Factoring Trinomials: Worded problems

A rectangle has an area of

$m^2+4m-12$ cm²

and a length of $m+6$ cm.

Is the rectangle's perimeter 16 cm?

First, let's consider the rectangle $ABCD$:

(Sketch - indicating the given information about AB on it)

Let's continue and write down the information about the area of the rectangle and the length of the given side in mathematical form:

$\begin{cases} \textcolor{red}{S_{ABCD}}= m^2+4m-12 \\ \textcolor{blue}{AB}=m+6\\ \end{cases}$(We'll use colors here for clarity of the solution later)

Now let's remember the fact that the area of a rectangle whose side lengths (adjacent) are:

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

$S_{\boxed{\hspace{6pt}}}=a\cdot b$and therefore the area of the rectangle in the problem (according to the sketch we established at the beginning of the solution) is:

$S_{ABCD}=AB\cdot AD$We can now substitute the previously mentioned data into this expression for area to get the equation (for understanding - refer to the marked colors and the data mentioned earlier accordingly):

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

Now, let's pause for a moment and ask what our goal is?

Our goal is, of course, to obtain the algebraic expression for the perimeter of the rectangle (in terms of m), for this we remember that the perimeter of a rectangle is the sum of the lengths of all its sides (let's denote by $P_{ABCD}$ the perimeter of the rectangle),

$$In addition, we remember that the opposite sides in a rectangle are equal to each other and therefore the perimeter of the given rectangle is:

$P_{ABCD}=AB+AD+BC+DC\\ \begin{cases} AB=DC\\ AD=BC \end{cases}\\ \downarrow\\ \boxed{\textcolor{purple}{P_{ABCD}=2\cdot(AB+AD)} }$However, we already have the algebraic expression for the rectangle's side $AB$ (from the given information in the problem, previously marked in blue), and therefore all we need in order to calculate the perimeter of the rectangle is the algebraic expression (in terms of m) for the length of side $AD$,

Let's return then to the equation we reached in the step before considering the perimeter (highlighted with a square around the equation) and isolate $AD$ from it, this we'll do by dividing both sides of the equation by the algebraic expression that is the coefficient of $AD$, that is by:$(m+6)$:

$\boxed{ \textcolor{red}{\textcolor{blue}{(m+6)}\cdot AD= m^2+4m-12}} \hspace{4pt}\text{/:}(m+6)\\ \downarrow\\ AD=\frac{m^2+4m-12}{m+6}$Let's continue and simplify the algebraic fraction we got, we'll do this easily by factoring the numerator of the fraction:

$m^2+4m-12$We'll use quick trinomial factoring for this (to review the rules of quick trinomial factoring) and we get:

$m^2+4m-12\leftrightarrow\begin{cases} \boxed{?}\cdot\boxed{?}=-12\\ \boxed{?}+\boxed{?}=4\ \end{cases}\\ \downarrow\\ (m+6)(m-2)$and therefore (we'll return to the expression for $AD$):

$AD=\frac{m^2+4m-12}{m+6} \\ \downarrow\\ AD=\frac{(m+6)(m-2)}{m+6}\\ \downarrow\\ \boxed{AD=m-2}$In the final stage, after we factored the numerator of the fraction, we simplified the fraction,

(Sketch - with the found length of AD)

Let's return then to the expression for the perimeter of the rectangle, which we got earlier and substitute into it the algebraic expressions for the lengths of the rectangle's sides that we got, then we'll simplify the resulting expression:

$\boxed{\textcolor{purple}{P_{ABCD}=2(AB+AD)} } \\ AB=m+6\\ AD=m-2\\ \downarrow\\ P_{ABCD}=2(m+6+m-2) \\ P_{ABCD}=2(2m+4) \\ \boxed{P_{ABCD}=4m+8}$(length units)

We have thus found the expression for the perimeter of the rectangle in terms of m,

Now let's try to answer the question asked:

Is it possible that the perimeter of the rectangle is 16 length units?

In other words, mathematically- does there exist an m for which:

$P_{ABCD}=16$?

To answer this question, we'll need to perform two steps:

a. Find the value of m for which the stated condition is met.

b. Check if the value of m that meets the requirement is logical in terms of all the problem data (which may be limited in terms of m).

Let's start with a:

Let's find the value of m that meets the stated requirement,

For this, we'll recall the expression for the perimeter that we found in the previous step (enclosed in a frame), and the stated requirement, then we'll demand that the requirement is met and solve the resulting equation:

$\begin{cases} \boxed{P_{ABCD}=4m+8}\\ P_{ABCD}=16 \end{cases}\\ \downarrow\\ 4m+8=16\\ 4m=8\hspace{6pt}\text{/:}4\\ \boxed{m=2}$We have thus found the value of m that meets the requirement regarding the perimeter of the rectangle.

b. Now let's check if for the value of m we found (which meets the requirement regarding the perimeter) we get logical values,

We'll consider that the realistic values in the problem that depend on m are the area and the lengths of the sides, and therefore can only receive positive values (i.e., greater than 0),

For this, we'll recall the various expressions for the area and the lengths of the sides and we'll also state the question asked here mathematically:

$m\stackrel{?}{= }2\leftrightarrow \begin{cases} \textcolor{red}{S_{ABCD}}=(m+6)(m-2) \\ \textcolor{blue}{AB}=m+6\\ AD=m-2 \end{cases}$Where we wrote the expression for the area in its expanded form,

We'll use substitution and note that for: $m=2$, the length of side $AD$becomes zero, and additionally, as expected - the area of the rectangle also becomes zero, that is mathematically:

$m=2\leftrightarrow \begin{cases} \textcolor{red}{S_{ABCD}}=(\underline{2}+6)(\underline{2}-2)=0 \\ \textcolor{blue}{AB}=\underline{2}+6=8\\ AD=\underline{2}-2=0 \end{cases}$Therefore, we can conclude that the solution m=2, which was obtained from the requirement that the perimeter of the rectangle be equal to: 16 length units,

is not possible in the problem, since for this value of m the length of the rectangle's side (and therefore also its area) becomes zero,

But this value is the only value for which the requirement regarding the perimeter of the rectangle is met,

That is - there does not exist an m for which the perimeter of the rectangle is 16 length units,

Therefore, it is not possible that the perimeter of the given rectangle is: 16 length units.

Therefore, the correct answer is answer b.

No

An orthohedron has a volume of

$b^3+5b^2+6b=v$ cm³.

Factorise the above expression.

$b(b+2)(b+3)$

A rectangle has an area equal to

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

What is the length of the width of the rectangle?

$m-2$

Given the rectangle whose area is

$m^2+4m-12$ cm².

and its length is $m+6$ cm.

What is the perimeter of the rectangle?

$4m+8$

A rectangle has an area of

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

What is the length of the width of the rectangle?

$m+6$