# MSC #19: The Product of the Sums is the Sum of the Product

Our MATH-Inic Special for today Dec. 19 is one of my favorites because of its numerous applications. I first wrote about using the “Product of the Sums is the Sum of the Product(PSSP) ” in my book “Algebra Made Easy as Arithmetic” as a means of checking the results of polynomial multiplication and in factoring polynomials.

Because of this, my daughter Veronica and I were invited by the Institute for the Advancement of Vedic Mathematics(IAVM) to conduct a two-part online webinar to explain our techniques.

Although I first learned about the use of digit sums during my elementary grades in the early 1960s to check the answers to arithmetic operations, it was only recently, when I began to study Vedic Mathematics(VM) that I learned about the PSSP sub-Sutra and that it can be applied to coefficients of polynomials.

But what excited me more is when I used PSSP in factoring polynomials. I came across a Youtube video explaining the use of the “AC Method” of factoring using the trinomial 6x^{2} + 59x + 144 as an example. Of course, it took some time before the factor pair of 6 x 144 or 864 which would add up to 59 is identified.

But because of my VM training, I quickly noticed that 1) the digit sum of the coefficients is 209(144+ 6 + 59) by “completing the whole”; 2) 209 is divisible by 11 (2 – 0 + 9 = 11) and 3) 209 = 19 x 11 using “the First by the First and the Last by the Last”.

209 is the digit sum of the product, so it must be equal to the product of the digit sums of the two factors. 209 = 209 x 1 = 11 x 19. We can quickly discard the 209 x 1 combination because it is not possible for a factor to have a digit sum of 1. So the sums of the coefficients of the factors must be 11 and 19.

6x^{2 }can be factored into
6x and x or 2x and 3x.

Now if we consider 6x as one of the factors, it needs a constant term of 13 so that its digit sum will be 19 or 5 to have a sum of 11. Obviously 13 and 5 are not factors of 144, so we have to rule out the 6x and x combination as a solution.

So it leaves us the 2x by 3x option.

2x needs a 17 to have a digit sum of 19. But 17 is also not a factor of 144.

To have a digit sum of 11 2x needs to
have a constant term of 9 and 9 is a factor of 144. We also note that 3x needs
a 16 to have a sum of 19. So the factors of the polynomial 6x^{2} + 59x
+ 144 would be (2x + 9) and (3x + 16)

In our featured example 15x^{2}
+ 14 x – 8, the digit sum is 21 which can be factored as 1 x 21 or 3 x 7. Also the first term 15x^{2} can be
factored into 15x(x) and 5x(3x)

This time we will start with the middle values, 3 x 7 and 3x(5x). The factor containing 3x cannot have a digit sum of 3 since the constant term would have to be 0. So 3x must have a constant term of 4 to have a digit sum of 7 and 5x needs -2 to have a digit sum of 3.

More examples and other applications of FFLL and PSSP can be found in Chapter 19 of “30 Master Strategies in Computing which will be published early next year.