Algebra and Son of Algebra

As promised on Facebook, here are some of our current algebra problems. Some humor was also promised, but there is little funny about algebra on a Saturday morning. I shall leave it to Mensley to supply the humor.

We have been factoring polynomials, learning the differences of two squares, how to factor a perfect square trinomial, how to factor ax²+bx+c, and how to factor out the greatest possible monomial.

Our strategy for factoring ax²+bx+c was summed up with a King Julien quote, "Now start pulling things out of other things and putting them into new things and it should fix my smoothie maker."

The last lesson we did was titled "factoring completely" and combined lots of factoring techniques. We did the sample problems together, and then attempted the assessment problems independently...which is our standard algebra approach.

Here are the problems, for those of you who'd like to work them. I'd also suggest a 12 step program for those same people, but I really, really need their help with #46. So, don't go looking for a cure for yourselves yet.

#10. 3xy² - 27x³

#20. 180x²y -108xy² -75x³

#28. a² - b² + ac - bc

#36. 8a³ + 4a²b - 2ab² - b³

#38. 3x⁵ + 15x³ - 108x

#46. (x-2)(x²-1)-6x-6

And here are my solutions. My answers line up with the teacher guide answer key, so I know they are correct. (Or the text is also wrong, one of those two outcomes.)

#10. 3xy² - 27x³

That's easy. we just factor out a 3x and are left with the difference of two squares, which we further factor tp get
3x(y-3x)(y+3x)

#20. 180x²y -108xy² -75x³

Again, we factor out a -3x and then are left with a perfect square trinomial to factor down for a final
-3x(5x-6y)²

#28. a² - b² + ac - bc

The first two terms in this are the difference of two squares, so we factor down those and then factor out a c from the last two terms.
That leaves us with (a-b)(a+b)+c(a-b)
and from that we factor out (a-b) and get
(a-b)(a+b+c)

#36. 8a³ + 4a²b - 2ab² - b³

I had to make three attempts at this one. Eventually, I rearranged the terms to 8z³ - 2ab² + 4a²b - b³
Then I factored a 2a from the first two terms and a b from the last two terms. So,
2a(4a²-b²) + b(4a² - b²)
Which can be written as (2a + b)(4a² - b²)
And since the second factor is the difference of two squares, our final factoring is (2a+b)(2a-b)(2a+b) or (2a+b)²(2a-b)

#38. 3x⁵ + 15x³ - 108x

We factor out a 3x and then factor the remaining trinomial to get
3x(x²+9)(x²-4)
The last factor is the difference between two squares, so the final factoring is 3x(x²+9)(x+2)(x-2)

Which leads me to my nemisis. I was rather hoping that the solution would magically occur to me as I was typing in the other problems. Especially since I was using html tags for my superscripts. Using html tags should automatically boost my brain power, right?

Alas, the html brain vitamins failed me.

#46. (x-2)(x²-1)-6x-6

And in a final flashback to highschool, please show your work. (Cause I know what the answer is supposed to be, I just can't figure how to get there.)