The air is getting thinner, so we must be near the peak of Polynomial Mountain. There are an unusual number of cubed terms around here. Two of them are getting rather close to each other. Here's the sum of two cubes.
a3 + b3 = (a + b)(a2 – ab + b2)
Whoa, they just split into a binomial multiplied by a trinomial. That must be a hit at polynomial parties.
Try and put them back together by multiplying things out:
(a + b)(a2 – ab + b2)
= a(a2 – ab + b2) + b(a2 – ab + b2)
= (a3 – a2b + ab2) + (a2b – ab2 + b3)
= a3 + b3
There we go, good as new. We noticed that for the sum of the cubes, the binomial has some addition in it, while the trinomial has a negative middle term. The difference of two cubes looks a little different.
a3 – b3 = (a – b)(a2 + ab + b2)
The binomial shares their negative sign, while the trinomial has a totally positive outlook.
Let's keep a positive outlook ourselves while we work a...
Sample Problem
Factor x3 + 27.
Shmooper, can we see that both terms are cubes? Aye-aye, we can see it.
We can think of it as:
(x)3 + (3)3
We saw before that this will factor to a binomial with positive terms, and a trinomial with a negative term in the middle. In this case, a = x and b = 3.
(x + 3)(x2 – 3x + 9)
We can multiply this out to check our work.
(x3 – 3x2 + 9x) + (3x2 – 9x + 27)
= x3 + 27
Checkmate, sample problem.
Sample Problem
Factor a6 – b3.
Don't be fooled, mathronaut. This polynomial just came from a costume party at the top of the mountain. It really is a difference of two cubes.
(a2)3 – (b)3
Exponents raised to another exponent multiply together, remember? This wasn't that good of a costume if we can see through it so easily. Sorry, but we tell it like it is.
Now, about factoring this out. Our cubes are being subtracted, so the binomial is too, while the trinomial is full of pluses.
(a2 – b)[(a2)2 + a2b + b2]
Simplifying, we get:
(a2 – b)(a4 + a2b + b2)
We suggest that next year, little polynomial, you go as Batman. Actually, we suggest that every year.
Sample Problem
Factor 64(x + y)3 + 27y3.
We have another costumed crusader here. This time, they've disguised their variables to look more complicated than they actually are. It's still a sum of two cubes, though.
Rewrite as:
[4(x + y)]3 + (3y)3
That makes it more obvious. Our terms are 4(x + y) and 3y. We can work with this; it's just going to require a full corset of parentheses, brackets, and braces.
[4(x + y) + 3y]{[4(x + y)]2 – 4(x + y)(3y) + (3y)2}
Simplifying this sounds like a good idea.
= (4x + 4y + 3y)[16(x2 + 2xy + y2) + (-12xy – 12y2) + 9y2]
= (4x + 7y)(16x2 + 32xy + 16y2 – 12xy – 3y2)
= (4x + 7y)(16x2 + 20xy + 13y2)
That's simple enough, we think. We wanted to factor it, after all.
Something interesting to note from these last few problems. Even though we had a trinomial come out of all of them, none of those trinomials could be factored. This will, in fact, almost always be the case.
That trinomial is a fast-talking rookie who prefers to work alone. He's been partnered with binomial, a long-time veteran of the force who plays by the rules and is only a few days away from retirement. Together, they fight crime.