{ (x + 3)(x - 7)}$

We cancel out common factors to get,

$\frac{12x + 36}{ {x}^{2} - 4x - 21 } = \frac{12}{x - 7}$

where

$x\ne -3,x\ne7$

Answer 2

The restrictions on the variable are 12 over quantity x minus 7, x ≠ 7, x ≠ −3.

Given :

$\dfrac{12x+36}{x^2-4x-21}$

Solution :

We know that,

$x^2-4x-21\neq 0$

$x^2-7x+3x-21\neq 0$

$(x-7)(x+3)\neq 0$

$\rm x\neq 7\;and\;x\neq -3$

Now simplify the given equation:

$\dfrac{12x+36}{x^2-4x-21} = \dfrac{12(x+3)}{(x-7)(x+3)}$

$\dfrac{12x+36}{x^2-4x-21} = \dfrac{12}{(x-7)}$

Therefore the correct option is B).

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https://brainly.com/question/2263981