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

Our Math Special for today 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” which was published in January 2017 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 technique.

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 6x2 + 59x + 144 as an example. Of course, we know that it will take some time before the factor pair of 864 (6 x 144) which would add up to 59 is identified.

Because of my VM training, I quickly noticed that 1) the digit sum of the coefficients is [(144+ 6) + 59] = 209 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 even the biggest factors of 6 and 144 will not add up to 209. So, the sums of the coefficients of the factors must be 11 and 19.

6x2 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 their digit sum will be 19 or 5 to have a sum of 11. Obviously 13 and 5 are not factors of 144, so we need to rule out the 6x and x combination as a solution.

It leaves us the 2x by 3x option.

2x needs a 17 to have their digit sum be equal to 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 so that their sum of 19. So, the factors of the polynomial (6x2 + 59x + 144) would be (2x + 9) and (3x + 16)

In our featured example, we want to find the sum of the coefficients A and B given that (2x2 + Ax – 4) (3x2 -2x + B) = 6x4 + 11x3 – 16x2 + 23x – 12.

Using The technique we discussed yesterday, we can use “the Last by the Last” to find B. We multiply the last terms of the trinomial and equate the product to the last term of the 4th degree polynomial: – 4 x B = -12. B, then is -12/ -4 = 3.

Next, we apply PSSP. Using only the coefficients and substituting 4 for B, we have:

(2 + A – 4)(3 – 2 + 3) = (6 + 11 – 16 + 23 – 12)

(A – 2)(4) = 12

4A – 8 = 12

4A = 20; A = 5

A + B = 5 + 3 = 8