Question

Suppose the area of a rectangle is 9x3 + 27x2 + 36x + 18 and the length is 3x + 3. What is the width of the rectangle?• Width:

Answer 1

We know that the area of a rectangle is the width multiplied by its length, basically

$A=b\cdot h$

But here, we have the area, which is a polynomial, same for the length, then, to find out the width we must do a division of polynomial, I'll call the unknown polynomial as P, we have

$\begin{gathered} A=P\cdot l \\ \\ P=\frac{A}{l} \\ \\ P=\frac{9x^3+27x^2+36x+18}{3x+3} \end{gathered}$

So now we got to solve

$P=\frac{9x^3+27x^2+36x+18}{3x+3}$

There is a lot of ways to solve it, first, I'll factor 3 in the denominator

$P=\frac{9x^3+27x^2+36x+18}{3(x+1)}$

As well the numerator

$P=\frac{3(3x^3+9x^2+12x+6)}{3(x+1)}$

Then we can already simplify 3, and get an easier expression

$P=\frac{3x^3+9x^2+12x+6}{x+1}$

Now I'll factor using the grouping method, I want the numerator to be something like (x+1)*(another polynomial), doing that we can "cut" the (x+1).

One way to do it is to expand it in some simple sums, look at 3x³, we want a 3x² to match with him, but look that 9x² is 3x²+6x², let's write that way

$P=\frac{3x^3+3x^2+6x^2+12x+6}{x+1}$

It doesn't seem to help us at all, but look that now we can factor 3x², then

$P=\frac{3x^2(x+1)+6x^2+12x+6}{x+1}$

Look, we have our first (x+1) in the numerator. Let's use the same idea but now for the term 6x², we want a 6x to match with him, we have 12x, so let's write it as 6x + 6x.

$P=\frac{3x^2(x+1)+6x^2+6x+6x+6}{x+1}$

Similar to what we did before, factor 6x, and again

$P=\frac{3x^2(x+1)+6x(x+1)+6x+6}{x+1}$

Now we have one more (x+1).

Now the easiest one, factor 6

$P=\frac{3x^2(x+1)+6x(x+1)+6(x+1)}{x+1}$

As we can see we have (x+1) multiplied by something in all terms in the numerator, then we can factor (x+1) now

$P=\frac{(x+1)(3x^2+6x+6)}{x+1}$

Now we can cancel (x+1), and our final expression will be

$P=3x^2+6x+6$

Therefore, the width is

$3x^2+6x+6=\text{ width}$