Factoring a binomial: 125^2 - n^3 q^2

[subdocwife731]

I am a bit confused on the question that our class has been asked to answer.
The directions simply say Factor, and that is it.
Here is the problem.

125^2-n^3q^2

I don't really know where to begin.

 

[stapel]

II don't really know where to begin.

That's awkward. Your class was supposed to cover factoring, in depth and at length, before having assigned any exercises on it. :?

Since we cannot reasonably teach the week or so of class sessions that your course skipped, you'll need to try self-study online (or else hire a qualified local tutor to teach you the missing information). Try here:

. . . . .Google results for "simple factoring GCF"

. . . . .Google results for "factoring quadratics"

. . . . .Google results for "factoring trinomials"

. . . . .Google results for "special factoring squares cubes"

Once you have studied some lessons (at least two from each link!), the following will make sense:

The GCF is 1, so there's nothing to take out front. This isn't a trinomial quadratic, so you're left with differences of squares or differences of cubes. But the first term isn't a cube, and the second term isn't a square or a cube. Since nothing applies, this is prime.

Eliz.

 

[wjm11]

Factor 125^2-n^3q^2

Eliz. has given you good guidance.

However, if you MUST factor, consider this: there won’t be any “pretty” answers to this. If we apply either the “difference of two squares” or the “difference of two cubes” formulas, we’ll end up with some messy (fractional) exponents. But here goes:

Difference of two squares: a^2 – b^2 = (a-b)(a+b)

125^2-n^3q^2 = 125^2 – ([n^(3/2)][q])^2 = (125 – qn^(3/2))(125 + qn^(3/2))

Difference of two cubes: a^3 – b^3 = (a-b)(a^2 + ab + b^2)
Note that
125^2 = (5^3)^2 = (5^2)^3 so,

125^2-n^3q^2 = (5^2)^3 – (nq^(2/3))^3 = ?

 

[Mrspi]

I am a bit confused on the question that our class has been asked to answer.
The directions simply say Factor, and that is it.
Here is the problem.

125^2-n^3q^2

I don't really know where to begin.

I wonder if you typed the original problem correctly. Could it have been THIS?

125 q^2 - n^3 q^2

If it was, then you could begin by removing a common factor of q^2, to get this:

q^2 (125 - n^3)

Then, you could factor 125 - n^3 as a difference of two cubes, because 125 is 5^3....5^3 - n^3.

Just asking, because the problem as originally typed looks really strange.