Find Rectangle Width: Area (m²+4m-12) with Length (m+6)

A rectangle has an area equal to

$m^2+4m-12$ cm² and a length of $m+6$ cm.

Determine the length of the width of the rectangle:

Observe the rectangle $ABCD$:

(Drawing - marking the given data regarding AB on it)

Continue and write down the data regarding the rectangle's area and the given side length in mathematical form:

$\begin{cases} \textcolor{red}{S_{ABCD}}= m^2+4m-12 \\ \textcolor{blue}{AB}=m+6 \\ \end{cases}$

(we'll use colors for greater clarity )

Remember that the area of a rectangle whose side lengths (adjacent) are:

$a,\hspace{2pt}b$is:

$S_{\boxed{\hspace{6pt}}}=a\cdot b$

Therefore the area of the rectangle in the problem (according to the drawing we established at the beginning of the solution) is:

$S_{ABCD}=AB\cdot AD$

Now we are able to insert the previously mentioned data into this expression for area to obtain the equation (for understanding - use the marked colors and the data mentioned earlier accordingly):

$\textcolor{red}{ S_{ABCD}}=\textcolor{blue}{AB}\cdot AD \\ \downarrow\\ \boxed{ \textcolor{red}{ m^2+4m-12}=\textcolor{blue}{(m+6 )}\cdot AD}$

Let's pause for a moment to determine our goal:

Our goal is of course to obtain the algebraic expression for the side adjacent to the given side in the rectangle (denoted by m), meaning we want to obtain an expression for the length of side $AD$,

Let's return then to the equation that we previously obtained and proceed to isolate $AD$. This can be achieved by dividing both sides of the equation by the algebraic expression that is the coefficient of $AD$, that is by:$(m+6 )$:

$\boxed{ \textcolor{red}{\textcolor{blue}{(m-2)}\cdot AD= m^2+4m-12}} \hspace{4pt}\text{/:}(m+6 )\\ \downarrow\\ AD=\frac{m^2+4m-12}{m+6 }$

Let's continue to simplify the algebraic fraction that we obtained. We can do this easily by factoring the numerator of the fraction:

$m^2+4m-12$

Apply quick trinomial factoring as shown below:

$m^2+4m-12\leftrightarrow\begin{cases} \boxed{?}\cdot\boxed{?}=-12\\ \boxed{?}+\boxed{?}=4\ \end{cases}\\ \downarrow\\ (m+6)(m-2)$

Therefore (returning to the expression for $AD$):

$AD=\frac{m^2+4m-12}{m+6 } \\ \downarrow\\ AD=\frac{(m+6)(m-2)}{m+6 }\\ \downarrow\\ \boxed{AD=m-2}$(length units)

In the final stage, after we factored the numerator of the fraction and reduced the fraction,

(Drawing - with the found AD length)

Therefore the correct answer is answer D.