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Our topic is factoring polynomials, and I can't seem to solve this question:

Express the area and perimeter of the shaded region in factored form.

rectangular_region

We've discussed how to solve for the perimeter given the area, although I really don't understand it. We're done with factoring using the common monomial factor, difference of two squares, perfect square trinomial, and general quadratic trinomial. We recently discussed factoring the sum and difference of two cubes.

How do I express the area and perimeter of the shaded region in factored form?

Thanks

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    $\begingroup$ Can you figure out what the side lengths of the rectangle are? $\endgroup$ – Joey Zou Jul 16 '16 at 8:35
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How wide is the shaded region? If the length of the larger rectangle is $a^3$, and the length of the right is $b^3$...

The width of the shaded region is $a^3-b^3$

How tall is the shaded region?

I assume the picture is indicating that the height is $a^2+ab+b^2$

What is the area of a rectangle described as in terms of width and height?

width times height

And what is that in this case?

$(a^3-b^3)(a^2+ab+b^2)$

What is the perimeter of a rectangle described as in terms of width and height?

twice the sum of width and height

And what is that in this case?

$2(a^3-b^3 + a^2+ab+b^2)$

Now... can we express any of these above in a more compact way via factoring? That is debatable, but I do recognize at least one simplification you can do for area:

Knowing that $a^3-b^3=(a-b)(a^2+ab+b^2)$ we have that the area can be described as $(a-b)(a^2+ab+b^2)^2$. Is that really more preferred than $(a^3-b^3)(a^2+ab+b^2)$? I don't see a big difference between the two, so personally I wouldn't care which. They are the same to me.

Using the same for perimeter, we can do the following:

$2(a^3-b^3 + a^2+ab+b^2) = 2((a-b)(a^2+ab+b^2)+1(a^2+ab+b^2)) = 2(a-b+1)(a^2+ab+b^2)$. Again, is this more useful than the first way of writing it? That is debatable...They are not fundamentally different.

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  • $\begingroup$ Hi, thanks for the answer. Can I know the method of factoring used in "Knowing that $a^3-b^3=(a-b)(a^2+ab+b^2)$ we have that the area can be described as $(a-b)(a^2+ab+b^2)^2$?" $\endgroup$ – Ron Jul 16 '16 at 8:52
  • $\begingroup$ @heneral The area, we already found to be $(a^3-b^3)(a^2+ab+b^2)$. Replacing the $(a^3-b^3)$ by the equivalent factored expression, we have: $(a^3-b^3)(a^2+ab+b^2) = ((a-b)(a^2+ab+b^2))(a^2+ab+b^2)=(a-b)(a^2+ab+b^2)^2$ $\endgroup$ – JMoravitz Jul 16 '16 at 8:54
  • $\begingroup$ Oh, right. Sorry. But how did $2((a-b)(a^2+ab+b^2)+1(a^2+ab+b^2))$ become $2(a-b+1)(a^2+ab+b^2)$? I'm sorry If I have too many questions. I'm a slow learner. $\endgroup$ – Ron Jul 16 '16 at 9:02
  • $\begingroup$ @Heneral You have $\color{red}x\color{green}z +\color{blue}y \color{green}z=(\color{red}x+ \color{blue}y) \color{green}z$ by the distributivity property (read in reverse). Here, $(a-b)$ plays the role of $x$ and $1$ plays the role of $y$, meanwhile $(a^2+ab+b^2)$ plays the role of $z$. Hence $\color{red}{(a -b)} \color{green}{(a^2 +ab+ b^2)}+ \color{blue}{1}\color{green}{(a^2+ ab+b^2)}=(\color{red}{a-b}+ \color{blue}1)\color{green}{(a^2 +ab+b^2)}$ $\endgroup$ – JMoravitz Jul 16 '16 at 9:06
  • $\begingroup$ Oh, I think I was able to understand; but please correct me if I'm wrong. $(a-b)(a^2 + ab + b^2)$ and $1(a^2 + ab + b^2)$ has a something in common which is $(a^2 + ab + b^2)$. From what I learned in Common Binomial Factoring, we divide the terms by that common binomial thing. But this time it's a trinomial so I cancel $(a^2 + ab + b^2)$ on both of the terms. Leaving me with $(a-b) + 1$ and $1$. After that, I express the polynomial as the product of the GCF ($(a^2 + ab + b^2)$) and the quotient making it $[(a-b) + 1](a^2 + ab + b^2)$ Is that correct? $\endgroup$ – Ron Jul 16 '16 at 9:20

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