# Problem of the Day: 3/8/13

What is the greatest prime factor of 14! (factorial)?

Solution to yesterday’s problem:

In order to find the perimeter of this rectangle, we should first find the side lengths of the rectangle. Knowing that the area of the rectangle is $3x^2-11x+6$, we can factor this to find the expressions that will multiply to get this area. On to factoring!

Factoring $3x^2-11x+6$

Now, some students at my school like to use these modern techniques they call the ZBox, or some such thing, but on this blog we’re going to be old-fashioned and use plain old grouping. I guess the way the textbook explains grouping is a bit complicated and that’s why they don’t appreciate this wonderful technique!

Anyway, first we need to find factors of 18 (the first coefficient times the last coefficient) that will add up to the middle coefficient, 11. Factors of 18 are (1,18), (2,9),… Stop! We have our factors, 2 and 9. Let’s rewrite the $-11x$ as the sum of those factors:

$3x^2-9x-2x+6$

Now we need to do the grouping part. Notice that the first two and the last two terms look like they could share some factors. (Sometimes these might be out of order, in which case you would rearrange them.) Note that I’ve put the negative INSIDE the grouping because otherwise it would cause some problems…

$(3x^2-9x)+(-2x+6)$

The terms in the first group have a common factor of 3x, so let’s pull that out and divide each term by 3x.

$3x(x-3)+(-2x+6)$

The second group has a common factor of 2, so we’ll pull that out as well.

$3x(x-3)+2(-x+3)$

At this point we would be able to pull out a common factor between these two remaining terms, but we can’t because there’s a subtle difference: (x-3) and (-x+3). We need to change the signs of the second term so that the two factors look the same. If we factor out a -1 from (-x+3), we’ll get

$3x(x-3)-2(x-3)$

Now let’s factor out an (x-3) from both terms. We’ll be left with

$(3x-2)(x-3)$

There are our factors and our side lengths! So to find the perimeter of the rectangle, we just take twice each term.

$2(3x-2)+2(x-3)$

$6x-4+2x-6$

$8x-10$